Question

A rope is wound through a hollow cylinder of mass 3 kg and radius 40 cm. If the rope is pulled with a force of 30N, the angular acceleration of the cylinder will be
A) $10rad{s^{ - 2}}$
B) $15rad{s^{ - 2}}$
C) $20rad{s^{ - 2}}$
D) $25rad{s^{ - 2}}$

Answer 1

Hint:Torque is defined as the product of force applied and the perpendicular distance. Moment of inertia is defined as the tendency of the body to resist any kind of angular motion. The relation of torque and moment can be used to solve this problem.

Formula used:The formula of the torque is given by $T = F \times r$ where T is the torque F is the force and r is the distance from the fixed point to the point where the force is applied. The formula of torque is also represented as$T = I\alpha $Where T is the torque I is the moment of inertia and $\alpha $ is the angular acceleration.

Step by step solution:
The mass of the hollow cylinder is given to be 3 kg and the radius of the hollow cylinder is 40 cm. The rope wound on the hollow cylinder is pulled by a force of 30N.
So the torque applied on the hollow cylinder is given by,
$ \Rightarrow T = F \times r$
Where T is the torque F is the force and r is the distance from the fixed point to the point where the force is applied.
Replace the value of the force and the radius of the hollow cylinder.
$ \Rightarrow T = F \times r$
$ \Rightarrow T = 30 \times \left( {\dfrac{{40}}{{100}}} \right)$
$ \Rightarrow T = 12N$………eq. (1)
The value of the torque can also be represented as,
$ \Rightarrow T = I\alpha $………eq. (2)
Where T is the torque I is the moment of inertia and $\alpha $ is the angular acceleration.
The moment of inertia of the hollow cylinder is equal to$I = M{R^2}$.
The moment of inertia is given by,
$ \Rightarrow I = M{R^2}$
$ \Rightarrow I = 3 \cdot {\left( {0 \cdot 4} \right)^2}$
$ \Rightarrow I = 0.48kg{m^2}$………eq. (3)
Replace the value of the moment of inertia and torque in the equation (2) from equation (1) and (3).
$ \Rightarrow T = I\alpha $
$ \Rightarrow 12 = \left( {0 \cdot 48} \right)\alpha $
$ \Rightarrow \alpha = \dfrac{{12}}{{0 \cdot 48}}$
$ \Rightarrow \alpha = 25rad{s^{ - 2}}$
The angular acceleration is given by$\alpha = 25rad{s^{ - 2}}$.

The correct answer for this problem is option D.

Note: It is advisable to remember the formula of torque and also understand so that you can use it whenever it is required and also it is advisable to member the moment of inertia of the standard solids which can be asked in the problem like cylinder or cuboid.