JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Rotational Mechanics

A uniform circular disc of mass

50 k g

and radius

0.4 m

is rotating with an angular velocity of

10 r a d s^{- 1}

about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius

0.2 m

, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in

r a d s^{- 1}

) of the system is

Answer: 8

Solution

Angular momentum about the vertical axis is conserved because the rings are placed on gently and exert no external torque about that axis. The disc has

I_{disc} = \frac{1}{2} M R^{2} = \frac{1}{2} (50) (0.4)^{2} = 4

kg m

^{2}

. Each ring lies flat with its centre a distance

0.2

m from the disc's axis, so its moment of inertia about that axis is its own

m r^{2}

plus the parallel-axis term

m d^{2}

, that is

6.25 (0.2)^{2} + 6.25 (0.2)^{2} = 0.5

kg m

^{2}

; the two rings together add

1.0

kg m

^{2}

. Conserving

I ω

gives

ω_{f} = \frac{4 \times 10}{4 + 1.0} = 8

rad s

^{- 1}

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Integer answer

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JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Rotational Mechanics

Solution

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