r otational m otion - santa rosa junior collegesrjcstaff.santarosa.edu/~lwillia2/20/20ch7hw.pdfr...

7-1

ROTATIONAL MOTION 7 Q7.1. Reason: By convention, clockwise rotations are negative and counterclockwise rotations are positive. As a result, an angular acceleration that decreases/increases a negative angular velocity is positive/negative. In like manner, an angular acceleration that decreases/increases a positive angular velocity is negative/positive. Knowing this we can establish the situation for each figure. Figure (a) the pulley is rotating clockwise (! = –), however since the large mass is on the left it is decelerating (" = +). Figure (b) the pulley is rotating counterclockwise (! = +) and since the large mass is on the left it is accelerating (" = +). Figure (c) the pulley is rotating clockwise (! = –) and since the large mass is on the right it is accelerating (" = –). Figure (d) the pulley is rotating counterclockwise (! = +), however since the large mass is on the right it is decelerating (" = –). Assess: It is important to know the sign convention for all physical quantities that are vectors. This is especially important when working with rotational motion.

Q7.2. Reason: Since we are dealing with a stubborn nut, we want a large torque. For a given force the torque is increased by increasing the moment arm. Slipping a length of pipe over the wrench handle increases the moment arm and hence the applied torque. Assess: You can increase the torque by either increasing the force or the moment arm. In this case we chose to increase the torque by increasing the moment arm.

Q7.3. Reason: The question properly identified where the torques are computed about (the hinge). Torques that tend to make the door rotate counterclockwise in the diagram are positive by convention (general agreement) and torques that tend to make the door rotate clockwise are negative. (a) + (b) # (c) + (d) # (e) 0

Assess: Looking at the diagram we see that F

!"

a and F

!"

c are parallel and are both creating a negative or

counterclockwise torque. But since F

!"

c is farther from the hinge, its torque will be greater. A similar argument can

be made for F

!"

b and F

!"

d . Note that F

!"

e causes no torque since it has no moment arm.

Q7.4. Reason: The screwdriver with a thick handle has a larger radius and so provides a larger moment arm for the force you exert when turning the driver. The larger moment arm leads to a larger torque on the screw. This can be seen from Equation 7.4. Assess: For a given force a larger moment arm leads to a larger torque.

Q7.5. Reason: The reason for large-diameter steering wheels in trucks is that more torque is needed to turn the wheels due to the greater mass of the truck. Making the steering wheel larger means that more torque is exerted on the steering shaft for the same force from the driver’s hands. Assess: Most light cars employ a rack-and-pinion steering system, while larger SUVs and trucks often employ a recirculating-ball steering system; however both systems can be assisted by pressurized hydraulic fluid (power steering), so steering, even in trucks, can be much easier than in the old days.

7-2 Chapter 7

Q7.6. Reason: When put farther from the hinge of the door, the torque exerted on the door by the doorstop is larger than when the doorstop is placed nearer the hinge. This is due to the larger moment arm for the force the doorstop exerts on the door. Assess: Torque is proportional to the moment arm of a force, from Equation 7.4.

Q7.7. Reason: The torque you exert on the branch is your weight multiplied by the lever arm, or the distance along the branch from the trunk, so the farther out you are, the greater the torque you exert on the branch. Assess: It is a good thing that tree branches themselves get thinner away from the trunk, so the weight of the branch itself doesn’t break it away from the trunk.

Q7.8. Reason: (a) The torque the student exerts turns the ball and rod in a clockwise direction as seen from the top. Clockwise torques are negative. The student exerts a negative torque. (b) After the push has ended, there is no force on the ball besides gravity and the force of the pivot and rod.The rod and pivot constrain the ball to move in a horizontal circle. Gravity acts in the vertical direction. Since gravity has no component that acts in the horizontal direction it does not exert any torque on the ball that would affect its rotation in a horizontal plane. So after the push has ended, the ball’s angular velocity is constant, from Equation 7.11. (c) Since the force producing the rotation is zero, the torque is zero. Assess: Note that gravity does produce a torque on the ball tending to rotate the ball in the vertical direction, though this torque is balanced by the vertical torque of the rod and pivot keeping the ball moving in a horizontal plane.

Q7.9. Reason: As suggested by the figure, we will assume that the larger sphere is more massive. Then the center of gravity would be at point 1, because if we suspend the dumbbell from point 1 the counterclockwise torque due to the large sphere (large weight times small lever arm) will be equal to the clockwise torque due to the small sphere (small weight times large lever arm). Assess: Look at the figure and mentally balance the dumbbell on your finger; your finger would have to be at point 1. The sun-earth system is similar to this except that the sun’s mass is so much greater than the earth’s that the center of mass (called the barycenter for astronomical objects orbiting each other) is only 450 km from the center of the sun.

Q7.10. Reason: In a sitting position, your center of gravity is located above the chair. When you lean forward, your center of gravity moves closer to the pivot point, which is your knees. You decrease the moment arm for the gravitational torque on your body by leaning forward. Your legs need to exert a smaller torque to overcome the gravitational torque if you lean forward. Assess: You can feel the effects of the larger gravitational torque by trying to rise without leaning forward.

Q7.11. Reason: Spin them. Because they would have different moments of inertia (2/5MR

2 for the solid

sphere and 2/3MR

2 for a thin-walled hollow sphere, with something in between if the wall of the hollow sphere is not particularly thin) their angular accelerations would be different with the same torque acting on them

(remember the rotational version of Newton’s second law: ! = "

net/I ). The solid one (with the smaller moment of

inertia) would accelerate quicker given the same torque. Assess: The two balls would have the same linear inertia (having the same mass) so dropping them in free fall, for example, and observing their linear accelerations would not distinguish between them; but the rotational inertia (moment of inertia) is different, and so exerting a net torque on them and measuring the resultant angular acceleration can distinguish between them.

Q7.12. Reason: The more distant an object’s mass is from an axis of rotation, the larger the moment of inertia of the object. Mass is distributed farther from the axis of rotation for a rod rotated about one end than a rod rotated about its center. Assess: From Equation 7.14, the contribution to moment of inertia is proportional to the square of the distance from the axis of rotation.

Q7.13. Reason: We will assume that the table has enough friction that the ends of the rods don’t slip. We will also ignore air resistance. The angular acceleration is the quantity of interest in this question. If the two rods have the same angular acceleration, they will hit the table at the same time if released at the same time from the same angle.

Rotational Motion 7-3

However, the angular acceleration is dependent on the torque and the moment of inertia: ! = "net/I. The net torque about the end of the rod on the table is the torque due to the weight of the rod (w = Mg) acting at the center of mass (L/2 from the end), because the normal and friction forces of the table on the end of the rod will produce a zero torque since the lever arm is zero. Both the heavy steel rod and the pencil will be modeled as a thin rod so that the moment of inertia (given in

Table 7.4) is I =

1

3ML

2, but of course neither M nor L is the same in the two cases.

Now it is important to watch the mass (as well as one L) cancel out of the calculation for

!.

! ="

net

I=

r(F ) sin #1

3ML

2=

r(w) sin #1

3ML

2=

(L

2)( Mg) sin #

1

3ML

2=

(1

2)g sin #

1

3L

=3g sin #

2L

What do we notice about the result? 1. M canceled, so the mass of the rod does not affect the angular acceleration. This is reminiscent of free fall

where the mass also cancels in a = F

net/m = mg/m.

2. Unlike free fall, since! (the angle the rod makes with the vertical) changes as the rod falls, the acceleration is

not constant. In our case! started out at 15° and increases to 90°; sin! also increases over that interval so the

acceleration increases as the rod falls. 3. One L canceled, but one is still left in the denominator; the longer the rod the smaller the angular

acceleration. This is the point that answers the question. Since the 1.0 m rod is longer than the 0.15 m pencil, it will have less angular acceleration and hence hit the table last.

The pencil hits first. Assess: The intuition we’ve developed about Galileo’s law of falling bodies (all bodies in free fall have the same acceleration) doesn’t quite apply in this case. Of course the rods aren’t in free fall; it is the way the L didn’t

cancel in !

net/I that makes the accelerations not the same.

Q7.14 Reason: A review of Section 7.6 in the text will prepare you to answer this question. Let’s separate the rolling motion into translation and rotation. Figure (a) below shows the translational velocity of the points of interest. Figure (b) below shows the velocity due to rotation at all points of interest. Finally, Figure (c) below

combines the velocity vectors due to both translation and rotation.

Looking at the magnitude of the resulting velocity vectors we can write the following:

!v

1>!v

2>!v

3>!v

4>!v

5= 0

Assess: We know that

!v

2 and

!v

3 are greater than

!v

4 because their horizontal component is equal to the

magnitude of

!v

4.

Q7.15. Reason: The bottom of the tire, the point in contact with the road, has (relative to the ground) an instantaneous velocity of zero, but the top of the tire has an instantaneous forward velocity twice as fast as the car’s forward velocity. See Figure 7.36. When the pebble works loose it flies off tangentially forward 60 mph faster than the car is going. So it hits the wheel well with a relative speed of about 60 mph. Assess: Yes, gravity pulls the pebble down and the trajectory is parabolic, but it is going so fast that it hits the wheel well before it falls far.

7-4 Chapter 7

Q7.16. Reason: Assuming all the forces have the same magnitude, the largest torque will be exerted on the nut by choice C. Choice A has a lesser torque than C because the moment arm of the force is smaller than in choice C. This can be seen from Equation 7.4. Choice D has a lesser torque than choice C because the component of the force perpendicular to the radial line is smaller. This can be seen from Equation 7.3. In choice B there is no torque exerted on the nut since the radial line and the force are parallel. The torque in this case is zero from Equation 7.5. Assess: The torque created by a given force can be increased by increasing the moment arm of a force or by increasing the component of the force perpendicular to the radial line.

Q7.17. Reason: We are given that r = 0.15 m and that

sin ! = 1.

Solve Equation 7.5 for F.

! = rF sin "

F =!

r sin "=

20 N #m

(0.15 m)(sin90°)= 133 N $ 130 N

The correct answer is C. Assess: This force is not unreasonable; it is like lifting 30 lbs. Notice the units worked out in the equation.

Q7.18. Reason: Since the center of gravity of piece two is to the right of the center of gravity of piece one, the horizontal position of the center of gravity of the two pieces should be between the center of gravity of the two pieces. The same argument applies to the vertical position of the center of gravity of the pieces. The only point that is located between the two centers of gravity is point D. Assess: Our solution to the problem is based on the fact that we can replace piece 2 with a single mass point (with the same mass as piece 2) located at the center of gravity of piece 2 and we can replace piece 1 with a single mass point (with the same mass as piece 1) located at the center of gravity of piece 1. We now need to find the center of gravity of these two mass points and our knowledge of the Physics involved makes us aware that it must be somewhere on the line connecting them. Only point D satisfies this condition.

