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}

 

Answers (1)

As we learnt in

Torque -

underset{	au }{
ightarrow}= underset{r}{
ightarrow}	imes underset{F}{
ightarrow}   

 

- wherein

This can be calculated by using either  	au=r_{1}F; or; 	au=rcdot 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{	au }=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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