Holt Physics Chapter 8

Rotational Equilibrium and Dynamics

Apply two equal and opposite forces acting at the center of mass of a stationary meter stick.

Does the meter stick move?

F2

F1

Fext = 0.

F1=F2

Apply two equal and opposite forces acting on a stationary meter stick.

Does the meter stick move?

F

F

The center of mass of the meter stick does not accelerate, so it does not undergo translational motion.

However, the meter stick would begin to rotate about its center of mass.

A torque is produced by a force acting on an extended (not point-like) object.

FThe torque depends on how strong the force is, and where it acts on the object.

Torques cause changes in rotational motion.

Torque is a vector. It is not a force,* but is related to force.

*So never set a force equal to a torque!

A

You must always specify your reference axis for calculation of torque. By convention, we indicate that axis with the letter “A” and a dot.

TorqueTorque is a quantity that measures the ability of a force to rotate an object around some axis.

Torque depends on force and the lever arm.Lever arm (moment arm)is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force.See figure 8-3, page 279

The most torque is produced when the force is perpendicular to the object.Formula=Fd(sin)d is the lever arm and is the angle between the lever arm and the force. (If 90º, then sin=1).See figure 8-5, page 280.

Force is POSITIVE if the rotation is counterclockwise.If there is more than one force, add the two resultant torques, using the appropriate signs.EX: Wishbone…sum the two torques

A

F2F1

+ or - ?-+ or - ?+

net = = 1+2 = F1d1 + (-F2d2)The sign of the net torque will tell you which direction the object will rotate.

A

F2F1

d1 d2

Example 8A, page 281A basketball is being pushed by two players during tip-off. One player(to the right) exerts a downward force of 11 N at a distance of 7.0 cm from the axis of rotation. The second player(to the left) applies an upward fore of 15 N at a perpendicular distance of 14cm from the axis of rotation. Find the net torque acting on the ball.Direction and sign??? Units???

F2=11 N

F1=15 N

+ or - ?-

+ or - ?-

7cm14cm

F1 = -15 N F2 = -11N

d1 = 0.14m d2 = 0.070m

net=?

net= 1+ 2= F1d1+ F2 d 2(-15N X 0.14m)+(-11N X 0.070m) =

-2.9Nm

Example with an Angle

An upward 34N force is exerted on the right side of a meter stick 0.30 m from the axis of rotation at an angle of 35 degrees. A second downward force of 67N is exerted at an angle of 49 degrees to the meter stick 0.40m to the left of the axis of rotation.

What is the net torque? Draw a diagram!!!

34N

67N

35

49

0.30m

0.40m

Signs???

BOTH are POSITIVE!!

F1 = 34N F2 = 67N

d1 = 0.30m d2 = 0.40m

1= 35 2= 49

net=?

net= 1+ 2= F1d1 sin1+ F2d2 sin2

(34NX.30m)sin35+(67NX.40m)sin49=

26Nm

Practice 8A, page 282

Section Review #3 and #4Same page

Happy Thursday!

Please take out 8a for me to check.

8B EquilibriumComplete equilibrium requires zero net force and zero net torque.Translational equilibrium: net force in x and y direction = 0Called 1st condition of equilibrium∑Fx = 0, ∑Fy = 0

Rotational equilibrium: net torque=0Called 2nd condition of equilibrium∑ = 0

A 45.0m beam that weighs 60.0N is supported in the

center by a cable. The beam is in equilibrium and supports

three masses. A 67.0kg mass is on one end, an 89.0kg mass is on the other. A fish is hanging 10.0m from the 67.0 kg mass. What is the mass (kg!) of the fish and what is the tension

(force!) in the cable.

67kg

89kg60N

45m

10m

?

FT

67kg

89kg60N

Convert to Newtons!

45m

10m

?

FT

657N

873N60N

45m

10m

?

FT

657N

873N60N

Choose AxisChoose center to eliminate a

variable!!

45m

10m

?

FT

657N

873N60N

How far is the fish from the axis?

45m

10m

?

12.5m

FT

657N

873N60N

Calculate Torques and sum to zero

45m

10m

?

12.5m

FT

657N

873N60N

Assign sign to torques

45m

10m

?