Q7.19. Reason: We look up the formula for the moment of inertia of a disk about its axis of symmetry in Table 7.4. Neglect the small hole in the center. The radius is half the diameter.

I =1

2MR

2=

1

2(0.015 kg)(0.060 m)

2= 2.7 !10

"5 kg #m

2

The correct answer is A. Assess: The answer is a small number, but CDs are small and light. An old vinyl LP record would be both larger and more massive so it would have a larger moment of inertia, but modeled as a disk would still

be I =

1

2MR

2.

Q7.20. Reason: Since the diameter of the disk increases by a factor of two, the radius of the disk increases by a factor of two. Assume that the density of the material used is the same and that the hole in the center of the disk is negligible. The total mass of the disk goes up by a factor of four since the volume of the disk increases by a factor of four. The moment of inertia of a disk is given in Table 7.4 and is proportional to the mass of the disk and the square of the radius of the disk. Since the mass goes up by a factor of four and the radius goes up by a factor of two, the moment of inertia of the disk goes up by a total factor of sixteen. The correct choice is D. Assess: Note that the increase of the total mass of the disk needed to be included in the calculation.

Q7.21. Reason: Both doors are slabs and the moment of inertia about an edge (the hinge) is given by

Table 7.4: I =

1

3Ma

2 where a is the width of the door. We are told that the width of door 1 is a while the width of

door 2 is 2a:

! ="

net

I=

r(F ) sin #

I

and since Bob and Barb push “straight against” the door ! = 90° so

sin ! = 1:

!2

!1

=(2a)F/

1

3M (2a)

2

a(2F )/1

3Ma

2=

1

4


This means that !

1= 4!

2so the correct choice is A.

Assess: The torque exerted by Bob, (2a)F , is the same as the torque exerted by Barb,

a(2F ), but the moment of

inertia of door 2 is four times the moment of inertia of door 1.

Q7.22. Reason: Consider the angular acceleration of the bat as it is just about to tip, as in Example 7.12. The center of gravity of the bat will be nearer the barrel because most of the bat’s mass is concentrated in the barrel. With the barrel up, the moment arm for the gravitational torque is larger than with the barrel down. The moment of inertia of the bat is also larger when the barrel is up. Torque is proportional to the length of the moment arm of the applied force. Moment of inertia depends on the square of the distance to the particles that make up the object. The moment of inertia of the bat with the barrel up is much larger than with the handle down, though the gravitational torque is only slightly larger. The bat is easier to balance with the barrel up, so the correct choice is A. Assess: This result makes sense.

Q7.23. Reason: If we consider the motion of a piece of the tread to be a combination of the forward translational motion (which is constant) and the rotational motion, then we will simply compute the acceleration for the rotational part to be a = v2/r,where v is the speed of the center of mass, 30 m/s:

a =v2

r=

(30 m/s)2

0.37 m= 2432 m/s

2! 2400 m/s

2

The correct choice is C. Assess: The piece of tread moves through the cycloid path shown in Figure 7.35, but the acceleration is due to the change in velocity, and the translational velocity isn’t changing, only the rotational velocity is changing.

Problems

P7.1. Prepare: We’ll assume a constant acceleration during the one revolution. We’ll use the second and third

equations for circular motion in Table 7.2, the third to find! and then the second to find !

f.

Known

r =

1

2D = .90 m

!t = 1.0 s

!" = 1.0 rev = 2# rad Find

v

f= r!

f

Solve: The third equation in Table 7.2 allows us to solve for

!. That !

0= 0 makes it easier.

!" =1

2#(!t)

2

! =2"#

("t)2

=2(2$ rad)

(1.0 s)2

= 12.6 rad/s2

The second equation in Table 7.2 gives !

f:

!

f= !

0+"#t = 0 rad/s + (12.6 rad/s

2)(1.0 s) = 12.6 rad/s

Finally, we compute v

f= r!

f= (.90 m)(12.6 rad/s) = 11 m/s.

Assess: This speed seems reasonable, about 1/4 of a baseball fast pitch. The hammer throw is similar to the discus, but the weight is on a wire so the radius of the circular motion is a bit longer than the arm and the release speed is a bit larger, hence the distance it goes before landing is a few meters more.

P7.2. Prepare: The motion has two parts: the disk accelerates for some time until it reaches a constant angular velocity and then rotates at this constant angular velocity for the remainder of the time. We can use the equations in Table 7.2 to calculate the number of turns the disk has made.

7-6 Chapter 7

Solve: One additional significant figure has been kept in each of the intermediate results. The disk starts from rest, so !i = 0 rad/s. The disk’s final angular velocity converted to rad/s is

!f

= 7200 rpm =7200 rev

min

"

#$%

&'2( rad

rev

"

#$%

&'1 min

60 s

"

#$%

&'= 754 rad/s

We can find the time the disk takes to accelerate to this angular velocity using the second equation in Table 7.2.

!taccelerating

="

f#"

i

$=

754 rad/s # 0 rad/s

190 rad/s2

= 3.97 s

Using the third equation in Table 7.2, the total angular displacement of the disk during this time is

!"accelerating

=1

2#(!t

accelerating)

2=

1

2(190 rad/s

2)(3.97 s)

2= 1500 rad

After it has reached its final angular velocity !

f, the disk spins with that angular velocity for the remainder of the

time. The angular displacement during this time is given by the first equation in Table 7.2.

!"

constant#= #

f!t

constant#= (754 rad/s)(10.0 s $ 3.97 s) = 4550 rad

The total angular displacement of the disk is !" = 1500 rad + 4550 rad = 6050 rad.

Converting this result to revolutions, we have !" = 960 revolutions.

Assess: It’s important to realize that there are two different parts to this motion.

P7.3. Prepare: We assume constant angular acceleration; then we can use the equations in Table 7.2.

Known

!t = 2.0 s

!

0= 0

!

f= 3000 rpm

Find ! !"

Convert !" to rad/s.

!" = "f#"

0= 3000

rev

min

2$ rad

1 rev

%

&'(

)*1 min

60 s

%

&'(

)*= 314 rad/s

Solve: (a)

! ="#

"t=

314 rad/s

2.0 s$ 160 rad/s

2

(b) We’ll use the last equation in Table 7.2, using !

0= 0.

!" = #0!t +

1

2$(!t)

2=

1

2(157 rad/s

2)(2.0 s)

2= 314 rad

Finally, convert to revolutions:

314 rad = 314 rad1 rev

2! rad

"

#$%

&'= 50 rev

Assess: 50 rev seems like a lot in 2 s, but it is reasonable with the large angular acceleration and the final angular velocity of 3000 rpm.

P7.4. Prepare: We can use Equation 7.3 to calculate the torque in this case.


Solve: Consider the diagram in Example 7.5 with the radius of the wrench increased to 35 cm.

As in Example 7.5, the torque Luis exerts is given by ! = "rF

#= "rF sin$. The angle and the torque Luis exerts

is the same as in the example. Solving for the force required for a wrench with a longer handle,

F =!"

r sin #=

!(!17 N $m)

(0.35 m)sin 120°= 56 N

Assess: This result makes sense. The radius has been increased by almost a factor of two, so only about one half of the force is required to exert the same torque.

P7.5. Prepare: The magnitude of the torque in each case is ! = rF because

sin ! = 1.

Solve: 1 2 3 4

2 2 2 2rF r F rF r F! ! ! != = = =

Examining the above we see that 1 2 3 4

< <! ! ! != . Since for each case = rF! (because sin ! = 1 ), in order to

determine the torque we have just kept track of each force (F), the magnitude of the position vector (r) which locates the point of application of the force, and finally the product (rF). Assess: As expected, both the force, and the lever arm contribute to the torque. Larger forces and larger lever arms make larger torques. Case 4 has both the largest force and the largest lever arm, hence the largest torque.

P7.6. Prepare: Refer to Figure P7.6. We will calculate the torques !1 through !6 about the hinge using Equation 7.5, that is, ! = rFsin".

Solve: !1 = (L/2)F sin 90° = LF/2; !2 = (L/4)(2F) sin 90° = !1; !3 = (L/2)F sin 45° =1/ 2;! !4 = (L/2)F sin 135°

=1/ 2;! !5 = L(2F) sin 90° = 4!1; and !6 = LF sin 0° = 0. So, from smallest to largest, the ranking order of the

torques is !6,!3 = !4, !1 = !2, !5.

Assess: Note that " is the angle between r!

and .F!

P7.7. Prepare: Torque by a force is defined as ! = Fr sin " [Equation 7.5], where " is measured counterclockwise

from the r!

vector to the F!

vector. The radial line passing through the axis of rotation is shown below by broken line. We see that the 20 N force makes an angle of +90° relative to the radius vector r2, but the 30 N force makes an angle of –90° relative to r1.

Solve: The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 20 N force:

(30 N)r1

sin !1+ (20 N)r

2sin !

2= (30 N)(0.02 m)sin("90°) + (20 N)(0.02 m)sin(90°)

= ( " 0.60 N #m) + (0.40 N #m) = "0.20 N #m

Assess: A negative torque will cause a clockwise rotation of the pulley.

P7.8. Prepare: Torque by a force is defined through Equation 7.5. The radial line passing through the

axis of rotation is shown below by broken line. We see below that the force F!

applied by your pull makes an angle of –120° relative to the radial line.

7-8 Chapter 7

Solve: The net torque on the spark plug is

! = Fr sin " = #38 N $m = F(0.25 m)sin(#120°) % F = 180 N

That is, you must pull with a force of 180 N to tighten the spark plug. Assess: The force applied on the wrench leads to its clockwise motion. That is why we have used a negative sign for the net torque.

P7.9. Prepare: The height, thickness, and mass of the door are all irrelevant (for this problem, but the mass is important for Problem 7.31). If the door closer exerts a torque of 5.2 N ! m, then you need to also apply a torque of 5.2 N ! m in the opposite direction. The way to do that with the least force is to make r as big as possible (the entire width of the door), and make sure the angle 90 .! = ° Solve: From = sinrF! " solve for .F Then we see that the needed torque is produced with the smallest force by maximizing r and sin .!

F =!

r sin "=

5.2 N #m

(0.91 m)sin 90°= 5.7 N

Assess: It is good to have problems where more than the required information is given. Part of learning to solve real-world problems is knowing (or learning) which quantities are significant, which are irrelevant, and which are negligible. Of course, in this case the mass is used later in Problem 7.31. The answer of 5.7 N seems like a reasonable amount of force, which might be supplied, say, by a doorstop. If your doorstop is a simple wedge of wood inserted under the door (as they are at my college), you can see that it should be positioned near the outside edge of the door so the friction force will produce enough torque to keep the door open.

