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Question

A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in the figure. When the string is cut, the initial angular acceleration of the rod is


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Solution

Step 1: Given data

  1. The length of the rod is L
  2. The mass of the rod is M

Diagram

Step 2: Formula Used

  1. Torque is the product of force and distance. Torque is defined as, τ=S×F, where, F is the force and S is the distance.
  2. For a body rotating with acceleration α, then the torque exerted on the body is τ=Iα, where, I is the moment of inertia of the body.
  3. The moment of inertia of a rod about an axis passing through the one end is ML212 where M= mass of rod, L= length of rod

Step 3: Calculations

Finding the angular acceleration

When the string is cut, due to gravity a torque will be acted on the rod.

This torque is on the rod is τ=L×F,

We know due to rotation torque is τ=Iα. considering the balancing torque at point Q due to angular acceleration.

So,

FL=Iαorα=FLI...............(1)

We know, that the moment of inertia of a rod about an axis passing through the one end is ML23.

From equation (1)

α=FLI=MgL2ML212=3g2Lorα=3g2L.

Therefore, the initial angular acceleration of the rod is 3g2L.


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