a) I = 1/2 m r2
Iturn = 1/2 * 1.9 * 0.24^2= 0.05472 kg.m2
Idisc = 1/2 * 0.2 * 0.24^2 = 0.00576kg.m2
Itot = 0.05472 + 0.00576 = 0.162 kg.m2
angular momentum = 0.05472 * 3.4 = 0.186048 kg.m2/s
Itot * w = 0.186048
w = 0.186048 / 0.162
new angular velocity = 1.148 rad/s
b) change in KE = 1/2 If wf2 - 1/2 Ii wi2 = 1/2 ((0.162 * 1.148^2) - (0.186048 * 3.4^2))
= -0.968 J
c) T = I * alpha
= 0.162 * [3.4 - 1.148 / 3.4]
constant torque = 0.107 N.m
A circular disk of mass 0.2 kg and radius 24 cm, initially not rotating, slips down a thin spindle onto a turntable (disk) of mass 1.7 kg and the same radius, rotating freely at 3.4 rad/s.
Find the new angular velocity of the combination.
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