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A string is wound round a hollow cylinder of mass 3 kg and radians 40 cm. The string is pulled with a force of 50 Newton. The angular acceleration of cylinder will be
1. 25rd s-2
2. 10rd s-2
3. 15rd s-2
4. 35rd s-2

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Correct Answer - Option 1 : 25rd s-2

Concept:

Torque: 

  • It is a type of twisting force that causes the rotation of an object.
  • Torque is the measure of force that can cause an object to rotate.
  • It is also called the moment of force.
  • Torque is the product of the force (F) and the perpendicular distance of force from point of rotation(S).
  • It is given by the formula:
  • Torque is also defined as the product of a moment of inertia and angular acceleration. 

τ = F × S

Where τ is t

or

τ = I × α 

I is moment of inertia, α is torque

Moment of Inertia:

  • Moment of inertia is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation. 
  • It is defined for an axis.
  • The moment of inertial of a particle m at distance r from the axis is 

I = mR2

  • Moment of inertia of hollow sphere is mr2

Angular acceleration:

  • The angular acceleration is defined as rate of change of angular momentum. 
  • It is the same as acceleration in linear motion.

Calculation:

Force applied F = 50 N

Mass m = 3 kg

radius r = 40 cm = 0.4 m

Torque τ = F × r =50 N × 0.4m = 20 N m

Moment of inertia I = mr2 = 3 kg × 0.4 m × 0.4 m = 0.48 kg m2

Angular acceleration α = τ \ I = 20 N m / 0.48 = 41.66 rad / s2

Note: According to official answer of UP PGT, the answer is 25rd s-2 which will come if we consider force as 30 N. 

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