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Physics 221 Chapter 10

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**Problem 1 . . . Angela’s new bike**

The radius of the wheel is 30 cm and the speed v= 5 m/s. What is the rpm (revolutions per minute) ?

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**Solution 1 . . . Angela’s rpm r = radius circumference = 2 r**

f = revolutions per second v = d/t v = 2 f r 5 = (2 )(f)(0.3) f = 2.6 revolutions per second or 159 rpm

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What is a Radian? A “radian” is about 60 degrees which is 1/6 of the circle (360 degrees). To be EXACT, the “radian pie” has an arc equal to the radius.

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**Problem 2 What EXACTLY is a Radian?**

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**Solution 2 What EXACTLY is a Radian?**

If each pie has an “arc” of r, then there must be 2 radians in a 3600 circle. 2 radians = 3600 6.28 radians = 3600 1 radian = 57.30

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**Angular Velocity = radians / time**

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**Problem 3 . . . Angular Velocity**

The radius of the wheel is 30 cm. and the (linear) velocity, v, is 5 m/s. What is the angular velocity?

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**Solution 3 . . . Angular Velocity**

We know from problem 1 that : f = 2.6 rev/s But 1 rev = 2 radians So = / t =(2.6)(2 ) /(1 s) = 16.3 rad/s

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**v = r V and Linear (m/s) Angular (rad/s) V d / t / t**

2 r f / t f / t v = r

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**a = r a and Linear (m/s2) Angular (rad/s2) a **

( Vf - Vi ) / t ( f - i ) / t a = r

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Problem Your CD player A 120 mm CD spins up at a uniform rate from rest to 530 rpm in 3 seconds. Calculate its: (a) angular acceleration (b) linear acceleration

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**Solution 4 . . . CD player = ( f - i ) / t**

= (530 x 2 / ) / 3 = 18.5 rad/s2 a = r a = 0.06 x 18.5 a = 1.1 m/s2

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Problem CD Music To make the music play at a uniform rate, it is necessary to spin the CD at a constant linear velocity (CLV). Compared to the angular velocity of the CD when playing a song on the inner track, the angular velocity when playing a song on the outer track is A. more B. less C. same

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Solution CD Music v = r When r increases, must decrease in order for v to stay constant. Correct answer B Note: Think of track races. Runners on the outside track travel a greater distance for the same number of revolutions!

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Angular Analogs d v a

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**Problem 6 . . . Angular Analogs**

d = Vi t + 1/2 a t ?

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**Solution 6 . . . Angular Analogs**

d = Vi t + 1/2 a t = i t + 1/2 t2

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Problem Red Corvette The tires of a car make 65 revolutions as the car reduces its speed uniformly from 100 km/h to 50 km/h. The tires have a diameter of 0.8 m. At this rate, how much more time is required for it to stop?

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** = - 4.4 rad/s2 f = i + t Solution 7 . . . Corvette**

100 km/h = 27.8 m/s = 69.5 rad/s since v = r Similarly 50 km/h = 34.8 rad/s (f)2 = (i)2 + 2 (34.8)2 = (69.5)2 + (2)()(65)(6.28) = rad/s2 f = i + t 0 = t t = 7.9 s

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**Torque Torque means the “turning effect” of a force.**

SAME force applied to both. Which one will turn easier?

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**Torque = distance x force**

= r x F Easy!

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**Which one is easier to turn?**

Torque Which one is easier to turn?

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**Torque . . . The Rest of the Story!**

= r F sin Easy!

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**Problem 8 . . . Inertia Experiment**

SAME force applied to m and M. Which one accelerates more?

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**Solution 8 . . . Inertia Experiment**

Since F = ma, the smaller mass will accelerate more

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**Problem 9 Moment of Inertia Experiment**

SAME force applied to all. Which one will undergo the greatest angular acceleration?

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**Solution 9 Moment of Inertia Experiment**

This one will undergo the greatest angular acceleration.

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**What is Moment of Inertia?**

F = m a Force = mass x ( linear ) acceleration = I Torque = moment of inertia x angular acceleration

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I = mr2 The moment of inertia of a particle of mass m spinning at a distance r is I = mr2 For the same torque, the smaller the moment of inertia, the greater the angular acceleration.

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All about Sarah Hughes . . . Click me!

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Problem Sarah Hughes Will her mass change when she pulls her arms in? Will her moment of inertia change?

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Solution Sarah Hughes Mass does not change when she pulls her arms in but her moment of inertia decreases.

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Problem Guessing Game A ball, hoop, and disc have the same mass. Arrange in order of decreasing I A. hoop, disc, ball B. hoop, ball, disc C. ball, disc, hoop D. disc, hoop, ball

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**Solution 11 . . . Guessing Game**

I (moment of inertia) depends on the distribution of mass. The farther the mass is from the axis of rotation, the greater is the moment of inertia. I = MR I = 1/2 MR I = 2 /5 MR2 hoop disc ball

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Problem K.E. of Rotation What is the formula for the kinetic energy of rotation? A. 1/2 mv2 B. 1/2 m2 C. 1/2 I2 D. I

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**Solution 12 . . . K.E. of Rotation**

The analog of v is The analog of m is I The K.E. of rotation is 1/2 I2

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Problem Long, thin rod Calculate the moment of inertia of a long thin rod of mass M and length L rotating about an axis perpendicular to the length and located at one end.

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**Solution 13 . . . Long, thin rod I = mr 2**

However, r is a variable so we need to integrate. (ain’t that fun!) A small mass m of length dr must = M/L dr I = M/L r2 dr I = (M/L)(L3 / 3 ) I = 1/3 ML2

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**Problem 14 . . . In the middle ID = ICM + MD2**

Suppose the rod spins about its C.M. One can use the Parallel Axis Theorem to calculate ICM ID = ICM + MD2 D is the distance between the C.M. and the other axis of rotation

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**Solution 14 . . . In the middle ID = ICM + MD2 1/3 ML2 = ICM + M(L/2)2**

ICM = 1/3 ML2 - 1/4 ML2 ICM = 1/12 ML2

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**Problem 1 The race of the century!**

Will it be the hoop or the disc?

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**Solution 1 . . . Race of the Century Hoop Loses ! ! !**

P.E. = K.E. (linear) + K.E. (angular) mgh = 1/2 mv2 + 1/2 I2 mgh = 1/2 mv2 + 1/2 I (v/r)2 For the disc, I = 1/2 mr2 So mgh = 1/2 mv2 + 1/2 (1/2 mr2)(v/r)2 Disc v = (4/3 g h)1/2 Similarly Hoop v = (g h)1/2

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