Question: A sanding disk with rotational inertia$\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{k}\mathit{g}{\mathit{m}}^{\mathbf{2}}$is attached to an electric drill whose motor delivers a torque of magnitude 16 Nm about the central axis of the disk. About that axis and with the torque applied for 33 ms,

(a) What is the magnitude of the angular momentum? (b) What is the magnitude of the angular velocity of the disk?

1. The rotational inertia of the sanding disk is$1.2\times {10}^{-3}kg{m}^2$
2. The magnitude of the torque is T = 16 Nm
3. The time for the applied torque is$\text{t}=33\text{ms}=33\times {10}^{-3}\text{s}$

Use the concept of Newton’s second law in angular form and use the expression of angular momentum in terms of rotational inertia and angular velocity.

Formula:

$$\begin{gathered} T=\frac{dL}{dt} \\ L=l\omega \end{gathered}$$

(a)

According to Newton’s second law in angular form, the sum of all torques acting on a particle is equal to the time rate of the change of the angular momentum of that particle.

$$\begin{gathered} \frac{d\overrightarrow{L}}{dt}={\overrightarrow{T}}_{net} \\ dL=Tdt \end{gathered}$$

Integrating this equation, we have

$$\begin{gathered} \int dL={\int }_0^{t}tdt \\ L=T{{\left[t\right]}_0}^t \end{gathered}$$

At initial time t =0 s, the initial angular momentum of the disk is also zero, hence

$$\begin{gathered} \Rightarrow \text{L}=16\text{N}.\text{m}\times 33\times {10}^{-3}\text{s} \\ \Rightarrow \text{L}=0.528\text{kg}\frac{{\text{m}}^2}{\text{s}} \\ \Rightarrow \text{L}=0.53\text{kg}\frac{{\text{m}}^2}{\text{s}} \end{gathered}$$

(b)

The expression of the angular momentum of the disk is

$$\begin{gathered} L=I\omega \\ \omega =\frac{L}{I} \\ \omega =\frac{0.528\text{kg}\frac{{\text{m}}^2}{\text{s}}}{1.2\times {10}^{-3}{\text{kgm}}^2} \\ \omega =440rad/s \end{gathered}$$