Question

Question: A grindstone increases in angular speed from 4.00 rad/s to 12.00 rad/s in 4.00 s. Through what angle does it turn during that time interval if the angular acceleration is constant? (a) 8.00 rad (b) 12.0 rad (c) 16.0 rad (d) 32.0 rad (e) 64.0 rad

Short answer

Answer

The grindstone will make an angular displacement of during the time interval of when its angular speed increases from $4.00 \text{rad/s}to12.00 \text{rad/s}$ .Hence, option (d) is the correct answer.

Step-by-step solution

Defining angular speed and angular acceleration

Angular speed can be defined as the rate of change of angular displacement. It can be expressed as follows:

$\mathit{\omega }\mathbf{=}\frac{\mathbf{\theta }}{\mathbf{t}}$

where

‘ $\mathbf{\theta }$’ is the angular displacement and ‘t ’ is the time.

‘$\mathit{\omega }$ ’ is the angular speed

Angular acceleration is defined as the time rate of change of angular velocity. It is expressed as follows:

$\mathit{\alpha }\mathbf{=}\frac{\mathbf{\Delta }\mathbf{\omega }}{\mathbf{\Delta }\mathbf{t}}$

where

‘ $\mathit{\alpha }$’ is angular acceleration

‘ $\mathbf{\Delta \omega }$’ is change in angular velocity and

‘ $\mathbf{\Delta t}$’ is change in time.

Calculating the angular displacement of grindstone

The formula for angular displacement is given by

$$\begin{gathered} \Delta \theta ={\omega }_{avg}\Delta t=\left(\frac{{\omega }_f+{\omega }_i}{2}\right)\Delta t \\ \Delta \theta =\left(\frac{12.00+4.00}{2}\right)\left(4.00\right) \\ \Delta \theta =32.0 \text{rad} \end{gathered}$$

So, the angle which the grindstone will make in 4.00 s is 32.0 rad when its angular speed increases from $4.00 \text{rad/s}to12.00 \text{rad/s}$ . Hence, the correct answer is option (d).