Question

a solid sphere is rolling without slipping on a horizontal surface. the ratio of its rotational kinetic energy to its translation kinetic energy is​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \large{ \sf{ \red{ \bold{ q = ( - \dfrac{ Q}{4} })}}}}$

Answer 2

Explanation:

PHYSICS

A solid sphere of radius r is rolling on a horizontal surface. The ratio between the rotational kinetic energy and total kinetic energy is:

A .

75

B .

72

C .

21

ANSWER

Moment of inertia of solid sphere about an axis passing through its centre is given by I=52MR2

Velocity of centre of mass is given by v=Rω

Hence, translational kinetic energy is given by, KEL=21Mv2

∴KEL=21MR2ω2

Rotational kinetic energy is given by KER=21×52MR2ω2

Total kinetic energy, KET=KEL+KER=21×57

PLEASE PLEASE Please please mark as brilliant answer