Question

A disk, initially rotating at $\mathbf{120}\mathbf{}\frac{\mathbf{rad}}{\mathbf{s}}$, is slowed down with a constant angular acceleration of magnitude $\mathbf{120}\mathbf{}\frac{\mathbf{rad}}{{\mathbf{s}}^{\mathbf{2}}}$.

(a) How much time does the disk take to stop?

(b) Through what angle does the disk rotate during that time?

Short answer

1. The time required for a disk to come to rest,$t=30.0\text{\hspace{0.17em}s}$
2. The angular displacement of the disk while it comes to rest,$\mathrm{\theta }=1.8\times {10}^3\text{rad}$

Step-by-step solution

Determine the given quantities

The initial angular velocity of the disk,${\mathrm{\omega }}_0=120\frac{\text{rad}}{\text{s}}$

The angular acceleration of the disk,$\mathrm{\alpha }=-4.0\frac{\text{rad}}{{\text{s}}^2}$

Determine the concept of kinematic equations

Use the kinematic equation for constant angular acceleration to solve for time and angular displacement.

$\mathrm{\omega }={\mathrm{\omega }}_0+\mathrm{\alpha t}$

$\mathrm{\theta }=\mathrm{\omega t}-\frac{1}{2}{\mathrm{\alpha t}}^2$

(a) Calculate the time required by the disk to take stop.

Kinematic equation for angular motion is:

$\begin{array}{l}\mathrm{\omega }={\mathrm{\omega }}_0+\mathrm{\alpha t}\\ \mathrm{t}=\frac{\mathrm{\omega }-{\mathrm{\omega }}_0}{\mathrm{\alpha }}\end{array}$

Substitute the values and solve further as:

$\begin{array}{c}\mathrm{t}=\frac{0-120}{-4.0}\\ =30.0\text{s}\end{array}$

The time required for a disk to come to rest, $\mathrm{t}=30.0\text{\hspace{0.17em}s}$.

(b) Calculate the rotation angle of disk

Consider the formula for the rotation angle as:

$\begin{array}{c}\mathrm{\theta }=\mathrm{\omega t}-\frac{1}{2}{\mathrm{\alpha t}}^2\\ =0\times 30.0-\frac{1}{2}\times (-4.0)\times {30.0}^2\\ =1.8\times {10}^3\text{\hspace{0.17em}rad}\end{array}$

The angular displacement of the disk while it comes to rest,$\mathrm{\theta }=1.8\times {10}^3\text{rad}$.