P7.10. Prepare: We can use Equations 7.5 and 7.3 to calculate the torques due to each force and then set the net torque to zero to solve for F2. Solve: Using Equation 7.4 to calculate the torque due to

1F with

145 ,! = °

1 1 1 1 1sin( ) (20.0 N)sin(45 )r F r! "= + = °

At this point, we can’t calculate the numerical value of this torque, since1r is not given. Since

2F is perpendicular

to the radial line the torque due to F2 is

1

2 2 2 2 2 2( )

2

rr F r F F! " #

= $ = $ = $% &' (

A negative sign has been inserted since F2 tends to turn the rod clockwise. Setting the net torque to zero gives the equation

1

1 2 1 2(20.0 N)sin(45 ) 0

2

rr F! ! " #

+ = ° $ =% &' (

Solving for2

F gives

F

2= 2(20.0 N)sin 45° = 28.3 N

Assess: This answer makes sense. F1 is twice as far from the pivot but its contribution to the torque is diminished by almost a factor of two due to the angle it makes with the radial line. The force F2 should be a larger than the force F1.


P7.11. Prepare: For both Tom and Jerry 1.5 mr = (or half the diameter).

Compute the magnitude of each torque.

!

Tom= (1.50 m)(40.0 N)sin 60° = 51.9615 N "m

!

Jerry= (1.50 m)(35.2 N)sin 80° = 51.9978 N "m

We keep extra significant figures in the intermediate calculations because we will be subtracting two nearly equal numbers. Solve: (a) The torque due to Tom will be positive because it will tend to produce counterclockwise rotation, while the torque due to Jerry will be negative because it will tend to produce clockwise rotation.

Tom Jerry51.9615 N m ( 51.9978 N m) 0.0363 N m! ! != + = " + # " = "

(b) With both Tom’s and Jerry’s torques tending to produce counterclockwise rotation, they are both positive and so the net torque would be their sum:

Tom Jerry51.9615 N m 51.9978 N m 104 N m! ! != + = " + " = "

Assess: As the difference between the two parts of the problem demonstrates, the direction of the forces really matters. But we can produce torques of the same magnitude with forces of different magnitude by adjusting the angles—even with the same .r Though we used extra significant figures in the intermediate calculations, we still report the answer to three significant figures.

P7.12. Prepare: Assume the bar is weightless. We can calculate the torques due to both forces using Equation 7.3. Solve: Refer to the diagram below.

For

1F

1 1 1( ) (0.75 m)(10 N) 7.5 N mr F!

"= # = # = # $

This torque is negative since1

F tends to turn the bar clockwise.

For2

F

2 2 2( ) (0.75 m)(0 N) 0 N mr F!

"= = = #

The net torque about the pivot is1 2

7.5 N m.! !+ = " #

Assess: The torque due to F2 is zero because it has no component perpendicular to the radial line.

P7.13. Prepare: Knowing that torque may be determined by ! = rF

", that counterclockwise torque is

positive and clockwise torque negative we can determine the net torque acting on the bar. Solve:

!clockwise

= " (0.25 m)(8.0 N) = "2.0 N #m

!counterclockwise

= (0.75 m)(10 N) = 7.5 N #m

!net

= !counterclockwise

+ !clockwise

= 7.5 N #m " 2.0 N #m = 5.5 N #m

Since the net torque is + 5.5 N !m, the bar will rotate in the counterclockwise direction around the dot.

Assess: The counterclockwise torque had both the larger r and the larger F, so the net torque was also counterclockwise. The numbers also seem reasonable, and the units work out.

7-10 Chapter 7

P7.14. Prepare: We can use Equation 7.3 to calculate the net torque on the bar. Recall that clockwise torques are negative and counterclockwise torques are positive. Solve: Refer to the diagram below.

Calculate the torque associated with

1F

1 1 1 1 1 1( ) sin( ) (0.75 m)(10 N)sin(30 ) 3.75 N mr F r F! "

#= = = ° = $

An additional significant figure has been kept in the intermediate result above. Calculation the torque associated with

2F

2 2 2 2 2 2( ) sin( ) (0.25 m)(8.0 N)sin(40 ) 1.29 N mr F r F! "

#= $ = $ = $ ° = $ %

The net torque is

!

1+!

2= 3.75 N "m–1.29 N "m = +2.46 N "m

Since all information is given to two significant figures, the result should be reported to two significant figures as

!

net= 2.5 N "m

Assess: This result makes sense. Force1

F is at a much larger distance from the axis than2,F while both forces

are close in magnitude and in the angle they make with the radial line. The net torque should act to turn the bar counterclockwise.

P7.15. Prepare: We’ll set up the coordinate system so the barbell lies along the x-axis with the origin at the left end and use Equation 7.7 to determine the x -coordinate of the center of gravity. We’ll model each weight (and the barbell itself in part (b)) as a particle, as if the mass were concentrated at each item’s own center of gravity. Solve: (a)

1 1 2 2

cg

1 2

(0.0 m)(20 kg) (1.7 m)(35 kg)1.08 m 1.1m

20 kg 35 kg

x m x mx

m m

+ += = = !

+ +

(b) In this part we include the barbell (call it particle 3), whose own center of gravity is at its geometrical center

3( 1.7 m/2 0.85 m),x = = since we assume it has uniform density.

1 1 2 2 3 3

cg

1 2 3

(0.0 m)(20 kg) (1.7 m)(35 kg) (0.85 m)(8.0 kg)1.25 m 1.2 m

20 kg 35 kg 8.0 kg

x m x m x mx

m m m

+ + + += = = !

+ + + +

Assess: To two significant figures the answers to both parts are close. Taking the barbell into account didn’t move the center of gravity much for two reasons: It wasn’t very massive, and its center of mass was already near the center of mass of the system.

P7.16. Prepare: The procedure in Tactics Box 7.1 can be used to calculate the center of gravity of the coins. Solve: Refer to the figure below.


The coordinates for the three masses are

1 1

2 2

3 3

0 m 0 m

0 m 0.1m

0.1m 0 m

x y

x y

x y

= =

= =

= =

All the coins have the same mass since they are identical. The x and y coordinates of the center of gravity of the coins may be determined by

1 1 2 2 3 3 1 2 3 3

cg 3

1 2 3

1 1 2 2 3 3 1 2 3 2

cg 2

1 2 3

1 1(0.1m) 0.03 m

3 3 3

1 1(0.1m) 0.03 m

3 3 3

x m x m x m x m x m x m x mx x

m m m m m m m

y m y m y m y m y m y m y my y

m m m m m m m

+ + + += = = = = =

+ + + +

+ + + += = = = = =

+ + + +

Assess: Most of the mass lies near the x and y axes, so this answer makes sense.

P7.17. Prepare: How will you estimate the mass of your arm? Of course different people’s arms have different masses, but even different methods of estimating the mass of your specific arm will produce slightly different results. But it is a good exercise, even if we have only one significant figure of precision. One way would be to make some rough measurements. Because the density of your arm is about the density of water, you could fill a garbage can to the brim with water, insert your arm, and weigh the water that overflows. Or you could look at Figure 7.24, which indicates that a guess of m = 4.0 kg for a person whose total mass is 80 kg is good. The gravitational force on a 4.0 kg object is w = mg (4.0 kg)(10 m/s2) = 40 N. I know it is about 1 yd from the tip of my outstretched arm to my forward-facing nose, but a meter is a bit bigger than a yard, so I would estimate that from fingertip to shoulder would be about 0.7 m. If I model my arm as a uniform cylinder of uniform density then the center of gravity would be at its center—0.35 m from the shoulder joint (the “hinge”). If I further refine the model of my arm, it is heavier nearer the shoulder and lighter toward the hand, so I will round the location of its center of gravity down to 0.30 m from the shoulder. Solve: Since the arm is held horizontally sin 1.! =

! = rF sin " = (0.30 m)(40 N)(1) = 12 N #m

Assess: Your assumptions and estimates (and, indeed, your arm) might be different, but your answer will probably be in the same order of magnitude; you probably won’t end up with an answer 10 ! bigger or 10 ! smaller.

P7.18. Prepare: The procedure in Tactics Box 7.1 provides a guide for determining the center of gravity of an object and/or combination of objects. It is important to note that the x component of the center of center of gravity is on the line of symmetry. Solve: The y component of the center of gravity is determined by:

ycm

=y

1m

1+ y

2m

2

m1+ m

2

=(2.5 cm)(800 g) + (7.5 cm)(400 g)

800 g + 400 g= 4.2 cm

Assess: Note that it was not necessary to convert the units to the MKS system. Since the mass units cancel, any appropriate unit may be used. Since the number is small, cm is an appropriate unit. Note that the center of gravity of the two objects is inside the cube and beneath the cylinder. Since they have the same height but the cube has twice the mass, this is a reasonable answer.

P7.19. Prepare: First let’s divide the object into two parts. Let’s call part #1 the part to the left of the point of interest and part #2 the part to the right of the point of interest. Next using our sense of center of gravity, we know the cm of part #1 is at 12.5 m and the cm of part #2 is at +37.5 cm. We also know the mass of part #1 is one fourth the total mass of the object and the mass of part #2 is three fourths the total mass of the object. Finally, we can determine the gravitational torque of each part using any of the three expressions for torque as shown below:

Equation (7.3) ! = rF

" is straightforward to use because the forces are perpendicular to the position vectors,

which locate the point of application of the force.

7-12 Chapter 7

Equation (7.4) ! = r

"F is straightforward to use because the position vectors that locate the point of application

of the forces are also the moment arms for the forces.

Equation (7.4) ! = rF sin " is straightforward to use because the angles are either 90

! or 270!

Solve: Using equation (7.4) we obtain the following

!

net= !

1+ !

2= m

1gr

1sin(90

!) + m

2gr

2sin("90

!)

= (0.5 kg)(9.8 m / s

2)(!0.125 m)(1) + (0.75 kg)(9.8 m / s

2)(0.375 m)(!1) = !2.1 N "m

Assess: According to this answer, if released the object should rotate in a clockwise direction. Looking at the figure this is exactly what we would expect to happen.

P7.20. Prepare: The beam is a solid rigid body. The steel beam experiences a torque due to the construction

worker’s weight

!w

Cand the beam’s weight

!w

B. The normal force exerts no torque since the net torque is

calculated about the point where the beam is bolted into place. The net torque on the steel beam about point O is the sum of the torque due to

Cw!

and the torque due toB.w!

The weight of the beam acts at the center of mass.

Solve:

! = (wC)(4.0 m) sin (!90°) + (wB)(2.0 m) sin (!90°)

= ! (70 kg)(9.80 m/s2)(4.0 m) – (500 kg)(9.80 m/s2)(2.0 m) = –12,000 N !m

The magnitude of the torque is 12,000 N !m.