12.5m

++ -

0

FT

∑=0=657N(22.5m)+Wf(12.5m)–873N(22.5m)=0

Wf=4860Nm/(12.5m) = 388.8N

mf=388.8N/9.81m/s2=39.6kg

657N

873N60N

45m

10m

39.6kg

12.5m

FT

∑Fy = 0, ∑Fdown = ∑Fup , change all to forces

Fish is 39.6kgX9.81m/s2= 389N

-657N-873N-60N

45m

10m

-389N

12.5m

FT

39.6kg

∑Fy = 0, ∑Fdown = ∑Fup

FT= -∑Fdown

=-(-657N- 389N -60N-873N)

=1980N

-657N-873N-60N

45m

10m

-389N

12.5m

FT

F, pillar 2/

person 2

Fw,

bridge

or board

How to draw a bridge and a car with pillars…or a board and a weight lifted by two people…

F, pillar 1/

person 1

Fw, car or

object

8B (2,3 and 4)page 288

Happy Thursday!

Write this down…

A uniform 5.00 m long horizontal beam that weighs 315N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53 degrees with the horizontal, and a 545 N person is standing 1.5 m from the pin. Find the force in the cable, FT, and the force exerted on the beam by the wall, R, if the beam is in equilibrium.

L=5.00 m Fg,beam=315 N =53Fg,person=545 N d=1.50mFT=? R=?Record distances, put weight of object at center of mass and position all forces.

RFT

315 N

1.50 m53

L=5.00 m

545 N

The unknowns are R ( Rx, Ry), and FT

Because we have equilibrium,

∑Fx = 0, ∑Fy = 0

Rx - FTcos = 0

Ry + FTsin - Fg,p – Fg,b = 0

R

FT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

FTy=FTsin

FTx=FTcos

Rx - FTcos = 0

Ry + FTsin - Fg,p – Fg,b = 0

Because there are too many unknowns, pause (don’t panic) and go to the second condition of equilibrium.

RFT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

Choose an axis and sum the torques…remembering signs for direction of rotation!!

RFT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

A

Why choose the pin

for A?

This eliminates R as

a variable!!

0

+

--

=FT L(sin) – Fg,bL/2 –Fg,p d=0

Now substitute and solve for FT.

FT = (Fg,bL/2 +Fg,p d)/(Lsin )

FT=315N(2.5m) + 545N(1.5m)

5.0mSin53

RFT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

A

FT= 1605Nm/4.0m

FT= 4.0X102 N

RFT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

A

We’re not done yet!! Go back to the first condition of equilibrium to solve for Rx, Ry and R.

RFT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

A

Rx - FTcos = 0 Rx = FTcos

Rx = 400N X cos53 = 240N

Ry+FTsin-Fg,p–Fg,b=0 Ry= -FTsin53+ Fg,p+Fg,b

Ry = -3.2X102N + 545N + 315N = 540 N

R = (Rx2 + Ry

2) = (240N2 + 540N2) =590 N

RFT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

A

Torque Work Packet

Finish Work packet!

Happy Friday!

Moment of Inertia

The moment of inertia is the resistance of an object to changes in rotational motion about some axis.Similar to mass…mass ( simple inertia) is the measure of resistance to translational motion…

Moment of Inertia depends on the object’s mass and the distribution of mass around the axis of rotation.The farther the center of mass from the axis of rotation, the more difficult it is to rotate the object, and therefore, the higher the moment of inertia.Use Table 8-1, page 285

Which will have the higher average velocity when allowed to roll down a ramp from rest? (refer to pg. 285)9kg solid ball r=0.50m

8kg hollow ball r=0.40m

Solid Ball I = 0.9kgm2, hollow ball I = 0.85 kgm2

Newton’s 2nd Law

F=ma can be translated to rotational motion.net = I = Iat/rI = moment of inertia = angular acceleration net = net torqueat=tangential accelerationr=radius

Example 1

An athlete tosses tennis ball using only the rotation of his forearm to accelerate the ball. The forearm and tennis ball have a combined moment of inertia of 0.175 kgm2, and the forearm length is 0.29 m. If the ball has a tangential acceleration of 55m/s2 just before it is released, what is the torque on the arm and ball?

