Question

An external torque is applied to a flywheel which is a solid cylinder of mass m = 100 kg and radius

Answer 1

The angular acceleration is $6.25 rad/s^2$

Explanation:

To solve this problem we can use the equivalent of Newton's second law for rotational motions:

$\tau = I \alpha$ (1)

where

$\tau$ is the torque acting on the body

I is the moment of inertia of the body

$\alpha$ is the angular acceleration

In this problem we have:

$\tau = 450 Nm$ is the torque

The moment of inertia of a solid cylinder about its axis is

$I=\frac{1}{2}MR^2$

where

M = 100 kg is the mass

R = 1.2 m is the radius

Substituting,

$I=\frac{1}{2}(100)(1.2)^2=72 kg m^2$

And solving eq.(1) for $\alpha$, we find the angular acceleration:

$\alpha = \frac{\tau}{I}=\frac{450}{72}=6.25 rad/s^2$

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