Question

A drum rotates around its central axis at an angular velocity of $\mathbf{12}\mathbf{.}\mathbf{60}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{s}$. If the drum then slows at a constant rate of $\mathbf{4.20}\mathbf{}\mathbf{rad}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$.

(a) how much time does it take and

(b) through what angle does it rotate in coming to rest?

Short answer

a. The time required to slow down is$3.00\mathrm{s}$

b. The angle of the rotation when it coming to rest is$18.9\mathrm{rad}$

Step-by-step solution

Listing the given quantities

Angular velocity of drum $12.60\mathrm{rad}/\mathrm{s}$

Rate of slowdown α is$4.20\mathrm{rad}/{\mathrm{s}}^2$.

Understanding the concept of angular velocity

We can use the kinematic equations

Formula:

$\begin{array}{l}\mathbf{\omega }\mathbf{=}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{+}\mathbf{\alpha t}\\ \mathbf{\theta }\mathbf{=}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{}{\mathbf{\alpha t}}^{\mathbf{2}}\mathbf{}\end{array}$

Calculation of time required to slow down

(a)

$\begin{array}{c}\mathrm{t}=\frac{-{\mathrm{\omega }}_0}{\mathrm{\alpha }}\\ =3.00\mathrm{s}\end{array}$

The time required to slow down is$3.00\mathrm{s}$

Calculation of angle of the rotation when it coming to rest

(b)

$\begin{array}{c}\mathrm{\theta }-{\mathrm{\theta }}_0=-\frac{{{\mathrm{\omega }}_0}^2}{2\mathrm{\alpha }}\\ =18.9\mathrm{rad}\end{array}$

The angle of the rotation when it coming to rest is$18.9\mathrm{rad}$