91Ó°ÊÓ

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

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:

$$\begin{gathered} T=\frac{dL}{dt} \\ L=l\mathrm{Ó\neg } \end{gathered}$$

(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.

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

Integrating this equation, we have

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

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

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

(b)

The expression of the angular momentum of the disk is

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