A record player turntable (M = 1.5 kg, R = 0.25 m) takes 30 seconds to accelerat
ID: 1324384 • Letter: A
Question
A record player turntable (M = 1.5 kg, R = 0.25 m) takes 30 seconds to accelerate to final velocity of 45 rpm. You may assume that the turntable is a solid disk with I = 1/2MR2.
a) What was the angular acceleration of the turntable?
b) What force was required to generate this acceleration?
After the turntable has reached its final speed of 45 rpm, a spider (m = 0.001) drops onto the outer rim of the turntable. You may treat the spider as a point of mass.
c) What was the initial angular momentum of the turntable?
d) What is the final angular speed of the turntable and spider combination?
Explanation / Answer
a) w1 = 0,
w2 = 45 rpm = 45*2*pi/60 = 4.71 rad/s
alfa = (w2-w1)/t
= (4.71-0)/30
= 0.157 rad/s^2
b) I = 0.5*M*R^2
= 0.5*1.5*0.25^2
= 0.046875 kg.m^2
now use, T = I*alfa
F*R = I*alfa
F = I*alfa/R
= 0.046875*0.157/0.25
= 0.03 N
c) L = I*w
= 0.046875*4.71
= 0.22 kg.m^2/s^2
d) I1*w1 = I2*w2
w2 = I1*w1/I2
= 0.046875*4.71/(0.046875 + 0.001*0.25^2)
= 4.7 rad/s