holt physics chapter 7

Holt Physics Chapter 7

Rotational Motion

Measuring Rotational Motion

Spinning objects have rotational motion

Axis of rotation is the line about which rotation occurs

A point that rotates around an axis undergoes circular motion

s=arc length, r=radius=angle of rotation (measured

in degrees or radians) corresponding to s

r

s

(rad) = s/rIf s = r, then = 1 radianIf s = 2r (circumference), then (rad) = 2r/r= 2 radians

Therefore 360 degrees = 2 radians, or 6.28

See figure 7-3, page 245

Equation to convert between radians and degrees (rad) = [/180º] X (deg)

Sign convention:clockwise is negative, counterclockwise is positive!

Angular Displacement () describes how much an object has rotated = s/r

Book Example 1: While riding on a carousel that is rotating clockwise, a child travels through an arc length of 11.5m. If the child’s angular displacement is 165 degrees, what is the radius of the carousel? = -165 degrees but…we need radians!

rad= [/180º] Xdeg = (/180º)(-165º) = -2.88 radians-2.88 radians

s = -11.5 m r=? = s/r so, r=s / r = -11.5/-2.88 = 3.99m3.99m

Example 2. A child riding on a merry-go-round sits at a distance of 0.345 m from the center.Calculate his angular displacement in radians, as he travels in a clock-wise direction through an arc length of 1.41 m.

Given:

Find:

Solution:

r = 0.345 m s = -1.41 m

rad 4.09- m 0.345

m 1.41-

r

sΔθ

The negative sign in our answer indicates direction.

Homework Practice 7A, page 247

Angular Speed (ave) is equal to the change in angular displacement(in radians!) per unit time.ave= / t , in radians/sec

Remember, One revolution = _______ radians,

so the number of revolutions =/2rad

Book Example 3: A child at an ice cream parlor spins on a stool. The child turns counterclockwise with an average angular speed of 4.0 rad/s. In what time interval will the child’s feet have an angular displacement of 8.0 radians.=8 radave= 4.0 rad/s t=?

ave= / t, so t=/ ave

t= 8 rad / 4.0 rad/s = 2.0 s = 6.3 sec

Example 4. A man ties a ball to the end ofa string and swings it around his head. Theball’s average angular speed is 4.50 rad/s.A) In what time interval will the ball movethrough an angular displacement of 22.7 rad?B) How many complete revolutions does thisrepresent?

Given:

Find:

Solution:

avg = 4.50 rad/s = 22.7 rad

t B) # of revolutions

s 5.04 d/sar 4.50

dar 22.7

rev. of #

ω

θΔt

avg

rev. 3.61 dar2

dar 22.7

rad2

θ

Angular Acceleration (ave) is the change in angular speed () per unit time(s).ave= /t = (2-1)/(t2-t1)Unit is rad/s2

Book Example 1: A car’s tire rotates at an initial angular speed of 21.5 rad/s. The driver accelerates, and after 3.5 sec the tire’s angular speed is 28.0 rad/s. What is the tire’s average angular acceleration during the 3.5 s time interval?

1=21.5 rad/s 2=28.0 rad/s

t=3.5 s ave= ?ave= 2-1/t =

ave =(28.0rad/s-21.5rad/s)/3.5s

ave=1.9 rad/s2

Example 2. The angular velocity of a rotatingtire increases from 0.96 rev/s to 1.434 rev/s with an average angular acceleration of 6.0 rad/s2 . Find the time required for given angular acceleration.

First change revolutions/sec to radians/sec by multiplying by 2.

(0.96rev/s)X 2 = 6.0 rad/sec (1.434rev/s)X 2 = 9.01 rad/sec

Given:

Find:

Solution:

1 = 6.0 rad/s 2=9.01 rad/savg=6.0rad/s2

t = ?

t = 9.01 rad/s – 6.0 rad/s =

6.0 rad/s2

0.50s

ω - ω

tavg

12

t

ω - ω 12

avg

Homework

Practice 7B (1,2, and 4) page 248 and 7C (1 and 3), pg 250

Angular Kinematics All points on a rotating rigid

object have the same angular acceleration and angular speed.

