Presentation on theme: "Physics 221 Chapter 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)"— Presentation transcript:
Physics 221 Chapter 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 B C D. 61 0
Solution 2 What EXACTLY is a Radian? If each pie has an “arc” of r, then there must be 2 radians in a circle. 2 radians = radians = radian =
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/s 2 ) Angular (rad/s 2 ) a ( V f - V i ) / 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/s 2 a = r a = 0.06 x 18.5 a = 1.1 m/s 2
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 = V i t + 1/2 a t 2 ?
Solution 6... Angular Analogs d = V i t + 1/2 a t 2 = i t + 1/2 t 2
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 ) (34.8) 2 = (69.5) 2 + (2)( )(65)(6.28) = rad/s 2 f = i + t 0 = 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 = mr 2 The moment of inertia of a particle of mass m spinning at a distance r is I = mr 2 For the same torque, the smaller the moment of inertia, the greater the angular acceleration.
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Problem Sarah Hughes Will her mass change when she pulls her arms in? Will her moment of inertia change?
Solution Sarah Hughes Mass does not change when she pulls her arms in but her moment of inertia decreases.
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
Solution 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 2 I = 1/2 MR 2 I = 2 /5 MR 2 hoop disc ball
Problem K.E. of Rotation What is the formula for the kinetic energy of rotation? A. 1/2 mv 2 B. 1/2 m 2 C. 1/2 I 2 D. I
Solution K.E. of Rotation The analog of v is The analog of m is I The K.E. of rotation is 1/2 I 2
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.
Solution 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 r 2 dr I = (M/L)(L 3 / 3 ) I = 1/3 ML 2
Problem In the middle Suppose the rod spins about its C.M. One can use the Parallel Axis Theorem to calculate I CM I D = I CM + MD 2 D is the distance between the C.M. and the other axis of rotation
Solution In the middle I D = I CM + MD 2 1/3 ML 2 = I CM + M(L/2) 2 I CM = 1/3 ML 2 - 1/4 ML 2 I CM = 1/12 ML 2
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 mv 2 + 1/2 I 2 mgh = 1/2 mv 2 + 1/2 I (v/r) 2 For the disc, I = 1/2 mr 2 So mgh = 1/2 mv 2 + 1/2 (1/2 mr 2 )(v/r) 2 Disc v = (4/3 g h) 1/2 Similarly Hoop v = (g h) 1/2