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Question

A solid sphere is in a rolling motion. In rolling motion, a body possesses translational kinetic energy Kt as well as rotational kinetic energy Kr simultaneously. The ratio Kt:(Kt+Kr) for the sphere is?


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Solution

Moment of inertia

  1. The moment of inertia of a body about a given axis in space is the sum of the products of the mass and square of the distance from the axis for each particle comprising the body.
  2. The moment of inertia is defined by the form, I=mx2, where, m is the mass of the body and x is the distance of the body from the axis of rotation.
  3. The moment of inertia of a solid sphere of mass m about the axis passing through its diameter is I=25mr2, r is the distance of the body from the axis of rotation.

Translational and rotational motion

  1. When a body rolls on a surface it has both translational and rotational motion.
  2. The translational kinetic energy of a body is given by Kt=12mv2, where, m is the mass of the body and v is the velocity of the body.
  3. The rotational kinetic energy of a body is given by Kr=12Iω2, where, I is the moment of inertia of the body and ω is the angular velocity of the body.

Step 1: Given data

The translational kinetic energy is Kt.

The rotational kinetic energy is Kr.

Step 2: Finding the energies

Let r be the radius of the sphere.

As we know, Kt=12mv2 and Kr=12Iω2.

So,

Kt+Kr=12mv2+12Iω2=12mv2+12Ivr2(since,ω=vr)Kt+Kr=12v2m+Ir2=12v2m+25mr2r2since,I=25mr2Kt+Kr=12v2m+25mKt+Kr=12v2×7m5Kt+Kr=7mv210

Step 3: Finding the ratio of energies

Again,

KtKt+Kr=12mv27mv210=57orKt:Kt+Kr=5:7

The ratio Kt:(Kt+Kr) for the sphere is:5:7.


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