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6.System of Particles and Rotational Motion
hard
Two identical circular loops are moving with same kinetic energy one rolls $\&$ other slides. The ratio of their speed is
A
$2 : 3$
B
$2 : \sqrt 2 $
C
$\sqrt 2 : 2 $
D
$\sqrt 5 : \sqrt 3$
Solution
For the rolling hoop, $\mathrm{KE}=\frac 12 \mathrm{mV}^{2}+\frac 12\left(\mathrm{mr}^{2}\right)(\mathrm{V} / \mathrm{r})^{2}=\mathrm{mV}^{2}$
whereas for the sliding hoop, simply $\mathrm{KE}=\frac 12 \mathrm{mv}^{2}$
If the KEs are equal, then $\mathrm{V}^{2}=\mathrm{v}^{2} / 2$
$\mathrm{V}^{2} / \mathrm{v}^{2}=\frac 12$
$\mathrm{V} / \mathrm{v}=\sqrt{1 / 2}=\sqrt{2} / 2$
Standard 11
Physics
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