Question

A solid sphere of mass M and radius R rolls on a horizontal surface without slipping. The ratio of rotational kinetic energy to total kinetic energy is –

• A. 1 : 2
• B. 2 : 5
• C. 2 : 7
• D. 7 : 10

Answer 1

Answer: (C)

2 : 7

Answer 2

$\frac { \mathrm { KE } _ { \text {Rot. } } } { \mathrm { KE } _ { \text {Total } } }$

$\frac { \frac { 1 } { 2 } \mathrm { mv } ^ { 2 } \frac { \mathrm { K } ^ { 2 } } { \mathrm { R } ^ { 2 } } } { \frac { 1 } { 2 } \mathrm { mv } ^ { 2 } \left( 1 + \frac { \mathrm { K } ^ { 2 } } { \mathrm { R } ^ { 2 } } \right) }$

$\frac { \mathrm { K } ^ { 2 } / \mathrm { R } ^ { 2 } } { 1 + \frac { \mathrm { K } ^ { 2 } } { \mathrm { R } ^ { 2 } } }$

= $\frac { 2 / 5 } { 1 + 2 / 5 }$