# A sanding disk with rotational inertia 2.7 x 10^{-3} kg.m^2 is attached to an electric drill...

## Question:

A sanding disk with rotational inertia 2.7 x 10{eq}^{-3} kg.m^2 {/eq} is attached to an electric drill whose motor delivers a torque of magnitude 30 N m about the central axis of the disk. About that axis and with torque applied for 28 ms, what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

## Torque:

Torque is the measure of force required to rotate an object from some distance . The SI unit of torque is N-m.

Now in mathematical form torque can be written as

$$\boxed{\vec\tau=\vec r\times\vec F}$$

Here,

• {eq}\vec r {/eq} is the distance.
• {eq}\vec F {/eq} is the force applied

Now from Newton's second law , Change in angular momentum is a torque

$$\boxed{\vec\tau=\frac{\partial \vec L }{\partial t}}$$

Given in the question inertia {eq}(I)=2.7\times10^{-3} \ kg-m^2 {/eq}

Torque {eq}\tau=30 \ N-m {/eq}

time {eq}(t)=28 \ ms=28\times10^{-3} \ s {/eq}

a) Now recall that torque {eq}\begin{align} (\tau)&=\frac{\partial L }{\partial t}\\ \end{align} {/eq}

So angular momentum

{eq}\begin{align} L&=\tau\times t\\ &=30\times28\times10^{-3}\\ &=0.840 \ kg-m^2/s\\ \end{align} {/eq}

b) Now recall that angular momentum {eq}(L)=I\omega {/eq}

So angular velocity

{eq}\begin{align} \omega&=\frac{L}{I}\\ &=\frac{0.840}{2.7\times10^{-3}}\\ &=311.111 \ rad/s\\ \end{align} {/eq}