Assess: The negative torque means these forces would cause the beam to rotate clockwise, as you can see without doing any calculations.

P7.21. Prepare: Model the arm as a uniform rigid rod. Its mass acts at the center of mass. The force in each case is the weight, .w mg=

Solve: (a) The torque is due both to the weight of the ball and the weight of the arm.

! =!ball

+!arm

= rb(m

bg)sin 90° + r

a(m

ag)sin 90°

= (0.70 m)(3.0 kg)(9.8 m/s2) + (0.35 m)(4.0 kg)(9.8 m/s

2

) = 34 N "m

(b) The torque is reduced because the moment arms are reduced. Both forces act at 45! = ° from the radial line, so


! =!ball

+!arm

= rb(m

bg)sin 45° + r

a(m

ag)sin 45°

= (0.70 m)(3.0 kg)(9.8 m/s2)(0.707) + (0.35 m)(4.0 kg)(9.8 m/s

2)(0.707) = 24 N "m

Assess: This problem could also have been done, as in Problem 7.17, by first finding the center of mass of the arm-ball system and computing the torque due to the combined weight acting at that point. The final answers would be the same.

P7.22. Prepare: We can use Equation 7.3 or 7.4 to calculate the torque. The weight of the beam acts at the center of gravity of the beam. Solve: The center of gravity of the beam is at the center of the beam, which is 1.0 m from the hinge. Each of the torques is negative since gravity tends to turn the bar in a clockwise direction in each case. The weight of the beam is w = mg (15 kg)(9.80 m/s2) = 147 N. An additional significant figure has been kept in this intermediate result. (a) A diagram of the beam in the upper position is below.

We can calculate the torque using Equation 7.3. The perpendicular component of the weight is cos( ).w ! The

torque is

! = "rw

#= "rw cos($ ) = "(1.0 m)(147 N)cos(20°) = "140 N %m

(b) A diagram of the beam in the middle position is below.

We can calculate the torque using Equation 7.5. The radial line is along the x-axis. The angle between the weight and the radial line is 90 .°

! = "rw sin# = "(1.0 m)(147 N)sin(90°) = "150 N $m

(c) A diagram of the beam in its lower position is below.

7-14 Chapter 7

Equation 7.4 can be used to calculate the torque. The moment arm for the force is shown in the diagram. It is the perpendicular distance from the hinge to the line of action of the force. The torque is

! = "r

#w = "(r cos$ )w = "(1.0 m)cos(45°)(147 N) = "100 N

Assess: Any one of Equations 7.3 through 7.5 can be used to calculate torque.

P7.23. Prepare: Set up a coordinate system along the beams with the origin at the left end. Assume each beam is of uniform density so its own center of gravity is at its geometrical center. Then use Equation 7.7. Solve: (a)

1 1 2 2

cg

1 2

(0.500 m)(10.0 kg) (1.00 m 1.00 m)(40.0 kg)1.70 m

10.0 kg 40.0 kg

x m x mx

m m

+ + += = =

+ +

(b) The gravitational torque on the two-beam system is the total weight acting at the center of gravity of the

system: 2(10.0 kg 40.0 kg)(9.80 m/s ) 490 N.w mg= = + =

sin (1.70 m)(490 N)sin90 833 N mrF! "= = ° = #

Because this torque is in the clockwise direction we report it as a negative torque: 833 N m.! = " #

Assess: This problem could also be done by computing the gravitational torque individually on each beam and adding them up. The final answer would be the same. For the beams to remain in equilibrium some other object must supply an equal torque in the opposite direction.

P7.24. Prepare: Equations 7.8 and 7.9 can be used to calculate the center of gravity of the joined beams. To calculate the gravitational torque we can use the fact that the weight of the entire structure acts on its center of gravity. Solve: (a) The center of gravity of each beam is at its center. Refer to the diagram below.

Using Equations 7.9 and 7.10, the position of the center of gravity is

v v cg h h cg h h cg

cg

v h v h

v v cg h h cg v v cg

cg

v h v h

(25.0 kg)(1.00 m)0.625 m

(25.0 kg) (15.0 kg)

(15.0 kg)(0.500 m)0.188 m

(25.0 kg) (15.0 kg)

m x m x m xx

m m m m

m y m y m yy

m m m m

+= = = =

+ + +

+= = = =

+ + +


(b) The gravitational torque about the origin is

2(0.625 m)(15.0 kg 25.0 kg)(9.80 m/s ) 245 N mr w!

"= # = # + = $

Assess: This result makes sense. The center of gravity should be close to the vertical and horizontal axes.

P7.25. Prepare: A table tennis ball is a spherical shell, and we look that up in Table 7.4. The radius is half the diameter. Solve:

2 2 7 22 2

(0.0027 kg)(0.020 m) 7.2 10 kg m3 3

I MR!

= = = " #

Assess: The answer is small, but then again, it isn’t hard to start a table tennis ball rotating or stop it from doing so. By the way, this calculation can be done in one’s head without a calculator by writing the data in scientific notation and mentally keeping track of the significant figures:

3 2 4 2 2 4 42 2 2

8 8 72 2 2

2 2 27(2.7 10 kg)(2.0 10 m) (27 10 kg)(2 10 m) 2 2 10 10 kg m

3 3 3

9 8 10 kg m 72 10 kg m 7.2 10 kg m

I! ! ! ! ! !

! ! !

= " " = " " = # # " " #

= # " # = " # = " #

P7.26. Prepare: Refer to Figure P7.26. We will calculate the moments of inertia I1 through I3 about axes through

the centers of the rods using Equation 7.14, that is, 2 2

1 1 2 2.I m r m r= +

Solve:

2 2 2

1

2 2 2

2 1

( /2) ( /2) /2

2 ( /4) ( /4) 3 /16 3 /8

I m R m R mR

I m R m R mR I

= + =

= + = =

and 2 2 2

3 1( /2) ( /2) 2I m R m R mR I= + = =

So, from smallest to largest, the ranking order of the moments of inertia is I2, I1, and I3. Assess: The moment of inertia is linearly proportional to mass and also to the second power of distance from the axis. Although the masses in case 3 are one-half the masses in case 1, the distance of the masses from the axis for case 3 is twice the distance for case 1. As a result I3 is larger than I1.

P7.27. Prepare: When the problem says the connecting rods are “very light” we know to ignore them (especially since we aren’t given their mass). Since all of the mass (seats plus children) is equally far from the axis of rotation we simply

use 2=I MR where M is the total mass.

Solve:

I = MR

2= [4(5.0 kg) +15 kg + 20 kg](1.5 m)

2= 120 kg !m

2

Assess: When all of the mass is the same distance R from the axis of rotation, the calculation of I is relatively simple.

P7.28. Prepare: We can obtain expressions for the moment of inertia for a cylinder and a sphere from Table 7.2. Given that the masses and the moments of inertia are the same, we can obtain a relationship between their radii and then determine the radius of the sphere. Solve: Knowing that the moment of inertia of the cylinder is equal to the moment of inertia of the sphere we may write

I

cylinder= I

sphere

Inserting expressions for the moment of inertia from Table 7.2

M

cR

c

2/ 2 = 2M

sR

s

2/ 5

Cancel the masses (recall they are equal) and solve for R

s

R

s= (5 / 4)

1/2R

c= 4.5 cm

7-16 Chapter 7

Assess: Knowing that the masses are equal and that the coefficient for the cylinder is 0.50 and for the sphere 0.40, we should expect the radius of the sphere to be greater then the radius of the cylinder in order for them to have the same moment of inertia.

P7.29. Prepare: Treat the bicycle rim as a hoop, and use the expression given in Table 7.4 for the moment of inertia of a hoop. Manipulate this expression to obtain the mass. Solve: The mass of the rim is determined by

m = I / R

2= 0.19 kg !m

2/ (0.65 m / 2)

2= 1.8 kg

Assess: Note that the units reduce to kg as expected. This amount seems a little heavy, but since we are not told what type of bicycle it is not an unreasonable amount.

P7.30. Prepare: We can calculate the angular acceleration of the balls using Equation 7.15. We can calculate the moments of inertia with Equation 7.14. Equation 7.3 can be used to calculate the torque. Assume the distance of each mass from the axis of rotation is

R.

Solve: Using Equation 7.14, the moment of inertia of the left object is I

L= MR

2+ MR

2= 2MR

2.

The moment of inertia of the right object is I

R= 2MR

2.

So, each object has the same moment of inertia. Using Equation 7.3, the torque on the left object is

L.rF RF!

"= # = #

The torque on the right object is R

.rF RF!"

= # = #

Since the torques and the moments of inertia are equal for both the left and right objects, the angular acceleration will be the same. Assess: This result makes sense. The torque is the same in each case since the distance from the axis and force applied is the same. The moments of inertia are equal though the configuration of masses is different because in each case the same amount of mass is at the same radius from the axis.

P7.31. Prepare: We will assume we are to compute I about an edge (the hinge of the door), so from Table 7.4

we have 21

3I Ma= where 0.91ma = and 25 kg.M =

Solve: (a)

2 2 21 1

(25 kg)(0.91m) 6.9 kg m3 3

I Ma= = = !

(b) After we let go of the door (or remove the doorstop) the net torque is the torque due to the door closer, which Problem 7.9 says is 5.2 N m.!

net 2

2

5.2 N m0.75 rad/s

6.9 kg mI

!"

#= = =

#

Assess: The results seem to be in a reasonable range. As long as the net torque of 0.52N !m continues to act, the

door will continue to accelerate. Usually there are dampeners to exert a slowing torque so that the door doesn’t slam.

P7.32. Prepare: We can calculate the angular acceleration using Equation 7.15. Solve: Using Equation 7.15 we find

5 2 2 3(4.0 10 kg m )(150 rad/s ) 6.0 10 N mI! "

# #= = $ % = $ %

Assess: This result is reasonable. The moment of inertia of the grinding wheel is small.

P7.33. Prepare: We will assume that the torque produced by the frictional force of the floor on your foot is the net torque (i.e., we will ignore frictional and other torques). We know that F = 7.0 N, r = 0.40 m, and ! = 1.8 rad/s2. We are also indirectly told (“in the direction that causes the greatest angular acceleration”) that 90 .! = ° We will solve for I from the rotational version of Newton’s second law:

net/ .I! "=

Solve:

I =!

net

"=

rF sin#

"=

(0.40m)(7.0 N)(sin 90°)

1.8 rad/s2

= 1.6 kg $m2


Assess: The answer is not particularly large, but your mass is distributed quite close to the axis of rotation,

so I is small. Carefully observe the units work out since N = kg !m/s

2.

P7.34. Prepare: We are given the moment of inertia I and the angular acceleration ! of the object. Since " = I! is the rotational analog of Newtons second law F = ma, we can use this relation to find the net torque on the object. Solve: " = (2.0 kg m2)(4.0 rad/s2) = 8.0 kg m2/s2 = 8.0

N !m Assess: The units are correct and the relatively small torque magnitude is consistent with the size of the moment of inertia and the angular acceleration.

