Achild is riding a merry-go-round that has an instantaneous angular speed of 12 rpm. if a constant friction torque of 12.5 nm is applied to the merry-go-round with a moment of inertia of 50.0 kg m2, what is the angular acceleration in rad/s2?

Explanation:

The equivalent of Newton's second law for rotational motions is:

where

is the net torque applied to the object

I is the moment of inertia

is the angular acceleration

In this problem we have:

(net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)

is the moment of inertia

Solving for , we find the angular acceleration:

true tan speed = ang vel x radius

the answer is c. the lower legs are levers, and the knees are fulcrums. the ankles hold the loads.