I = 0.175 kgm2 at=55m/s2

r=0.29m =?

=I, where =at/r

=I at/r

= 0.175 kgm2 X55m/s2/0.29m=

33Nm

Example 2

A 25 g CD (radius =7.0 cm) is rotating at 100 rev/min. If it stops in 8.5 sec, what is the angular acceleration of the CD? How much torque is required to stop the CD?

r=0.07m, m=0.025kg, t=8.5 sec,i = 100rev/min=100x260=10.5rad/s

f= 0 = /t = (0-10.5rad/s)/8.5s

= -1.2 rad/s2

=I…What is I? Look in table on page 285

Rotating disk is 1/2mr2

= -1.2 rad/s2

=I = 1/2mr2=

=1/2(.025kg)(.07)2(-1.2rad/s2)

=-7.35X10-5 Nm

8C, page 291

Be ready to use table 8-1 on page 285 to calculate I and use old chapter 7 formulas for quantities like , , s, at and .

Fire Drill and Lock down changes…

If there is a fire drill during a class change, report to the waiting area outside for the teacher you are going to.

Lock down: before hiding in the back, we will blockade the door with a table and then consider, if there is a breach, what we have to fight with.

Momentum and Rotation

Linear momentum can be translated to angular momentumL = I

L = angular momentum

I = moment of inertia – look in the table again (page 285!!)

= angular speed, and ω =vt/r (you will need to use ch. 7 formulas again)

Conservation of Angular Momentum

As in linear momentum, angular momentum is also conserved.

Li = Lf

Example page 293

A 65 kg student is spinning on a merry-go-round that has a mass of 5.25X102 kg and a radius of 2.00 m. She walks from the edge of the merry-go-round toward the center. If the angular speed of the merry-go-round is initially 0.20 rad/sec, what is its angular speed when the student reaches a point 0.50m from the center?

mm = 525 kg ms =65 kg

ri,s= rm =2.00m rf,s = 0.50 m

i =0.20 rad/s f =?

Use conservation of momentum

Li = Lf

Lm,i + Ls,i = Lm,f + Ls,f

Need moments of inertia!! Because L = I= Ivt/r

The merry-go-round is a _____

The Student is a _____

Merry-go-round (I = ½ MR2 )Student (I = MR2 )Merry-go-round (I = ½ MR2 )Student (I = MR2 )

Lm.i + Ls,i = Lm,f + Ls,f

½MmRm2i+MsRs,i

2i =½MmRm2f+MsRs,f

2f

½MmRm2i+MsRs,i

2i= ½MmRm2f+MsRs,f

2f

½525kg(2.00m)2(0.20rad/s)+65kg(2.00m)2(0.20rad/s)=

½525kg(2.00)2f +65kg(0.50m)2f

210+52=1050f +16.25f

Plug it into the calculator and solve for f

f =0.245rad/s

Example 2

A comet has a speed of 7.056 X 104 m/s at a distance of 4.95X1010 m. At what distance from the sun would the comet have a speed of 5.0278X104 m/s?

So, I = MR2

Mass is constant

ω= vt/rRi = 4.95X1010 m Vi=7.056 X 104 m/s Rf= ? Vf= 5.0278X104 m/s

MRi2 vi/ri = MRf

2 vf/rf

Ri vi= Rf vf

Rf =Ri vi/ vf

Li=Lf Point Mass

Ii = I f

Rf = 4.95X1010 m X 7.056 X 104 m/s

5.0278X104 m/s

Rf= 6.95X1010m

8 D, page 294

Kinetic EnergyRotational Kinetic energy (KErot) is the kinetic energy associated with their angular speed.Formula KErot = ½ I2 = ½ I(vt/r)2

Conservation of Kinetic Energy also applies…KEtrans + KErot + PEi = KEtrans + KErot + PEf

½ mvi2 + ½ Ii

2 + mghi = ½ mvf2 + ½ If

2 + mghf

Be sure to keep track of initial and final conditions as well as angular vs. translational speeds and moments of inertia.

Example page 296A solid ball with a mass of 4.10 kg and a radius of 0.050m starts from rest at a height of 2.00 m and rolls down a 30 slope. What is the translational speed of the ball when it leaves the incline?

