# A  rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angluar acceleration of the cylinder if the rope is pulled with a force of 30 N? Option 1) $0.25\: rad/s^{2}$ Option 2) $25\: rad/s^{2}$ Option 3) $5\: m/s^{2}$ Option 4) 25 $m/s^{2}$

As we learnt in

Torque -

$\underset{\tau }{\rightarrow}= \underset{r}{\rightarrow}\times \underset{F}{\rightarrow}$

- wherein

This can be calculated by using either  $\tau=r_{1}F\; or\; \tau=r\cdot F_{1}$

$r_{1}$ = perpendicular distance from origin to the line of force.

$F_{1}$ = component of force perpendicular to line joining force.

and,

Analogue of second law of motion for pure rotation -

$\vec{\tau }=I\, \alpha$

- wherein

Torque equation can be applied only about two point

(i) centre of motion.

(ii) point which has zero velocity/acceleration.

$T=I \alpha$

$FR \sin \Theta=I \alpha$

$FR=I \alpha= \alpha = \frac{FR}{MR^{2}}$

$\alpha= \frac{30 \times 0.4}{2 \times (0.4)^{2}}\:\:\:\Rightarrow \alpha 25 rad/s^{2}$

Option 1)

$0.25\: rad/s^{2}$

This option is incorrect.

Option 2)

$25\: rad/s^{2}$

This option is correct.

Option 3)

$5\: m/s^{2}$

This option is incorrect.

Option 4)

25 $m/s^{2}$

This option is incorrect.

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