Question

A string is wound around a hollow cylinder of mass $5\text{kg}$ and radius $0.5\text{m}$. If the string is now pulled with a horizontal force of $40\text{N}$ and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)

• A. $10{\text{rad/s}}^2$
• B. $16{\text{rad/s}}^2$
• C. $20{\text{rad/s}}^2$
• D. $12{\text{rad/s}}^2$

Answer 1

The correct option is B $16{\text{rad/s}}^2$

Given:

$m=5\text{kg}$ and $R=0.5\text{m}$

Horizontal force, $F=40\text{N}$

As, the cylinder is rolling without slipping. Hence, torque is producing rotation about centre $O$.


So, $\tau =rF\sin \theta$

Here, $r=R$ and $\theta ={90}^o$

So, $\tau =RF$

$\Rightarrow \tau =0.5\times 40=20\text{Nm}....\left(i\right)$

If $\alpha$ is the angular acceleration about $O$ then,

$\tau =I\alpha$

Where, $I=M{R}^2$

$\therefore \tau =M{R}^2\alpha ....\left(i\right)$

Comparing Eq. $\left(i\right)$ with Eq. $\left(ii\right)$,

$M{R}^2\alpha =20$

$\Rightarrow \alpha =\frac{20}{5\times (0.5{)}^2}$

$\Rightarrow \alpha =16{\text{rad/s}}^2$

Hence, $\left(B\right)$ is the correct answer.