# 2 doubts from Rotational Motion

Doubt 1 : [SOLVED]

A circular disc of radius R has MI equal to I about a perpendicular axis. The radius of the concentric circular part which should be removed so that remaining part has half the MI of disc about the same axis is ?

I tried it by taking MOI of whole disc then subtracting the MOI with distance as x but no answer is matching. Any Hints ?

Options for Doubt 1

A)
B) R/2
C)
D)

Doubt 2 : [ALMOST SOLVED]

Kinetic energy of a particle moving along a circle of radius r is given by B x^{2} where x is the distance travelled. The resultant acceleration acting on the particle is?

What should be the approach ?

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For the second oneâ€¦after getting vâ€“> radial velocity, in terms of x and then differentiating it will give tangential acceleration.
Also vÂ˛/r is the centripetal acceleration.

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Thank you so much, Iâ€™ll try this.

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Did you get the first one?

Not yet

Not yet but I will update here if I get it.

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You must have messed up with the mass of the disk. If we remove some part, mass will also be reduced.

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Ohh I was taking as M only. Let me recheck it now . Thnx for this.

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Thank you @LaveshGupta, itâ€™s done, the error was from mass.

@Amish_1706 Yes, I took a variable as mass per unit area and then while subtracting use that with area as Pie (x)^2 .
If you want I can post the solution here.
Option A is the correct answer.

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Can you post the solutionâ€¦?

Sure, standby.

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Also, my r and R are same ( small and caps R , are same, please donâ€™t get confused )
Lamda is mass per unit area which remains constant whether we cut the disc or make chips from it.

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Thanks Bro Got it
Actually I was thinking that the perpendicular axis was any arbitrary axis at some distance from the center of the disk because it wasnâ€™t mentioned in the question that it passes from the center. That way I wasnâ€™t able to eliminate one variable.

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Yeah maybe Allen wanted to save space and ink by not specifying about it.
Edit : Itâ€™s mentioned Concentric

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â€¦I donâ€™t think they are that poor

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