Question

A child 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?

Answer 1

Answer:

$-0.25 rad/s^2$

Explanation:

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

$\tau = I \alpha$

where

$\tau$ is the net torque applied to the object

I is the moment of inertia

$\alpha$ is the angular acceleration

In this problem we have:

$\tau = -12.5 Nm$ (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)

$I=50.0 kg m^2$ is the moment of inertia

Solving for $\alpha$, we find the angular acceleration:

$\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2$