Question

A body is rolling down an inclined plane. If kinetic energy of rotation is 40% of kinetic energy in translatory state, then the body is a

• A. Ring
• B. Cylinder
• C. Hollow ball
• D. Solid ball

Answer 1

Answer: (D)

Solid ball

Answer 2

Rotational kinetic energy is,

$\mathrm { K } _ { \mathrm { R } } = \frac { 1 } { 2 } \mathrm { I } \omega ^ { 2 } = \frac { 1 } { 2 } \mathrm { Mk } ^ { 2 } \left( \frac { \mathrm { v } } { \mathrm { R } } \right) ^ { 2 }$ ($\because \mathrm { I } = \mathrm { Mk } ^ { 2 }$ and v = R$\omega$) $= \frac { 1 } { 2 } \mathrm { Mv } ^ { 2 } \left( \frac { \mathrm { k } ^ { 2 } } { \mathrm { R } ^ { 2 } } \right)$

Translational kinetic energy is,

$\mathrm { K } _ { \mathrm { T } } = \frac { 1 } { 2 } \mathrm { Mv } ^ { 2 }$

As per question, $\mathrm { K } _ { \mathrm { R } } = 40 \% \mathrm { k } _ { \mathrm { T } }$

$\therefore \frac { 1 } { 2 } \mathrm { Mv } ^ { 2 } \left( \frac { \mathrm { k } ^ { 2 } } { \mathrm { R } ^ { 2 } } \right) = 40 \% \frac { 1 } { 2 } \mathrm { Mv } ^ { 2 }$ or $\frac { \mathrm { k } ^ { 2 } } { \mathrm { R } ^ { 2 } } = \frac { 40 } { 100 } = \frac { 2 } { 5 }$ For solid sphere, $\frac { \mathrm { K } ^ { 2 } } { \mathrm { R } ^ { 2 } } = \frac { 2 } { 5 }$

Hence, the body is solid ball.