3. A student stands initially at rest at the center of a turntable that can rota
ID: 1460281 • Letter: 3
Question
3. A student stands initially at rest at the center of a turntable that can rotate without friction. The student begins to rotate a heavy ball on the end of a 0.800m chain about her head, clockwise when viewed from above. The ball has a mass of 2.00kg, and it makes one revolution every 3.00s. The student and platform together have a moment of inertia of 0.900kgm2. (20pts) a. What is the angular velocity of the student? (It is a vector.) b. In the previous problem, what is the final total kinetic energy of the system, assuming the student, ball, and platform constitute the system?
Explanation / Answer
a. According to conservation of angular momentum
Initial angular momentum of system = Final angular momentum of system
0 = angular momentum of ball + angular momentum of student
0 = mwr^2 + Iw'
m = mass of ball = 2.00 kg
I = moment of inertia of student and platform = 0.900 kgm^2
w = angular velocity of ball =1 revolution per 3 seconds = 2.09 rad/s
r = distance of ball from axis of rotation = 0.800 m
w' = angular velocity of student = ?
0 = (2.00)(2.09)(0.800)^2 + (0.900)w'
w' = - 2.97 rad/s [ student rotates counter-clockwise]
b. Final total kinetic energy = rotational kinetic energy of ball + rotational kinetic energy of student & platform
= 0.5 mw^2r^2 + 0.5 Iw'^2
= 0.5(2.00)(2.09)^2(0.800)^2 + 0.5(0.900)(2.97)^2
= 6.79 J
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