Angular and linear quantities correspond if angular speed () is instantaneousinstantaneous and angular acceleration () is constant.See table 7-2, pg. 251 or your

formula sheet

Equations for Motion with Constant Acceleration

taif 2

2

1tatx i

xaif 222

Linear Rotational

tαωωf i

2tα2

1tωθ i

θ2ωω 22if

νf)Δt (ν Δx i21 Δt )ω (ω Δθ fi2

1

Book Example 1 (page 251)The wheel on an upside-down bicycle moves through 11.0 rad in 2.0 sec. What is the wheel’s angular acceleration if its initial angular speed is 2.0 rad/s?

=11.0 rad t=2.0 si= 2.00rad/s = ?

Use one of the angular kinematic equations from Table 7-2 to solve for … which one?

it ½ (t)2

Solve for =2(-it)/(t)2

=2[11.0rad-(2.00rad/s)(2.0s)]/(2.0s)2

= 3.5 rad/s2

Example 2. A CD initially at rest begins to spin, accelerating with a constant angular acceleration about its axis through its center,achieving an angular velocity of 3.98 rev/swhen its angular displacement is 15.0 rad.what is the value of the CD’s angular acceleration?

Given:

Find:

Original Formula:

i = 0.0 rad/s f = 3.98 rev/s X 2 = 25.0 rad/s = 15.0 rad

θ2ωω 2i

2f

The i = 0.0 rad/s, so

Given:

Find:

Formula:

i = 0.0 rad/s f = 25.0 rad/s = 15.0 rad

θ2ω 2f

Now, we isolate

Given:

Find:

Working Formula:

i = 0.0 rad/s f = 25.0 rad/s = 15.0 rad

θ2

ω α

2f

Given:

Find:

Solution:

i = 0.0 rad/s f = 25.0 rad/s = 15.0 rad

222

rad/s 20.8 rad) 0.15(2

rad/s) (25.0

θ2

ω α

f

Homework

Problems 7D, 1-5 on page 252

Tangential Speed

Although any point on a fixed object has the same angular speed, their tangential speeds vary based upon their distance from the center or axis of rotation.

Tangential speed (vt)– instantaneous linear speed of an object directed along the tangent to the objects circular path. (see fig. 7-6, pg. 253)

Formula vt = r

Make sure is in radians/s

r is in metersvt is in m/s

Book Example 1 (pg. 254): The radius of a CD in a computer is 0.0600m. If a microbe riding on the disc’s rim had a tangential speed of 1.88m/s, what is the disc’s angular speed?r=0.0600m vt=1.88m/s=?Use the tangential speed

equation (vt = r )to solve for angular speed.

=vt/r =1.88m/s/0.0600m=31.3 rad/s

Example 2. A CD with a radius of 20.0 cm rotates at a constant angular speed of 1.91 rev/s. Find the tangential speed of a point on the rim of the disk.

Given:

Find:

Solution:

r =20.0cm= 0.200 m = 2X1.91rev/s= 12.0 rad/s

t

m/s 2.40 rad/s) m)(12.0 (0.200 rω ν t

Tangential Acceleration (at)- instantaneous linear acceleration of an object directed along the tangent to the objects circular path.

Formulaat= r

Make sure that at is in m/s2

r is in meters is in radians/s2

Tangential Acceleration

Example 1(at): A spinning ride at a carnival has an angular acceleration of 0.50rad/s2. How far from the center is a rider who has a tangential acceleration of 3.3 m/s2?

=0.50rad/s2 at= 3.3 m/s2

r=?Use the tangential acceleration

equation (at= r) and rearrange to solve for r.

r=at/r = (3.3 m/s2 )/(0.50rad/s2)r=6.6m

Example 2. The angular velocity of a rotating CDwith a radius 20.0 cm increases from 2.0 rad/s to6.0 rad/s in 0.50 s. What Is the tangentialacceleration of a point on the rim of the disk during this time interval?

Hint: Start by finding the angular acceleration

Given:Find:

Solution:

r = 0.200 m i = 2.0 rad/s f= 6.0 rad/st = 0.50 s

2rad/s 8.0 s 0.50

rad/s 2.0 - rad/s 6.0

t

ω - ω if

Given:

Find:

Solution:

r = 0.200 m = 8.0 rad/s2 t = 0.50 s

at

22t m/s 1.6 )rad/s m)(8.0 (0.200 rα a

Problems 7E (1,2 and 4), page 255 and 7F (all) pg 257

Centripetal Acceleration

Centripetal acceleration is the acceleration directed toward the center of a circular path.Caused by the change in direction

Formula ac = vt2/r

And also……ac = r2

Example 1 pg. 258: A test car moves at a constant speed around a circular track. If the car is 48.2 m from the track’s center and has a centripetal acceleration of 8.05 m/s2. What is its tangential speed?r=48.2m ac = 8.05 m/s2 vt=?Use the first centripetal acceleration

equation (ac = vt2/r) to solve for vt.

vt2= acXr

vt= acXr

vt= (8.05m/s2X48.2m)

vt= 19.7m/s

Example 2. An object with a mass of 0.345 kg,Moving in a circular path with a radius of 0.25 m,Experiences a centripetal acceleration of 8.0 m/s2.Find the object’s angular speed.