2.00m

30

v

hi = 2.00 m m = 4.10 kg

R = 0.050 m vi = 0.0 m/s

= 30.0 hf = 0 m vf = ?

2.00m

30

v

I = 2/5MR2What will I be???

hi = 2.00 m m = 4.10 kgR = 0.050 m vi = 0.0 m/s

= 30.0 hf = 0 m vf = ?

½ mvi2 + ½ Ii

2 + mghi = ½ mvf2

+ ½ If2 + mghf

2.00m

30

v

WE have two variables

So we need to find one

In terms of the other…

REMEMBER!!!

=vt/r

hi = 2.00 m m = 4.10 kgR = 0.050 m vi = 0.0 m/s

= 30.0 hf = 2.00 m vf = ?

½ mvi2 + ½ Ii

2 + mghi = ½ mvt,f2

+½I f2 +mghf

2.00m

30

v

Substitute vt/r into

The equation for

f

(vt,f /r)

4.10kg(9.81m/s2) (2.00m) = ½ 4.10kgvf2

+½I (vf/0.050m) 2

I = 2/5MR2

4.10kg(9.81m/s2) (2.00m) = ½ 4.10kgvf2

+½(2/5MR2) (vf/0.050m) 2

PLUG IN THE NUMBERS

80.442kgm2/s2 = 2.05kg vf2 +

0.82kg vf2

2.87kg vf2 = 80.442kgm2/s2

vf2 =28 m2/s2

vf = 5.29 m/s

4.10kg(9.81m/s 2) (2.00m) =½ 4.10kg vf2

+½(2/5)(4.10kg)(0.050m) 2 (vf/0.050m) 2

Example 2

A 3.5 kg spherical potato (radius .070m) is kicked up a 30 degree slope at a speed of 5.4 m/s. What distance along the slope did the potato roll before it stopped?

mp= 3.5kg, vi,p= 5.4m/s θ=30°hf=? Hypotenuse (slope dist.)= ? ½ mvi

2 + ½ Ii2 +mghi = ½ mvf

2 + ½ If2 + mghf

*Substitute vt/r for ω

½ mvi2 + ½ Ii

2 = mghf

*Solve for hf

hf=½ mvi2 + ½ I(vt /r)2 =(51.03 + 20.412)/34.335

mg

hf=2.08m sin30=opp/hyp

Hyp=opp/sin30

D=2.08m/sin30=4.2m

(vt,i/r)2

(vt,i/r)2

heigh

t30

Kinetic Energy

Practice 8E, page 297

Review Problems: 6,7,8,9, 11,15,18,20,21,24,27,35,37,50

Simple MachinesA machine is any device that transmits or modifies force.

There are six simple machinesLever, inclined plane, wheel and axle, wedge, pulley and screw

Using Simple MachinesMachines usually give a mechanical advantage (ratio of output force to input force)Formula MA = output force = Fout

input force Fin

Because Torque input = Torque output, the ratio of the distances also gives MA. (see page 299)

MA = din/dout

Efficiency

Because Work in = Work out, if a machine is frictionless, F1d1=F2d2(see page 299…again)

Efficiency is a measure of how much input energy is lost (heat or sound because of friction) compared with how much energy is used to perform work on an object.eff = Wout/Win

Happy Friday!!

Lab Today!

Build a device using at least 3 simple machines (see page 298) that will lift a 200 gram mass at least 6 cm and determine the MA and the efficiency of the device (see page 301)…The most efficient device that meets the criteria wins!

Section Review page 301

Review for TestReview Problems page 305-308

1,6-9,11,12,14,15,18-20,21,24,27,31,32,34,35,37,40,43,72,75

8B Example with a different axis of rotation for the torque…

A uniform 5.00 m long horizontal beam that weighs 315N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53 degrees with the horizontal, and a 545 N person is standing 1.5 m from the pin. Find the force in the cable, FT, and the force exerted on the beam by the wall, R, if the beam is in equilibrium.

L=5.00 m Fg,beam=315 N =53Fg,person=545 N d=1.00mFT=? R=?Record distances, put weight of object at center of mass and position all forces.