P7.35. Prepare: A circular plastic disk rotating on an axle through its center is a rigid body. Assume the axis is perpendicular to the disk. Since " = I! is the rotational analog of Newton’s second law F = ma, we can use this relation to find the net torque on the object. To determine the torque (") needed to take the plastic disk from #i = 0 rad/s to #f = 1800 rpm = (1800)(2$)/60 rad/s = 60$ rad/s in tf – ti = 4.0 s, we need to determine the angular acceleration (!) and the disk’s moment of inertia (I) about the axle in its center. The radius of the disk is R = 10.0 cm. Solve: We have

I =1

2MR

2=

1

2(0.200 kg)(0.10 m)

2= 1.0 !10

"3kg #m2

$f

= $i+%(t

f" t

i) &% =

$f"$

i

tf" t

i

=60' rad/s " 0 rad/s

4.0 s= 15' rad/s

2

Thus, " = I! = (1.0 ! 10"3 kg !m2)(15$ rad/s2) = 0.047 N !m.

Assess: The solution to this problem required a knowledge of torque, moment of inertia, rotational dynamics and rotational kinematics. You should consider it an accomplishment to have mastered these concepts and then combined them to solve a problem.

P7.36. Prepare: Equation 7.15 can be used to find the initial angular acceleration once the torque is known. The torque can be calculated with Equation 7.4. Solve: Refer to the diagram below.

Gravity is the only force acting on the object. The object’s weight has a moment arm equal to cos .r r !

"= The

gravitational torque on the object is

2cos (0.17 m)(2.5 kg)(9.80 m/s )cos(25 ) 3.77 N mr w rw! "

#= $ = $ = $ ° = $ %

An additional significant figure has been kept in this intermediate calculation. The angular acceleration is then

! ="

I=

#3.77 N $m

0.085 kg $m2

= #44 rad/s2

Assess: Any one of Equations 7.3 through 7.5 can be used to calculate the torque. In this problem, using

Equation 7.4 avoids the need to determine the angle between the direction of the vectors r!

and W

!"!

.

P7.37. Prepare: What causes angular accelerations? (Net) torques. We’ll apply the rotational version of Newton’s second law. We’ll write the net torque as #" to emphasize that we are summing the two given torques; the 12 N force is producing a positive (counterclockwise) torque, while the 10 N force is producing a negative (clockwise) torque. For each torque R = 0.30 m.

7-18 Chapter 7

We will assume that the rope comes off tangent to the pulley on each side, so that 90! = ° and sin ! = 1.

Looking up the formula for I of a cylinder in Table 7.4, and using 0.80 kg,M = gives

2 2 21 1

(0.80 kg)(0.30 m) 0.036 kg m2 2

I MR= = = !

Solve:

2

2 2

(0.30 m)(12 N) (0.30 m)(10 N) (0.30 m)(12 N 10N)17 rad/s

0.036 kg m 0.036 kg mI

!"

# $ $= = = =

% %

Assess: This result answers the question. The proper units cancel to give! in rad/s2. Notice that the specific angles the ropes make with the vertical do not matter, as long as they are exerting torques in opposite directions and coming off of the pulley tangentially.

P7.38. Prepare: Equation 7.15 can be used to calculate the frictional torque once the angular acceleration is known. The second equation in Table 7.2 can be used to calculate the angular acceleration. Solve: The wheel stops spinning in 12 seconds from an initial angular velocity of 0.72 revolutions per second. The initial angular velocity in radians per second is

!i= 0.72 rev/s = (0.72 rev/s)

2" rad

rev

#

$%&

'(= 4.5 rad/s

The angular acceleration of the wheel is

24.5 rad/s

0.38 rad/s12 st

!"

# $= = = $

#

The only force acting on the wheel is friction. Using the results above in Equation 7.16 the magnitude of the frictional torque is

2 2(0.30 kg m )(0.38 rad/s ) 0.11 N mI! "= = # = #

Assess: This is a relatively large amount of friction. Assuming the torque of the wheel bearings acts at a radius of around a centimeter the frictional force exerted by the bearings is over 10 N. This shows you just one of the problems you will have if you do not properly maintain your bicycle.

P7.39. Prepare: With a constant string tension we will have constant angular acceleration, so we can use the equations in Table 7.2, but it is also clear that we will need to do a preliminary calculation to get !. The string will come off tangentially so that 90 .! = ° We will also interpret “to get the top spinning” as

o0.! = The radius is

half the diameter.

Known

5 2

o

0.025 m

0.30 N

90

3.0 10 kg m

5.0 rev 31.4 rad

0

r

F

I

!

"

#

$

=

=

= °

= % &

' = =

=

Find

! (preliminary) t!

net 2

5 2

sin (0.025 m)(0.30 N)(1)250 rad/s

3.0 10 kg m

rF

I I

! "#

$= = = =

% &


Solve: Solve the angular displacement equation for circular motion in Table 7.2 (witho

0)! = for :t!

2

2

1( )

2

2 2(31.4 rad)0.50 s

250 rad/s

t

t

! "

!

"

# = #

## = = =

Assess: As you may know from personal experience, it doesn’t take long for a toy top to complete five revolutions, and this bears that out.

P7.40. Prepare: This is an excellent capstone problem. In order to work it we need know to have mastered moment of inertia, rotation dynamics, and rotational kinematics. It should be noted the we know very little about the pot (only the radius). However if you go around your house and pick up pottery with about this radius, you will find it is very light compared to the massive potters wheel. In addition its radius is less than half that of the wheel. As a result, we will ignore the relatively small impact of the pottery and just use the potters wheel. Knowing the dimensions of the wheel we can determine its moment of inertia. Knowing the force of friction and radius of the wheel we can determine the torque due to friction that is slowing the wheel. Knowing the moment of inertia and the torque due to friction, we can establish the angular deceleration and then the stopping time. Solve: An expression for the moment of inertia of the potters wheel is

I = MR2

/ 2

The decelerating torque due to friction is

! = fR

! = fR

We can combine these to obtain the angular deceleration

! = "# / I = " fR / ( MR

2/ 2) = "2 f / MR

Using rotational kinematics, obtain the time to stop the wheel

t = (! "!o) /# = "("!

oMR / 2 f ) = 180

rev

min

$

%&'

()2* rad

rev

$

%&'

()min

60 s

$

%&'

()(20 kg)(0.17 m) / 2(2.3 N) = 25 s

Assess: Considering the size and mass of the wheel, this is a reasonable amount of time for the wheel to slow to a stop.

P7.41. Prepare: The rope is a constraint that makes the magnitudes of the accelerations of the blocks the same, and if the rope doesn’t slip that further implies that ablock = !pulleyR. See Equation 7.16.

We will assume that the pulley is a solid cylinder, so that I

pulley=

1

2MR

2. “Light rope” means we can assume that

the mass of the rope is zero. For the torque calculation r = R = 0.30 m.

Known

1

2

0.30 m

2.5 kg

1.5 kg

0.75 kg (pulley)

90

R

m

m

M

!

=

=

=

=

= °

Find a

7-20 Chapter 7

In order to keep track of directions, draw and label a figure.

We will need to set up a system of three equations in three unknowns, because we do not know the tensions T1 and T2 in the ropes (they are not equal to each other nor to the weights of the blocks). Solve: Use F ma! = for each block. Call up positive, so block 1 will have a negative (downward) acceleration because it weighs more than block 2.

1 1 1

T m g m a! = ! (1)

2 2 2

T m g m a! = (2)

Also write an equation for the rotational form of Newton’s second law for the pulley. The forces producing torques are

1T and

2.T

1 2

( )T T R I!" = (3)

These are the three equations with three unknowns. Now substitute /a R! = and divide both sides by R so that equation (3) becomes

1 2 2

aT T I

R! = (4)

Solve equations (1) and (2) for1

T and2

T and then subtract equation (2) from equation (1).

1 2 1 2 1 2

( ) ( )T T m m g m m a! = ! ! + (5)

Now set1 2

T T! from equations (4) and (5) equal to each other.

1 2 1 22

( ) ( )a

I m m g m m aR

= ! ! + (6)

All that remains is to solve equation (6) for a.

2

1 2 1 22

1 2

1 2

( ) ( )

( )

( )I

R

Ia m m m m g

R

m m ga

m m

! "+ + = #$ %

& '

#=

+ +

Substitute 21

2I MR= for the pulley.

1 2

2

2

2 2

1 2 1 2 2

0.75 kg

1 22 21 2

( ) ( ) (2.5 kg 1.5 kg)(9.8 m/s ) (1.0 kg)(9.8 m/s )2.2 m/s

( ) 0.375 kg 4.0 kg(2.5 kg 1.5 kg)( )MR M

R

m m g m m ga

m mm m

! ! != = = = =

+ + ++ ++ +

So block 2 (the lighter one) accelerates up at 2.2 m/s2. Assess: Because we had three unknowns, T1, T2, and a, we needed three equations. We are glad to get an answer less than 9.8 m/s2 since that is the limit that block 1 could accelerate down if m2 and M were zero. In the other limit, a ! 0 as (m1 " m2) ! 0, as we would expect. This set-up with blocks of unequal masses connected by a light rope over a pulley is called an Atwood Machine.


P7.42. Prepare: We can use the rolling constraint to find the speed of the dot and the angular speed of the tires. Solve: (a) Since the tires are rolling without slipping the angular velocity of the tires is given by Equation 7.18.

cm5.6 m/s

14 rad/sR 0.40 m

v! = = =

(b) The blue dot is undergoing translation and rotation. At the top of the tire, it has a translational velocity equal to the speed of the bike, and an additional velocity equal to R! due to the rotation of the tire. See Figure 7.36. The speed of the dot at this point is

cm cm2 2(5.6 m/s) 11 m/sv v R v!= + = = =

(c) See the diagram below.

The dot has a translational velocity equal to the velocity of the center of mass of the tire in the horizontal direction. The tangential velocity of the dot is in the vertical direction and has a magnitude equal to .R! The

velocity of the dot is equal to the sum of these two vectors. Since the tire is rolling without slipping,cm

.v R!=

See the vector diagram below.

The speed of the dot is equal to

2 2

cm cm( ) ( ) 7.9 m/sv v v= + =

Assess: Note that the speed of the dot during the downward motion for part (c) is equal to the speed of the dot during the upward motion shown in the diagram.

P7.43. Prepare: The component of the static friction force that keeps the pebble in circular motion is

centripetal and the centripetal force is determined by F

c= mv

2/ R .