Given:

Find:

Original Formula:

m = 0.345 kg r = 0.25 m ac = 8.0 m/s2

2

c rω a


Given:

Find:

Solution:

m = 0.345 kg r = 0.25 m ac = 8.0 m/s2

rad/s 5.7 /s32m 0.25

/sm 8.0

r

a ω 2

2c

Tangential and centripetal acceleration are perpendicular, so you can use Pythagorean theorem and trig to find total acceleration and direction, if asked for.

See figure 7-9, on page 259

Using the Pythagorean Theorem

ac

at

atotal

Homework: 7G pg. 258 and Section

Review pg. 259

Suppose a ball (mass m) moves with a constant speed V while attached to a string….

“Is there an outward force, pulling the ball out?”

NO!Tac

at(and vt)

“What about the “centrifugal force” I feel when my car goes around a curve at high speed?”

There is no such thing* as centrifugal force!

You feel the centripetal force of the door pushing you towards the center of the circle of the turn!

Your confused brain interprets the effect of Newton’s first law (inertia) as a force pushing you outward.*Engineers talk about centrifugal force when they attach a coordinate system to an accelerated object. In order to make Newton’s Laws work, they have to invent a pseudoforce, which they call “centrifugal force.”

Centripetal force is necessary for circular motion.

Without it, the object would go in a linear path tangent to the circular motion at that point.Then it would follow the motion

of a free-falling body (parabolic path toward the ground).

Centripetal Force

Causes of Circular Motion

Centripetal acceleration (ac) is caused by a FORCE directed toward the center of the circular path. This force is called centripetal force (Fc)

Formula Fc=mac Fc=mvt2/r or,

Fc=mr2

The unit is the Newton (kg·m/s2)

Book Example Pg. 261: A pilot is flying a small plane at 30.0m/s in a circular path with a radius of 100.0m. If a force of 635N is needed to maintain the pilot’s circular motion, what is the pilot’s mass?r=100.0m Fc= 635 N vt=30.0m/s

m=?

Formula Fc=mvt2/r solve for

m…m=Fc(r/vt

2)m=635NX[100.0m/(30.0m/s)2]m=70.6kg

Example 2: A raccoon sits 2.2 m from the center of a merry-go-round with an angular speed of 2.9 rad/s. If the magnitude of the force that maintains the raccoon’s circular motion if 67.0 N, what is the raccoon’s mass? r=2.2m Fc= 67.0 N =2.9 rad/s

m=?

Formula Fc=m r 2 solve for

m…m=Fc/r

2

m=67.0N/[(2.2m)(2.9rad/s)2]m=3.6kg

Newton’s Law and Gravity

Gravitational force depends on the mass of the objects involved.

Formula Fg=Gm1m2/r2

G is the constant of universal

gravitation = 6.673X10-11Nm2/kg2

Gravitational force is localized to the center of a spherical mass

Book Example pg. 264: Find the distance between a 0.300kg billiard ball and a 0.400kg billiard ball if the magnitude of the gravitational force is 8.92X10-11N.m1=0.300kg m2=0.400kg

Fg=8.92X10-11N r=?

Fg= Gm1m2/r2 G=6.673X10-11Nm2/kg2

r2=(G/Fg)m1m2

r=[(6.673X10-11Nm2/kg2)/8.92X10-11N](0.30kgX0.400kg)

r= 3.00X10-1m

This is where we get our value for Free Fall Acceleration

For the earth, Fg=mg, and Fg= mg= GmEm/rE

2

So, g= GmE/rE2

If the earth’s radius is 6.37X106m, and the earth’s mass is 5.98X1024kg, what is acceleration due to gravity?

9.83m/s2 …close!!

Assignments…

7H and I

Amusement Park Physics

Review Problems for Ch. 7

1,3,6,7,9,10,12,16,20,21,24,25,30,38,40

holt physics chapter 7

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