Post on 05-Jan-2016
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Physics 221Chapter 10
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) ?
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
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.
Problem 2 What EXACTLY is a Radian?
• A. 550
• B. 570
• C. 590
• D. 610
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
Angular Velocity
• Angular Velocity = radians / time
= / t
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?
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
V and
• Linear (m/s) Angular (rad/s)
• V
• d / t / t
• 2 r f / t 2 f / t
v = r
a and
• Linear (m/s2) Angular (rad/s2)
• a • ( Vf - Vi ) / t ( f - i ) / t
a = r
Problem 4 . . .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
Solution 4 . . . CD player
= ( f - i ) / t
= (530 x 2 /60 - 0) / 3
= 18.5 rad/s2
• a = r • a = 0.06 x 18.5• a = 1.1 m/s2
Problem 5 . . . 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
Solution 5 . . . 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!
Angular Analogs
• d
• v
• a
Problem 6 . . . Angular Analogs
• d = Vi t + 1/2 a t2 ?
Solution 6 . . . Angular Analogs
• d = Vi t + 1/2 a t2 = i t + 1/2 t2
Problem 7 . . . 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?
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)
= - 4.4 rad/s2
f = i + t0 = 34.8 - 4.4 t
t = 7.9 s
Torque
• Torque means the “turning effect” of a force.
• SAME force applied to both. Which one will turn easier?
Torque
Torque = distance x force
= r x F
Easy!
Torque
Which one is easier to turn?
Torque . . . The Rest of the Story!
= r F sin
Easy!
Problem 8 . . . Inertia Experiment
• SAME force applied to m and M. Which one accelerates more?
Solution 8 . . . Inertia Experiment
• Since F = ma, the smaller mass will accelerate more
Problem 9 Moment of Inertia Experiment
• SAME force applied to all. Which one will undergo the greatest angular acceleration?
Solution 9 Moment of Inertia Experiment
• This one will undergo the greatest angular acceleration.
What is Moment of Inertia?
F = m a Force = mass x ( linear ) acceleration
= I
Torque = moment of inertia x angular acceleration
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.
All about Sarah Hughes . . .
Click me!
Problem 10 . . . Sarah Hughes
• Will her mass change when she pulls her arms in?• Will her moment of inertia change?
Solution 10 . . . Sarah Hughes
• Mass does not change when she pulls her arms in but her moment of inertia decreases.
Problem 11 . . . 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
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 = MR2 I = 1/2 MR2 I = 2 /5 MR2
• hoop disc ball
Problem 12 . . . 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
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
Problem 13 . . . 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.
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
Problem 14 . . . In the middle
• 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
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
Problem 1 The race of the century!
Will it be the hoop or the disc?
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