We must remember that the top of the tire is going twice as fast as the car (see Figure 7.36); therefore the pebble will be released when it is at the top. We are given 0.0012 kg,m = /2 0.3 6 m,r D= = and the central frictional force 3.6 N.F =

Solve: At the point just before release2

v

rF=ma m .= Solve this for v (the speed of the top of the tire).

/ (3.6 N)(0.38 m)/0.0012 kg 33.8 m/sv Fr m= = =

The car is going half the speed of the pebble, or 17 m/s. Assess: This is equal to 38 mph, which is a reasonable speed.

7-22 Chapter 7

P7.44. Prepare: The particle is moving in a circle and a motion of 2! radians corresponds to one full rotation. Angular velocity at any given time is defined as ! = !"/!t, or the slope of the angular-position-versus-time graph. Solve: (a) From t = 0 s to t = 1 s the particle rotates clockwise from the angular position + 4# rad to "2# rad. Therefore, 2 ( 4 ) 6 rad! " " "# = $ $ + = $ in one sec, or 6 rad/s.! "= # From 1st = to 2 s,t = 0 rad/s.! = From t =

2 s to t = 4 s the particle rotates counterclockwise from the angular position –2# rad to 0 rad. Thus 0 ( 2 )! "# = $ $ =

2# rad and rad/s.! "= + (b)

Assess: Since we take positive angular displacements as counterclockwise and negative angular displacements clockwise, we know the particle is traveling around the circle in a clockwise direction. Knowing that the slope of the angular position-versus-time plot is the angular velocity, we can establish a plot of angular velocity-versus-time.

P7.45. Prepare: The crankshaft is a rotating rigid body. The crankshaft’s angular acceleration is given as $ = !!/!t , or slope of the angular-velocity-versus-time graph. Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s, maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up to 200 rad/s from t = 4 s to t = 7 s. The angular acceleration ($) graph is based on the fact that $ is the slope of the !-versus-t graph.

Assess: Knowing that the slope of the angular velocity-versus-time plot is the angular acceleration, we can establish a plot of angular acceleration-versus-time.


P7.46. Prepare: A computer disk rotating on an axle through its center is a rigid body. Assuming the axis of rotation to be perpendicular to the disk, the dot on the edge of the disk is in nonuniform circular motion. A pictorial representation of the problem is shown below. We will first find the angular acceleration of the disk from given information during the first 0.50 s, and then use this acceleration to find other quantities. While !0 = 0, the angular velocity at t = 0.50 s needs to be converted to SI units: 2000 rev/min (1 min/60 s)(2! rad/1 rev) = 209.4 rad/s.

Solve: (a) Determine the angular acceleration and then use Equation 7.2 to determine the tangential acceleration

2

f i f i( ) (209.4 rad/s) 0 (0.50 s 0 s) 418.9 rad/st t! ! " " "= + # $ = + # $ =

Because " is a constant for the first 0.50 s, the tangential acceleration at t = 0.25 s is

2 2 2

t(0.04 m)(418.9 rad/s ) 16.76 rad/s 17 rad/sa r!" = = = =

(b) The centripetal acceleration is 2,

ca r!= so we must first find ! at t = 0.25 s and then the centripetal

acceleration.

2

f i f i( ) 0 (418.9 rad/s )(0.25 s 0 s) 104.7 rad/st t! ! "= + # = + # =

Thus,

2 2 2(0.04 m)(104.7 rad/s) 440 m/s

ca r!= = =

(c) Using this result, we can determine the angular position

!1

= !0

+"0(t

1# t

0) +

1

2$(t

1# t

0)

2= 0.785 rad + 0 rad +

1

2(418.9 rad/s

2)(0.5 s # 0 s)

2= 53.15 rad = 8.46 rev

2

1 0 1 0( ) 0 (418.9 rad/s )(0.50 s 0s) 209.4 rad/st t! ! "= + # = + # =

For the next 0.5 s or at t2 = 1 s,

2 1 1 2 1( )t t! ! "= + # 53.15 rad/s (209.4 rad/s)(1s 0.5s) 157.8 rad 25 rev= + ! = =

So, the dot is at the same place it started, since the disk has rotated through an integral number of revolutions. (d) The speed of the reference dot is

2 2(0.04 m)(209.4 rad/s) 8.4 m/sv r!= = =

Assess: Given a little information and a thorough understanding of rotational kinematics and you can determine everything about the rotational motion of an object. Note that is each case the units reduced to the desired quantity and the results are reasonable.

P7.47 Prepare: Knowing the kinematic equations and the fact that the distance the car travels is some number of revolutions (or circumferences) of the tire, we can solve this problem. Solve: The acceleration of the car may be obtained from the expression:

v = v

o+ at which gives

a =v ! v

o

t=

v

t since

v

o= 0 m/s

7-24 Chapter 7

The distance traveled by the car during the time it is accelerating may be determined by:

2a!x = v

2" v

o

2 which gives

!x =v

2" v

o

2

2a=

v2

2a=

v2

2(v / t)=

vt

2

Finally, the number of times the tires rotate (i.e. the number of circumferences of the tires) may be determined by:

!x = N (2"r) = N"d which gives

N =!x

"d=

(vt / 2)

"d=

vt

2"d=

(20 m / s)(10 s)

2[" (0.58 m) / rotation]= 55 rotations

Assess: As with many kinematics problems, we can check our work by approaching the problem in a different manner. For example, since the acceleration is constant, the distance the car travels is just the average velocity times the time of travel. This may be expressed as follows:

!x = vave

t =v + v

o

2

"

#$%

&'t =

vt

2 which is the same as the expression obtained above for the distance traveled.

Note also that the final units are in rotations and that 55 is a reasonable number of rotations for a tire in 10 seconds.

P7.48. Prepare: This problem requires a knowledge of one-dimensional kinematics and the fact that as the cylinder goes around once, the elevator will advance an amount equal to the diameter of the cylinder. Solve: Knowing that the elevator advances an amount equal to the circumference of the cylinder allows us to write

!y = 2"R

From one dimensional kinematics we have

v

2! v

o

2= 2a"y

and

v = v

o+ at

Combine the first two equations, solve for a, and insert the result into the third equation to obtain

t = 4!R / v

0= 4! (0.50 m) / 1.6 m / s = 3.9 s

Assess: This is a rather long time for a passenger elevator but a reasonable time for a freight elevator.

P7.49. Prepare: The disk is a rotating rigid body and it rotates on an axle through its center. We will use Equation 7.5 to find the net torque.

Solve: The net torque on the axle is

! = FArA sin "A + FBrB sin "B + FCrC sin "C + FDrD sin "D

= (30 N)(0.10 m) sin (!90°) + (20 N)(0.05 m) sin 90° + (30 N)(0.05 m) sin 135° + (20 N)(0.10 m) sin 0°

= !3 N !m + 1 N ! m + 1.0607 N ! m = ! 0.94 N ! m

Assess: A negative net torque means a clockwise acceleration of the disk.


P7.50. Prepare: The torque is provided by the force of static friction between your fingers and the knob. Solve: See the diagram below.

Since the knob has no velocity in the horizontal direction the net force in the horizontal direction is zero, which means F = n. The maximum force of static friction is given by fs, max = μsn. The torque provided by the force of friction is given by Equation 7.3, ! = rμsn. Since there are two torques acting in the same direction, the net torque is

4

net s2 2 2(0.005 m)(0.12)(0.60 N) 7.2 10 Nr n! ! μ "

= = = = #

Assess: Note that though the net force on the knob is zero, the net torque on the knob is not zero.

P7.51. Prepare: Equation 7.9 tells us the center of gravity of a compound object. If we take all the rest of the body other than the arms as one object (call it the trunk, even though it includes head and legs) then we can write

trunk trunk arm arm

cg

2y m y my

M

+=

wheretrunk arm

70 kgM m m= + = (the mass of the whole body).

The language “by how much does he raise his center of gravity” makes us think of writingcg

.y!

Since we have modeled the arm as a uniform cylinder 0.75 m long, its own center of gravity is at its geometric center, 0.375 m from the pivot point at the shoulder. So raising the arm from hanging down to straight up would change the height of the center of gravity of the arm by twice the distance from the pivot to the center of gravity: (!ycg)arm = 2(0.375 m) = 0.75 m. Solve:

c trunk trunk c arm, up arm cg trunk trunk cg arm, down arm

cg body cg with arms up cg with arms down

arm arm

cg arm, up cg arm, down cg arm

( ) 2( ) ( ) 2( )( ) ( ) ( )

2 2 2(3.5 kg)(( ) ( ) ) ( ) (0.75 m) 0.075 m 7.5 cm

70 kg

g gy m y m y m y m

y y yM M

m my y y

M M

+ +! = " = "

= " = ! = = =

Assess: 7.5 cm seems like a reasonable amount, not a lot, but not too little. The trunk term subtracted out, which is both expected and good because we didn’t know

cg trunk( ) .y

P7.52. Prepare: We can find the center of gravity of the stack by using Equation 7.9 with the centers of gravity of stack of pennies and nickels.

Solve: Additional significant figures will be kept in the intermediate results below. The five nickels have a total mass of

all nickels nickel5( ) 5(5.7 g) 28.5 gm m= = =

The center of gravity of the stack of nickels will be in the center of the pile of nickels. The total height of the nickels is

all nickels5(1.9 mm) 9.5 mmh = =

7-26 Chapter 7

The center of gravity of the pile of nickels is at half this height.

cg, all nickels4.75 mmy =

The four pennies have a total mass of

all pennies penny4( ) 4(2.5 g) 10 gm m= = =

The center of gravity of the pile of pennies relative to the base of the entire stack is the half of the height of the pile of pennies plus the height of the pile of nickels.

cg, all pennies all nickels

1(4)(1.5 mm) 9.5 mm 3.0 mm 12.5 mm

2y h= + = + =

Using the results above in Equation 7.10, the center of mass of the stack is at

cg, all nickels all nickels cg, all pennies all pennies

cg

all nickels all pennies

(4.75 mm)(28.5 g) (12.5 mm)(10 g)6.8 mm

28.5 g 10 g

y m y my

m m

+ += = =

+ +

Assess: Most of the mass is concentrated near the table so this result is reasonable. We could have found the center of mass using Equation 7.9 and the individual centers of gravity of each of the coins, but that would have been a much longer calculation.

P7.53. Prepare: We follow the steps outlined in Tactics Box 7.1. We assume that both blade (b) and handle (h) are of uniform density (so that their respective centers of gravity will be at their geometric centers). The construction of the square is such that the blade extends all the way through the handle (i.e. the left end of the blade is at x = 0). Solve: (a)

xcg

=x

hm

h+ x

bm

b

mh

+ mb

=(2 cm)(40 g) + (8 cm)(80 g)

40 g + 80 g=

80 cm !g + 640 cm !g

120 g= 6.0 cm

ycg

=y

hm

h+ y

bm

b

mh

+ mb

=(5.5 cm)(40 g) + (9.5 cm)(80 g)

40 g + 80 g=

220 cm !g + 760 cm !g

120 g= 8.2 cm

So the center of gravity of the whole carpenter’s square is at the point (6.0 cm, 8.2 cm).


