Question

The wheel of a car accelerated uniformly from rest, rotates through $1.5\text{rad}$ during the first $\text{second}$. The angle rotated during the next $\text{second}$ will be :

• A. $7.5\text{rad}$
• B. $4.5\text{rad}$
• C. $1.5\text{rad}$
• D. $6.5\text{rad}$

Answer 1

The correct option is B $4.5\text{rad}$

The wheel is accelerating uniformly i.e., the angular acceleration $(\alpha$) is constant.

Applying kinetic equation,

$\Delta \theta ={\omega }_{i}t+\frac{1}{2}\alpha t^2$

$[{\omega }_i=0,\text{since wheel started from rest}$]

$\Rightarrow 1.5=0+\frac{1}{2}\alpha \times (1{)}^2$

$\therefore \alpha =3{\text{rad/s}}^2$

Again apply kinetic equation for angular velocity after $1.5\text{rad}$ rotation,

$\Rightarrow \omega ={\omega }_i+\alpha t$

$\Rightarrow \omega =0+\left(3\right)\left(1\right)=3\text{rad/s}$

Applying kinetic equation for angle rotated in next second ,

$\Delta \theta =\omega t+\frac{1}{2}\alpha t^2$

$\Rightarrow \Delta \theta =3\left(1\right)+(\frac{1}{2}\times 3\times 1)$

$\therefore \Delta \theta =4.5\text{rad}$

Hence, option $\left(b\right)$ is correct.

Alternative solution :

${\omega }_i=0,\alpha =\text{constant}$

Angular displacement during first second,

$\Delta \theta {}^{\prime }={\omega }_{i}t+\frac{1}{2}\alpha t^2$

$\Rightarrow 1.5=0+\frac{1}{2}\alpha \times (1{)}^2$

$\therefore \alpha =3{\text{rad/s}}^2$

Angular displacement during the first two seconds will be,

$\Delta \theta {}^{\prime \prime }={\omega }_{i}t+\frac{1}{2}\alpha t^2$

$\Rightarrow \Delta \theta {}^{\prime \prime }=0+\frac{1}{2}\left(3\right)(2{)}^2=6\text{rad}$

Thus angle rotated during the $2nd\text{second}$ is

$\Delta \theta =\Delta \theta {}^{\prime \prime }-\Delta \theta {}^{\prime }$

$\therefore \Delta \theta =6-1.5=4.5\text{rad}$