RFT

315 N

1.50 m53

5.00 m

545 N

The unknowns are Rx, Ry, and FT

Because we have equilibrium,

∑Fx = 0, ∑Fy = 0

Rx - FTcos = 0

Ry + FTsin - Fg,p – Fg,b = 0

R

FT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

FTy=FTsin

FTx=FTcos

Rx - FTcos = 0 Rx=FTcosRy+FTsin-Fg,p–Fg,b =0Ry=-FTsin+Fg,p+Fg,b

Because there are too many unknowns, pause and go to the second condition of equilibrium.

RFT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

Choose an axis and sum the torques…remembering signs for direction of rotation!!

R

FT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

A

We have to put the

axis at the center of

mass!!

This does not

eliminate R as a

variable!!

But it does

Eliminate the

315N force.

=FT L/2(sin) – RyL/2 +Fg,p d=0

Now substitute previous equation for Ry (Ry=-FTsin+Fg,p+Fg,b) into the torque eauation and solve for FT.

RFT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

A

=FT L/2(sin) – RyL/2 +Fg,p d=0=FT 5.00m/2(sin53) – Ry5.00m/2 +545N(1.0m)=0

RFT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

A

=FT5.00m/2(sin53)–(-FTsin+Fg,p+Fg,b)5.00m/2 +545N(1.0m)=0

FT5.00m/2(sin53)–(-FTsin53+545+315)5.00m/2 +545N(1.0m)=0

FT(1.997m) + FT(.799)(2.5m)-860N(2.5m) +545Nm=0

3.995m FT = 2150Nm-545NmFT = 1605Nm/3.995m=402N=4.0X102 N

RFT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

A

Now we can substitute force in the wire (FT) into the Rx and Ry

equations to find Rx and Ry, and then solve for R…

RFT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

A

Rx - FTcos = 0 Rx = FTcos

Rx = 400N X cos53 = 240N

Ry+FTsin-Fg,p–Fg,b=0 Ry= -FTsin53+ Fg,p+Fg,b

Ry = -3.2X102N + 860 N = 540 N

R = (Rx2 + Ry

2) = (240N2 + 540N2) =590 N

RFT

315 N

1.50 m53

5.00 m

545 N

Ry

Rx

A

A bridge 20.0 m long and weighing 4.00X105N is supported by two pillars located 3.00 m from each end. If a 1.96X104N car is parked 8.00 m from one end of the bridge, how much force does each pillar exert?

L=20.0 m Fg,bridge=4.00X105Ndpillars = 3.0m from endFg,car=1.96X104N dcar=8.00mFp1=? Fp2=?Record distances, put weight of object at center of mass and position all forces.

FP2 FP1

4.00X105N

8.0 m

1.96X104N

20.0 m

3.0 m3.0 m

The unknowns are Fp1 and Fp2.

Because we have equilibrium,

∑Fx = 0, ∑Fy = 0

Fp2 + Fp1 -Fcar -Fbridge= 0

Fp2 = Fc +Fb-Fp1 = 4.2X105 - Fp1

FP2 FP1

4.00X105N

8.0 m

1.96X104N

20.0 m

3.0 m3.0 m

Now lets look at torque…

Choose the axis to be a pillar to eliminate an unknown…

FP2 FP1

4.00X105N

8.0 m

1.96X104N

20.0 m

3.0 m3.0 m

A

Now lets look at torque…=Fp1 (14.0m) – Fc(5.0m) -Fb (7.0m)=0=Fp1 (14.0m) – (1.96X104N)(5.0m) -(4.00X105N) (7.0m)=0

Fp1 =2.07X105NFp2 = 4.2X105 - Fp1 = 4.2X105 - 2.07X105NFp2 = 2.13X105N

FP2 FP1

4.00X105N

8.0 m

1.96X104N

20.0 m

3.0 m3.0 m

A

Window washer….problem 3, page 288

∑Fx = 0, ∑Fy = 0

Fr1 + Fr2 -200N -700N= 0

=Fr2 (3.00m)–700N(1.0m)-200N(1.5m)=0

Fr2 = 333N

Fr1 = 900N- Fr2 = 900N – 333 = 567N

Fr1 Fr2

200N700N

3m1m