(b) Hanging the tool so that it can freely rotate about the point x = 0, y = 0 we get the result shown in the following figure.

Note that the square will hang such that the center of gravity will be below the point of suspension. (c) The angle the long side of the blade makes with the vertical is the same as the angle the line from the origin to (6.0 cm, 8.2 cm) makes with the horizontal in the original configuration. Maybe the best way to see this is to look at the original unrotated figure and draw a line from (0,0) to (6.0 cm, 8.2 cm) and remember that in the rotated figure this line becomes the vertical. That angle is

1 18.2 cm

tan = tan 546.0 cm

y

x! " "# $ # $

= = °% & % &' ( ' (

Assess: It is easily possible to have the center of gravity of an object lie outside the object’s physical boundary (think of a doughnut). However, in this case the center of gravity is inside the boundary of the object (close to the inner edge of the blade); hence we must be able to balance the carpenter’s square on our fingertip at the square’s center of gravity. The location we found seems like it is about the right place.

P7.54. Prepare: The moment of inertia of any object depends on the axis of rotation. In part (b), the rotation axis passes through mass A and is perpendicular to the page. The rotation axis for part (c) is the line joining B and D. Refer to Figure P7.54 for parts (a) and (b). Solve: (a)

xcg

=!m

ix

i

!mi

=m

Ax

A+ m

Bx

B+ m

Cx

C+ m

Dx

D

mA

+ mB

+ mC

+ mD

=(100 g)(0 m) + (200 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 m)

100 g + 200 g + 200 g + 200 g

= 0.057 m

ycg

=m

Ay

A+ m

By

B+ m

Cy

C+ m

Dy

D

mA

+ mB

+ mC

+ mD

=(100 g)(0 m) + (200 g)(0.10 m) + (200 g)(0.10 cm) + (200 g)(0 m)

700 g

= 0.057 m

7-28 Chapter 7

(b) The distance from the axis to mass C is 14.14 cm. The moment of inertia through A and perpendicular to the page is

IA

= mir

i

2

i

! = mAr

A

2+ m

Br

B

2+ m

Cr

C

2+ m

Dr

D

2

= (0.100 kg)(0 m)2

+ (0.200 kg)(0.10 m)2

+ (0.200 kg)(0.1414 m)2

+ (0.200 kg)(0.10 m)2

= 0.0080 kg " m2

(c) Refer to the figure below for this part of the problem. The moment of inertia about a diagonal that passes through B and D is

2 2

BD A A C CI m r m r= +

where rA = rC = (0.10 m) cos 45° = 7.07 cm and are the distances from the diagonal. Thus,

I

BD= (0.100 kg)r

A

2+ (0.200 kg)r

C

2= 0.0015 kg !m

2

Assess: Being on the axis of rotation, it is clear that mass A in part (b) and masses B and D in part (c) do not contribute to the moment of inertia.

P7.55. Prepare: Since the rods are “light” we will neglect their contribution. We will also have to neglect the contribution of the ball through which the axis goes because we do not know the radius of the balls. It is likely that the other two balls contribute vastly more to the total I than the one pierced by the axis does, unless the balls are large compared to the lengths of the rods. The other two balls will each contribute the same amount, and we will treat them as point particles (both because this is generally safe, and because we do not know their radius). See Equation 7.14. The distance r for each of them from the axis is 0.30 m; we also know the mass of each ball is 0.10 kg.m =

Solve: 2 2 2

(2) (2)(0.10 kg)(0.30 m) 0.018 kg mI mr= = = !

Assess: The balls are not very massive, nor are the rods particularly long, so we are satisfied with an answer that appears smallish. If the balls were five times as massive and they were 1 m away from the axis, then the answer would be 1 kg ! m2.

P7.56. Prepare: The three masses connected by massless rigid rods is a rigid body. Refer to Figure P7.56 for masses and their distance from the rotation axis. Solve: (a)

xcg

=!m

ix

i

!mi

=(0.100 kg)(0 m) + (0.200 kg)(0.06 m) + (0.100 kg)(0.12 m)

0.100 kg + 0.200 kg + 0.100 kg= 0.060 m

ycg

=!m

iy

i

!mi

=

(0.100 kg)(0 m) + (0.200 kg) (0.10 m)2" (0.06 m)

2( ) + (0.100 kg)(0 m)

0.100 kg + 0.200 kg + 0.100 kg= 0.040 m

(b) The moment of inertia about an axis through A and perpendicular to the page is

IA

= !mir

i

2= m

B(0.10 m)

2+ m

C(0.10 m)

2= (0.100 kg)[(0.10 m)

2+ (0.10 m)

2] = 0.0020 kg " m2


(c) The moment of inertia about an axis that passes through B and C is

IBC

= mA

(0.10m)2! (0.06m)

2( )2

= 0.0013 kg "m2

Assess: Note that mass mA does not contribute to IA, and the masses mB and mC do not contribute to IBC.

P7.57. Prepare: We can use Equation 7.14, and Table 7.4. Solve: Refer to the figure below.

(a) The moment of inertia of the skater will be the moment of inertia of her body plus the moment of inertia of her arms. The center of mass of each arm is at her side, 20 cm from the axis of rotation. The mass of each arm is half of one eighth of the mass of her body, which is 4 kg. The mass of her body is 64 kg – 8 kg = 56 kg. With her arms at her sides, her total moment of inertia is

2 2 2

body arm arm body body arm arm arm arm

2 2 2 2

1( ) ( ) ( )

2

1(56 kg)(0.20 m) (4 kg)(0.20 m) (4 kg)(0.20 m) 1.4 kg m

2

I I I I M R M R M R= + + = + +

= + + = !

(b) With her arms outstretched, the center of mass of her arms is now 50 cm from the axis of rotation. Her moment of inertia is now

I = Ibody

+ Iarm

+ Iarm

=1

2M

body(R

body)

2+ M

arm(R

arm)

2+ M

arm(R

arm)

2

=1

2(56 kg)(0.20 m)

2+ (4 kg)(0.50 m)

2+ (4 kg)(0.50 m)

2= 3.1 kg !m

2

Her moment of inertia has increased. Assess: Her moment of inertia with arms outstretched is almost twice as large as with them at her side. This is reasonable, since their distance from the axis of rotation is much larger.

P7.58. Prepare: The compact disk is a rigid body rotating about its center. The initial angular velocity is !i = 0 and the final angular velocity is !f = 2000 rpm, which must be converted to SI units of rad/s. Noting that 1 revolution corresponds to 2! radians, !f =

2

60(2000 rpm)( )rad/s 209.4 rad/s.

!=

Solve: (a) The rotational kinematics equationf i f i

( )t t! ! "= + # gives

2209.4 rad/s 0 rad (3.0 s 0 s) 69.81 rad/s! != + " # =

The torque needed to obtain this operating angular velocity is

! = I" = (2.5#10

$5kg %m

2)(69.81 rad/s

2) = 1.8#10

$3 N %m

7-30 Chapter 7

(b) From the rotational kinematics equation,

2 22

f i i f i f i

1 1( ) ( ) 0 rad 0 rad (69.81 rad/s )(3.0 s 0 s) 314.1 rad

2 2

314.1 revolutions 50 rev

2

t t t t! ! " #

$

= + % + % = + + % =

= =

Assess: Fifty revolutions in 3 seconds or 1000 rpm is a reasonable value.

P7.59. Prepare: This problem requires a knowledge of translational ( F

net= ma ) and rotational (

!

net= I" )

dynamics. Notice that the counter clockwise torque is greater than the clockwise torque, hence the system will rotate counterclockwise. Let’s agree to call any force that tends to accelerate the system positive and any force that tends to decelerate the system negative. Also let’s agree to call the small disk M1 and the large disk M2. Solve:

Write Newton’s second law equation for m2 and m1 as follows:

m

2g !T

2= m

2a

2= m

2R

2" or

T

2= m

2g ! m

2R

2"

and

T

1! m

1g = m

1a

1= m

1R

1" or

T

1= m

1g + m

1R

1!

The net torque acting on the system may be determined by

! = R

2T

2" R

1T

1= R

2(m

2g " m

2#R

2) " R

1(m

1g + m

1#R

1)

The moment of inertia of the system is

I = I

1+ I

2= ( M

1R

1

2/ 2) + ( M

2R

2

2/ 2)

Knowing

! = I"

We may combine the above to get

R

2(m

2g ! m

2"R

2) ! R

1(m

1g + m

1"R

1) = [( M

1R

1

2/ 2) + ( M

2R

2

2/ 2)]"

Which may be solved for ! to obtain

! =(m

2R

2" m

1R

1)g

R2

2(m

2+ M

2/ 2) + R

1

2(m

1+ M

1/ 2)

= 3.5 rad/s2

Assess: This angular acceleration amounts to speeding up about a half revolution per second every second. That is not an unreasonable amount.


P7.60. Prepare: The flywheel is a rigid body rotating about its central axis as shown below. The initial angular speed will be taken as !i = 0 and the final angular speed is !max = !f = 1200 rpm, which must be converted to SI units of rad/s. Noting that 1 revolution corresponds to 2! radians, !f = 1200 rpm

(2!/60) rad/s = 40" rad/s.

Solve: The radius of the flywheel is R = 0.75 m and its mass is M = 250 kg. The moment of inertia about the axis of rotation is that of a disk:

I =1

2MR2

=1

2(250 kg)(0.75 m)

2= 70.31 kg !m

2

The angular acceleration is calculated as follows:

!

net= I" #" = !

net/I = (50 N $ m)/(70.31 kg $ m

2) = 0.711 rad/s

2

Using the kinematics equation for angular velocity gives

2

f i f i f f( ) 1200 rpm 40 rad/s 0 rad/s 0.711 rad/s ( 0 s) 180 st t t t! ! " #= + $ = = = + $ % =

Assess: To solve this problem, you used your knowledge of moment of inertia, rotational dynamics, and rotational kinematics. As you move farther into Physics, you will find that you need mastery of multiple concepts in order to solve then problems. You should enjoy this process.

P7.61. Prepare: Two balls connected by a rigid, massless rod are a rigid body rotating about an axis through the center of gravity. Assume that the size of the balls is small compared to 1 m. We have placed the origin of the coordinate system on the 1.0 kg ball. Since

about cg,I! "= we need the moment of inertia and the angular

acceleration to be able to calculate the required torque.

