A body has equal amount of rotational kinetic energy and translational kinetic energy while rolling without slipping on a horizontal surface. Given body is an example of

Question

A body has equal amount of rotational kinetic energy and translational kinetic energy while rolling without slipping on a horizontal surface. Given body is an example of (A) disc (B) sphere (C) ring (D) cylinder

Tardigrade Admin · Accepted Answer

Translational Kinetic Energy = (1/2) m v2 Rotational Kinetic Energy = (1/2) I ω 2 Translational Kinetic Energy = Rotational Kinetic Energy (1/2)mv2=(1/2)Iω 2 Put v=rω , you will get I = M R2 which is inertia of ring.

Q. A body has equal amount of rotational kinetic energy and translational kinetic energy while rolling without slipping on a horizontal surface. Given body is an example of

disc

sphere

ring

cylinder

Translational Kinetic Energy $= \frac{1}{2} m v^{2}$
Rotational Kinetic Energy $= \frac{1}{2} I ω^{2}$
Translational Kinetic Energy = Rotational Kinetic Energy
$\frac{1}{2} m v^{2} = \frac{1}{2} I ω^{2}$
Put $v = r ω$ , you will get $I = M R^{2}$ which is inertia of ring.

Q. A body has equal amount of rotational kinetic energy and translational kinetic energy while rolling without slipping on a horizontal surface. Given body is an example of

disc

sphere

ring

cylinder

Translational Kinetic Energy =21​mv2 Rotational Kinetic Energy =21​Iω2 Translational Kinetic Energy = Rotational Kinetic Energy 21​mv2=21​Iω2 Put v=rω , you will get I=MR2 which is inertia of ring.

Translational Kinetic Energy $= \frac{1}{2} m v^{2}$
Rotational Kinetic Energy $= \frac{1}{2} I ω^{2}$
Translational Kinetic Energy = Rotational Kinetic Energy
$\frac{1}{2} m v^{2} = \frac{1}{2} I ω^{2}$
Put $v = r ω$ , you will get $I = M R^{2}$ which is inertia of ring.