Suppose a body of mass M and radius R is allowed to roll on an in… - Vidyadip

MCQ

Suppose a body of mass $M$ and radius $R$ is allowed to roll on an inclined plane without slipping from its topmost point $A$. The velocity acquired by the body, as it reaches the bottom of the inclined plane, is given by $\beta = 1 + \frac{I}{{M{R^2}}}$

A
$\sqrt {2gh} $
B
$\sqrt {\beta \times 2gh} $
✓
$\sqrt {\frac{{2gh}}{\beta }} $
D
${\frac{{2gh}}{\beta }}$

✓

Answer

Correct option: C.

$\sqrt {\frac{{2gh}}{\beta }} $

c
As the body rolls the inclined plane, it loses potential energy. Howerver, in rolling it acquires both linear and angular speeds and hence, gain in kinetic energy of translation and that of rotation. So by conservation of mechanical energy,

$\mathrm{Mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$

But as in rolling, $\mathrm{v}=\mathrm{R} \omega$

${\mathrm{Mgh}=\frac{1}{2} \mathrm{mv}^{2}\left[1+\frac{\mathrm{I}}{\mathrm{MR}^{2}}\right]}$

Let ${1+\frac{\mathrm{I}}{\mathrm{MR}^{2}}=\beta}$

Let $\mathrm{Mgh}=\frac{1}{2} \beta \mathrm{Mv}^{2}$

Hence $v=\sqrt{\frac{2 g h}{\beta}}$

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