Solve: The center of gravity and the moment of inertia are

xcm

=(1.0 kg)(0 m) + (2.0 kg)(1.0 m)

(1.0 kg + 2.0 kg)= 0.667 m and y

cm= 0 m

Iabout cm

= !mir

i

2= (1.0 kg)(0.667 m)

2+ (2.0 kg)(0.333 m)

2= 0.667 kg "m

2

We have !f = 0 rad/s, tf – ti = 5.0 s, and 2

320 rpm 20(2 rad/60 s) rad/s,

i! " "= = = so !f = ! i + #(tf – ti )

becomes

22 2

0 rad/s rad/s (5.0 s) rad/s3 15

! !" "# $

= + % = &' () *

Having found I and #, we can now find the torque $ that will bring the balls to a halt in 5.0 s:

! = Iabout cm

" =2

3kg #m2

$

%&'

()*

2+15

rad/s2

$

%&'

()= *

4+45

N #m = *0.28 N #m

7-32 Chapter 7

The magnitude of the torque is 0.28 N !m.

Assess: The minus sign with the torque indicates that the torque acts clockwise.

P7.62. Prepare: Assume the rope has no mass and does not slip on the pulley. Equation 7.15 and Newton’s second law for linear motion can be applied to find the acceleration of the mass. There is motion constraint on the rope and pulley. Solve: See the diagram below.

Choosing the upward direction as positive, Newton’s second law applied to the mass gives

2yma T mg= !

The normal force on the pulley and the pulley’s weight exert no torque. The moment arm for both tensions is the radius of the pulley. For the rotational motion Equation 7.15 gives

2 1.T R T R I!" =

Since upward motion gives clockwise acceleration, the motion constraint is .y

a R!= "

Using this constraint and solving for2

T in the torque equation we obtain 2

2 1/ .T T Ia R= !

Substituting this result into the equation for linear motion and solving for the acceleration gives

1

2/

y

T mga

m I R

!=

+

The moment of inertia of a cylinder is 21

2.I MR= Substituting this into the acceleration equation gives

2

1 210 N (1.5 kg)(9.80 m/s )

1.88m/s/2 1.5 kg 1.0 kg

y

T mga

m M

! != = = !

+ +

An additional significant figure has been kept in this intermediate result. Note that the acceleration is negative. The block is moving downward. Starting with zero initial velocity, the block moves !y = – 0.30 m. Using the second equation in Table 2.4, the time it takes the block to move 30 cm is

2

2 2( 0.30 m)0.56 s

1.88 m/sy

yt

a

! "! = = =

"

Assess: The expression for the acceleration makes sense. If the mass of the block or the moment of inertia of the pulley increases, the acceleration decreases. If the tension applied to the rope is smaller than the weight of the mass, the acceleration is negative instead of positive. Note that the tensions acting on either side of the pulley can’t be equal due to the moment of inertia of the pulley. If the tensions were equal there would be no torque on the pulley, and the pulley would not rotate.


P7.63. Prepare: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution of a disk and that the string does not slip. Because the pulley is not massless and frictionless, tension in the rope on both sides of the pulley is not the same. We will have to be careful with the appropriate masses when we write below Newton’s second law for the blocks and the pulley. A pictorial diagram of the problem and free-body diagrams for the two blocks are shown. We have placed the origin of the coordinate system on the ground.

Solve: Applying Newton’s second law to m1, m2, and the pulley yields the three equations:

T

1! w

1= m

1a

1!w

2+ T

2= m

2a

2T

2R !T

1R ! 0.50 N " m = I#

Noting that –a2 = a1 = a, 21

p2,I m R= and ! = a/R, the above equations simplify to

T1! m

1g = m

1a m

2g !T

2= m

2a T

2!T

1=

1

2m

pR

2"

#$%

&'a

R

"

#$%

&'1

R+

0.50 N (mR

=1

2m

pa +

0.50 N (m0.060 m

Adding these three equations,

2 1

2 1 1 2 p1

1 2 p2

2

2

1 ( ) 8.333 N( ) 8.333 N

2

(4.0 kg 2.0 kg)(9.8 m/s ) 8.333 N1.610 m/s

2.0 kg 4.0 kg (2.0 kg/2)

m m gm m g a m m m a

m m m

! " # ## = + + + $ =% &' ( + +

# #= =

+ +

We can now use kinematics to find the time taken by the 4.0 kg block to reach the floor:

yf

= yi+ v

i(t

f! t

i) +

1

2a

2(t

f! t

i)

2" 0 = 1.0 m + 0 +

1

2(!1.610 m/s

2)(t

f! 0 s)

2" t

f=

2(1.0 m)

(1.610 m/s2)

= 1.1 s

Assess: Compared to free fall where we would use a = –9.80 m/s2, a = – 1.61 m/s2, and a time of 1.1 s for the block to reach floor are reasonable.

P7.64. Prepare: Please refer to Figure P7.64. The disk is a rigid spinning body. The initial angular velocity of 300 rpm in SI units is (300)(2")/60 = 10" rad/s. After 3.0 s the disk stops, so #f = 0. Solve: Using the kinematic equation for angular velocity,

f i 2

f i f i

f i

(0 rad/s 10 rad/s) 10( ) rad/s

(3.0 s 0 s) 3t t

t t

! ! " "! ! # #

$ $ $= + $ % = = =

$ $

7-34 Chapter 7

Thus, the torque due to the force of friction that brings the disk to rest is

! = I" = # fR $ f = #I"R

= #1

2mR

2( )"

R= #

1

2(mR)" = #

1

2(2.0 kg)(0.15 m) #10

%3

rad/s2

&

'()

*+= 1.6 N

Assess: The minus sign with fR! = " indicates that the torque due to friction acts clockwise.

P7.65. Prepare: This is an excellent review problem. In order to solve this problem you will need a working

knowledge of rotational kinematics (!

f= !

o+"t), moment of inertia of a cylinder

(I = MR

2/ 2), rotational

dynamics (! = I" ), torque

(! = Rf

ksin"), and kinetic friction

( f

k= μ

kN ).

Solve: First determine an expression for the angular acceleration

! = ("

f#"

o) / $t

Next obtain an expression for the moment of inertias of the grindstone

I = MR2

/ 2

Then obtain an expression for the torque acting on the grindstone

! = I" =MR

2

2

#

$%&

'()

f*)

o

+t

#

$%

&

'(

Write a second expression for the torque in terms of the force of friction and then the normal force

! = f

kR = μ

kNR

Finally, equate the last two expressions for the torque and solve for N (the force with which the man presses the knife against the grindstone).

N = MR(!f"!

o) / 2μ

k#t =

(28 kg)(0.15 m)(180 rev/min)(2$ rad/rev)(min / 60 s)

2(0.2)(10 s)= 9.9 N

Assess: This is essentially the force equal to the weight of a 1 kg mass. This is a reasonable force (i.e. one that the man could easily exert and yet not grind the knife to a sliver in a matter of minutes).

P7.66. Prepare: The sacs are constrained by the stamens. Assume the mass of the stamen and sac is given to two significant figures. Solve: Refer to the diagram below.

(a) We can use the third equation in Table 7.2 to find the angular acceleration of a sac, and then use Equation 7.15 to find the corresponding tangential acceleration.

! =2"#

"t2

=

2(60°)2$ rad

360°( )(0.30 %10

&3s)

2= 2.3%10

7rad/s

2


The tangential acceleration is

a

t= !R = (2.3"10

7rad/s

2)(1.0 "10

#3m) = 2.3"10

4m/s

2

(b) Using the second equation in Table 7.2 with the result above gives

!f ="#t = 7.0 $103 rad/s

(c) The moment of inertia of the stamen and sac can be calculated using Equation 7.14.

2

2 6 4 2 6 3 2 12 2

stamen sac(1.0 10 kg)(5.0 10 m) (1.0 10 kg)(1.0 10 m) 1.25 10 kg m

2

RI m m R

! ! ! ! !" #= + = $ $ + $ $ = $ %& '

( )

An additional significant figure has been kept in this intermediate result. The “straightening torque” is

12 2 7 2 5(1.25 10 kg m )(2.3 10 rad/s ) 2.9 10 N mI! "

# #= = $ % $ = $ %

(d) The torque due to gravity can be calculated with Equation 7.3. See the diagram above. In its initial position, the stamen is horizontal, so the weight of both the stamen and the sac is perpendicular to the radial line. The net torque is

net stamen sac stamen sac

4 6 2 3 6 2

8

2

(5.0 10 m)(1.0 10 kg)(9.80 m/s ) (1.0 10 m)(1.0 10 kg)(9.80 m/s )

1.5 10 N m

Rw Rw! ! !

" " " "

"

# $= + = " "% &

' (= " ) ) " ) )

= " ) *

The gravitational torque is over 2000 times less than the straightening torque. It was reasonable to neglect the gravitational torque. Assess: These results seem reasonable. The accelerations are huge, while the torque is relatively small. This is due to the relatively low moment of inertia of the structure.

P7.67. Prepare: The center of gravity must follow a parabolic path, so the answer isn’t C. While it is possible to bend over and get one’s center of gravity outside the body (search the Web for the Fosbury flop technique in the high jump), in this case the center of gravity of the dancer stays in the body, so the answer isn’t D. Solve: The correct answer is B, with the reasoning being that, as shown in the diagram, the head can stay quite level by bringing up the arms and legs and then lowering them again. This allows the center of gravity to follow the parabola while the head stays about level. Assess: The whole point of movement is to raise and lower the center of gravity, and this is done by raising and lowering the arms and legs.

P7.68. Prepare: Your center of gravity will travel as if it were in free fall and so reaches the same height no matter how you move your arms. Solve: In order to get your head as high as possible, you should lower your center of gravity as much as possible. You should have your arms at your side at the top of the leap. The correct answer is A. Assess: Compare to the motion in a grand jeté. There, the dancer moves her center of gravity upward so her head appears to move in a straight line.

P7.69. Prepare: While the dancer is in the air, the gravitational force on her acts as if her mass were concentrated at her center of gravity. Solve: The correct answer is B; there is no lever arm to create a torque because the gravitational force is directly through the center of gravity. Even if the arms and legs move, the center of gravity moves and the gravitational force is still through the center of gravity. So if the axis of rotation is through her center of gravity then there can be no gravitational force. Assess: Of course, computing the gravitational torque about some axis other than through the center of gravity could produce a torque, but we were told to use an axis through the center of gravity.

7-36 Chapter 7

P7.70. Prepare: The basic expression used to determine the moment of inertia is Equation 7.14. As indicated in this expression, the moment of inertia depends on the mass and how the mass is distributed. Solve: As the dancer moves her arms and legs outward, she is moving the center of gravity of her arms and legs farther from the vertical axis through the center of her body and hence increasing her moment of inertia about that axis. The correct answer is A. Assess: Moment of inertia increases as the square of the distance of a mass from the axis of rotation.

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