Question

A wheel rotates with a constant angular acceleration of $3.6{\text{rad/s}}^2$. If angular velocity of the wheel is $4.0\text{rad/s}$ at time $t_0=0$, what angle does the wheel rotate in $1\text{s}$? What will be its angular velocity at $t=1\text{s}$?

• A. $\theta =2.2\text{rad},\omega =7.6\text{rad/s}$
• B. $\theta =2.2\text{rad},\omega =0.4\text{rad/s}$
• C. $\theta =5.8\text{rad},\omega =7.6\text{rad/s}$
• D. $\theta =5.8\text{rad},\omega =0.4\text{rad/s}$

Answer 1

The correct option is C $\theta =5.8\text{rad},\omega =7.6\text{rad/s}$

Given,
Angular acceleration $\alpha =3.6{\text{rad/s}}^2$
Angular velocity of wheel at $t_0=0,{\omega }_0=4.0\text{rad/s}$
Angle rotated in $1$ second from eq. $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$
$\theta =4\text{(rad/s)}\times 1\text{(s)}+\frac{1}{2}\times \left(3.6{\text{rad/s}}^2\right)\times 1\times {\text{(s)}}^2$
$\theta =5.8\text{rad}$
From eqn. $\omega ={\omega }_0+\alpha t$, angular velocity at $t=1\text{s}$
$\omega =4\text{(rad/s)}+3.6{\text{(rad/s}}^2)\times 1\text{(s)}$
$\omega =7.6\text{rad/s}$