The moment of inertia of a solid cylinder about its axis is given by (1/2)MR2 . If this cylinder rolls without shipping the ratio of its rotational kinetic energy to its translational kinetic energy is-

Question

The moment of inertia of a solid cylinder about its axis is given by (1/2)MR2 . If this cylinder rolls without shipping the ratio of its rotational kinetic energy to its translational kinetic energy is- (A) 1:1 (B) 2:2 (C) 1:2 (D) 2:3

Tardigrade Admin · Accepted Answer

KR=(1/2)Iω 2 ⇒ KR=(1/2)× (1/2)mR2(v2/R2)⇒ KR=(1/2).(1/2)mv2 KR=(1/2)KT ⇒ (KR/KT)=(1/2)

Q. The moment of inertia of a solid cylinder about its axis is given by $\frac{1}{2} M R^{2}$ . If this cylinder rolls without shipping the ratio of its rotational kinetic energy to its translational kinetic energy is-

$1 : 1$

$2 : 2$

$1 : 2$

$2 : 3$

$K_{R} = \frac{1}{2} I ω^{2} \Rightarrow K_{R} = \frac{1}{2} \times \frac{1}{2} m R^{2} \frac{v ^{2}}{R ^{2}} \Rightarrow K_{R} = \frac{1}{2} . \frac{1}{2} m v^{2}$
$K_{R} = \frac{1}{2} K_{T} \Rightarrow \frac{K _{R}}{K _{T}} = \frac{1}{2}$

Q. The moment of inertia of a solid cylinder about its axis is given by 21​MR2 . If this cylinder rolls without shipping the ratio of its rotational kinetic energy to its translational kinetic energy is-

1:1

2:2

1:2

2:3