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Question

Question: The radius of a wheel is R and its radius of gyration about its axis passing through its center and ...

The radius of a wheel is R and its radius of gyration about its axis passing through its center and perpendicular to its plane is K. If the wheel is rolling without slipping the ratio of its rotational kinetic energy to its translational kinetic energy is-

A

K2R2\frac { \mathrm { K } ^ { 2 } } { \mathrm { R } ^ { 2 } }

B

R2 K2\frac { \mathrm { R } ^ { 2 } } { \mathrm {~K} ^ { 2 } }

C
D
Answer

K2R2\frac { \mathrm { K } ^ { 2 } } { \mathrm { R } ^ { 2 } }

Explanation

Solution

= 12mv2k2R212mv2\frac { \frac { 1 } { 2 } \mathrm { mv } ^ { 2 } \cdot \frac { \mathrm { k } ^ { 2 } } { \mathrm { R } ^ { 2 } } } { \frac { 1 } { 2 } \mathrm { mv } ^ { 2 } }= k2R2\frac { \mathrm { k } ^ { 2 } } { \mathrm { R } ^ { 2 } }