Question

A wheel starting from rest is uniformly accelerated at $2{\text{rad/s}}^2$ for 20 seconds. It rotates uniformly for the next 20 seconds and is finally brought to rest in the next 20 seconds. Total angular displacement of the wheel in radians is

• A. $400\text{rad}$
• B. $800\text{rad}$
• C. $1600\text{rad}$
• D. $1200\text{rad}$

Answer 1

The correct option is C $1600\text{rad}$

Given, initial angular acceleration $\alpha =2{\text{rad/s}}^2$
At $t=20\text{s}$, angular speed is given by $\omega ={\omega }_0+\alpha t$
Here, initial angular speed given as ${\omega }_0=0$
$\Rightarrow \omega =0+2{\text{(rad/s}}^2)\times (20\text{s})=40\text{rad/s}$

From $t=20\text{s}$ to $t=40\text{s}$
$\omega$ has uniform value, $\omega =40\text{rad/s}$
In next 20 seconds, it comes to rest.
Hence, from $\omega ={\omega }_0+\alpha t$, taking $\omega =0$ and ${\omega }_0=40\text{rad/s}$
$0=40+\alpha \left(20\right)$
or $\alpha =-2{\text{rad/s}}^2$

Hence, angular displacement in first 20 seconds
${\theta }_1={\omega }_{0}t+\frac{1}{2}\alpha t^2$
${\theta }_1=0+\frac{1}{2}\times 2\left({\text{rad/s}}^2\right)\times {20}^2\left({\text{s}}^2\right)$
${\theta }_1=400\text{rad}$
Displacement in next 20 seconds
Here, uniform angular velocity, hence $\alpha =0$
${\theta }_2={\omega }_0\times t$
${\theta }_2=40\left(\text{rad/s}\right)\times 20\left(\text{s}\right)$
${\theta }_2=800\text{rad}$
Angular displacement in final 20 seconds
${\theta }_3={\omega }_{0}t+\frac{1}{2}\alpha t^2$
Here ${\omega }_0=40\left(\text{rad/s}\right)$
$\alpha =-2\left({\text{rad/s}}^2\right)$
${\theta }_3=40\left(\text{rad/s}\right)\times 20\text{s}+\frac{1}{2}\times (-2)\left({\text{rad/s}}^2\right)\times \left(20{)}^2\right({\text{s}}^2)$
${\theta }_3=400\left(\text{rad}\right)$

Therefore, total angular displacement in 60 seconds
$\theta ={\theta }_1+{\theta }_2+{\theta }_3$
$\theta =400\left(\text{rad}\right)+800\left(\text{rad}\right)+400\left(\text{rad}\right)$
$\theta =1600\text{rad}$