Q. When a disc rolls without slipping, the ratio of its translational kinetic energy to rotational kinetic energy is?

जब कोई डिस्क बिना फिसले घूमती है, तो उसकी स्थानांतरीय गतिज ऊर्जा और घूर्णी गतिज ऊर्जा का अनुपात क्या होता है?

1:1

2:1

1:2

3:1

For a rolling disc:

Translational kinetic energy $K E_{t} = \frac{1}{2} M v^{2}$
Rotational kinetic energy $KE_r = \frac{1}{2} I \omega^2$ , and for a disc $I = \frac{1}{2} M R^2$ .
Since rolling without slipping implies $v = \omega R$ ,

KE_r = \frac{1}{2} \times \frac{1}{2} M R^2 \times \frac{v^2}{R^2} = \frac{1}{4} M v^2

Therefore, ratio:

\frac{KE_t}{KE_r} = \frac{\frac{1}{2} M v^2}{\frac{1}{4} M v^2} = 2 : 1

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