Question

A hollow cylinder is rolling without slipping on a horizontal surface as shown in figure. Then, the angular acceleration of the cylinder will be -

• 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 $16{\text{rad/s}}^2$

In this case, rolling without slipping is not possible without friction, as external force is acting on the cylinder.
Hence, friction will act at the point of contact of the cylinder, in backward direction.
Let the angular acceleration be $\alpha .$


Taking instantaneous axis of rotation passing through the point of contact $P$ [since in pure rolling condition, $P$ will be at rest]
On applying equation of torque about instantaneous axis of rotation,
$\sum T=I\alpha$
${\tau }_f+{\tau }_N+{\tau }_{mg}+{\tau }_F=I\alpha ...\left(i\right)$
Here, $N,mg,\&f$ pass through the instantaneous axis of rotation, hence ${\tau }_N=0,{\tau }_{mg}=0,{\tau }_f=0$
${\tau }_F=F\times 2r,$ where $2r=1\text{m}$

Substituing in eq.$\left(i\right)$
$1\times 40=2(5\times (0.5{)}^2)\alpha$

$\left[\begin{array}{c}I=I_0+m{d}^2=m{r}^2+m{r}^2=2m{r}^2\\ I_0=m{r}^2\text{for hollow cylinder}\end{array}\right]$

$\Rightarrow \alpha =16{\text{rad/s}}^2$
Hence, angular acceleration of the hollow cyclinder is $16{\text{rad/s}}^2$