JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Rotational Mechanics

A ring of mass

M

and radius

R

is rotating with angular speed

ω

about a fixed vertical axis passing through its centre O with two point masses each of mass

\frac{M}{8}

at rest at O . These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is

\frac{8}{9} ω

and one of the masses is ata distance of

\frac{3}{5} R

from

O

. At this instant the distance of the other mass from O is :

(A) $\frac{2}{3} R$
(B) $\frac{1}{3} R$
(C) $\frac{3}{5} R$
(D) $\frac{4}{5} R$ ✓ Correct

Answer: (D)

Solution

By conservation of angular momentum:

Initial:

L_{i} = M R^{2} ω

Final:

L_{f} = (M R^{2} + \frac{M}{8} (\frac{3 R}{5})^{2} + \frac{M}{8} d^{2}) \frac{8 ω}{9}

Setting

L_{i} = L_{f}

M R^{2} ω = (M R^{2} + \frac{M}{8} \cdot \frac{9 R ^{2}}{25} + \frac{M}{8} d^{2}) \frac{8 ω}{9}

9 R^{2} = 8 R^{2} + \frac{9 R ^{2}}{25} + d^{2}

R^{2} = \frac{9 R ^{2}}{25} + d^{2}

d^{2} = R^{2} - \frac{9 R ^{2}}{25} = \frac{16 R ^{2}}{25}

d = \frac{4 R}{5}

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JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Rotational Mechanics

Solution

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