Question

if a spherical ball is under pure rolling on a table then the fraction of total kinetic energy

Answer 1

The moment of inertia of a sphere about its central axis and a solid spherical shell of mass M and radius R as shown in the figure above is Isphere=25Isphere=25MR2MR2

Kinetic energy of the rotation of the sphere KErotation=12rotation=12Iω2Iω2 where ωω is the angular velocity.

⇒⇒ KErotation=12rotation=12Iω2=1225Iω2=1225MR2ω2MR2ω2

Now, angular velocity ω=vRω=vR, where vv is the linear velocity of the sphere.

⇒⇒ KErotation=12rotation=12Iω2=1225Iω2=1225MR2MR2v2R2v2R2=15=15Mv2Mv2

Kinetic energy of the linear motion KElinear motion=12linear motion=12Mv2Mv2

Total Kinetic energy == KErotation+rotation+KElinear motion=15linear motion=15Mv2+12Mv2+12Mv2Mv2=710=710Mv2Mv2

Now, the fraction of its total energy associated with rotation =KE of RotationTotal KE=KE of RotationTotal KE=15Mv2710Mv2=15Mv2710Mv2

=27

Answer 2

Explanation:

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