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Question

A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v24g with respect to the initial position. The object is a


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Solution

Step1: Given data

  1. The initial velocity of the body is u=v.
  2. The maximum height of the body on the curved surface is H=3v24g.

Diagram

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Step2: Conservation of energy

  1. From the concept of conservation of energy, the sum of kinetic and potential energy is conserved in every point of motion. i.e., 12mv2+12Iω2=mgh
  2. The kinetic energy of a body moving with a velocity is Kt=12mv2 and the kinetic energy due to rotation is 12Iω2, where, m is the mass of the body, I is the moment of inertia, and ω is the angular velocity of the body.
  3. The potential energy of a body at height h is mgh, and g is the acceleration due to gravity.

Step4: Finding the moment of inertia

We know from the conservation of kinetic energy, that the sum of kinetic and potential energy is conserved at h=0 and h=h.

So, 12mv2+12Iω2=mgh.

or12mv2+12Iω2=mgh.or12mv2+12IvR2=mg3v24gsince,ω=vRandh=3v24gor12IvR2=mg3v24g-12mv2or12IvR2=3mv24-12mv2or12IvR2=14mv2orI=14mv212vR2=12mv2.R2v2=12mR2orI=12mR2.

Therefore, the moment of inertia of the body is I=12mR2, this is the moment of inertia of a disc about an axis passing through the center perpendicular to the plane. So, the object is disc.


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