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A net torque of 2.84 N*m is applied on a solid sphere for 5.32 seconds. If...

A net torque of 2.84 N*m is applied on a solid sphere for 5.32 seconds. If the solid sphere has a moment of inertia of 2.96 kg*m^2 and it starts from rest, what is its final rotational kinetic energy?

Solutions

Expert Solution

Moment of inertia of the solid sphere = I = 2.96 kg.m2

Net torque on the solid sphere = = 2.84 N.m

Angular acceleration of the solid sphere =

I =

(2.96) = 2.84

= 0.96 rad/s2

Initial angular speed of the solid sphere = 1 = 0 rad/s (At rest)

Final angular speed of the solid sphere = 2

Time period = T = 5.32 sec

2 = 1 + T

2 = 0 + (0.96)(5.32)

2 = 5.11 rad/s

Final rotational kinetic energy of the solid sphere = E

E = 38.6 J

Final rotational kinetic energy of the solid sphere = 38.6 J

genius_generous answered 3 months ago
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