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 accelaration of the cylider will be (Neglect the mass and thickness of the string):


40N

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

Answer 1

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

FBD of the cylinder,

From newton's second law,
$40+f=ma$
$\Rightarrow 40+f=m\left(R\alpha \right)......\left(i\right)$

Taking torque about the centre of the cylinder, we get,
$40\times R-f\times R=I\alpha$
$40\times R-f\times R=m{R}^2\alpha$
$40-f=mR\alpha .....\left(ii\right)$
Equating $\left(i\right)$ and $\left(ii\right)$, we get,
$40+f=40-f$
$\therefore f=0$
Putting this value in $\left(i\right)$, we get,
$40+0=5\times 0.5\times \alpha$
$40=2.5\alpha$
$\therefore \alpha =16{\text{rad/s}}^2$

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