Rotation 2 Question 4

4. 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 M8 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 89ω and one of the masses is at a distance of 35R from O. At this instant, the distance of the other mass from O is

(2015 Adv.)

(a) 23R

(b) 13R

(c) 35R

(d) 45R

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Answer:

Correct Answer: 4. (d)

Solution:

  1. Let the other mass at this instant is at a distance of x from the centre O. Applying law of conservation of angular momentum, we have

I1ω1=I2ω2

(MR2)(ω)=MR2+M835R2+M8x289ω

Solving this equation, we get x=45R.

NOTE If we take identical situations with both point masses, then answer will be (c). But in that case, angular momentum is not conserved.



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