A 15 kg flywheel has all its mass around its outer rim. A string is wrapped arou

Question

A 15 kg flywheel has all its mass around its outer rim. A string is wrapped around it and a m = 3.7 kg weight is hanging on the string. The flywheel has radius R = 0.25 m.

a) When the 3.7 kg weight is dropping with a speed of 1.7 m/s, what is the angular velocity of the flywheel?

w = ?? rad/s

b) When the 3.7 kg weight is moving with a speed of 1.7 m/s, what is the kinetic energy of the entire system?

K.E. = ?? J

c) If the system started from rest, how far has the weight fallen?

hfall = ?? m

d) What is the angular acceleration at this point?

a = ?? rad/s2

Explanation / Answer

a. Angular velocity: w = v/r v= 1.7 m/s r = .25 m w = 6.8 rads/s b. KE System KE wheel : 1/2 I w^2 I = m r^2 KE wheel = 1/2 m r^2 w^2 KE wheel = 1/2 (15 kg) (.25m)^2 (6.8 rad/s)^2 KE wheel = 21.675 J KE weigh= 1/2 (3.7kg) (1.7 m/s)^2 KE weight = 5.3465 J Total KE = 27.0215 J c. How far has weight fallen PE = KE mgh = 27.0215 h = 27.0215/(3.7)(9.81) h = .745 m d. Angular acceleration: First, find acceleration of weight a = g * 5.3465J/21.0215 J a = 2.417 m/s^2 angular acceleration = a/r = 9.67rads/s^2

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