Assume there is a person sitting on a stool that is free to rotate, the person h
ID: 1989553 • Letter: A
Question
Assume there is a person sitting on a stool that is free to rotate, the person holds a wheel in which its center of mass is directly above the center of mass of the person plus the stool, then the person rotates the wheel counterclockwise at 6 rad/s. Suddenly, the person flips the wheel over and the angular speed of the wheel changes to 6 rad/s, but in clockwise direction. Given that the moment of inertia of the wheel is 0.6 kgm2, and that of the person plus the stool is 12 kgm2
a. Find the angular speed of the person plus the stool when the person flips the wheel over.
b. How much work is done by the person on flipping the wheel?
Attempt to a solution:
a.
Moment of inertia of the person plus the stool
= Is+p = 12 kgm2
Iw = 0.6 kgm2
0 = -6 rad/s, f = 6 rad/s
Li = Lf
(Iw)(0) = (Iw)(f) + (Is+p)(s+p)
(0.6)(-6) = (0.6)(6) + (12)(s+p)
s+p = -1.88 rad/s
b.
W = (1/2)(I)(f)2 - (1/2)(I)(i)2
= (1/2)(12)(6)2 - (1/2)(12)(-6)2
= 0 J ????????????????????
I got these two answers above, but I think they will be most likely wrong
Please give me a correct solution, thank you very much
I may have written the question unclearly, if you have anything unclear, feel free to ask.
Explanation / Answer
I1 = 0.6 kgm^2 , w1 = 6 rad/s
I2 = 12 kgm2 , w2=?
from conservation of angular momentum
I1w1 = I2w2
solving we get
w2 = 0.942 rad/s
b)
work done = 1/2 (I2w22 - I1w12 ) = - 101.15 J