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Rotational Mechanics Problem

1. A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at the top of an incline and released. The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling. Least time will be taken in reaching the bottom by

(a) solid sphere

(b) hollow sphere

(c) disc

(d) all will take same time

2. In the previous question, the smallest kinetic energy at the bottom of the incline will be achieved by

(a) solid sphere

(b) hollow sphere

(c) disc

(d) all achieve the same kinetic energy

Please provide your basis for choosing an option. I would like to know the concepts involved as well.

Note by Nishant Sharma
4 years ago

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it just meant that slipping will take place Jatin Yadav · 4 years ago

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@Jatin Yadav Right. Nishant Sharma · 4 years ago

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solution:

1.as slipping would take place hence friction is same for all, also mgsin(theta) is same for all hence net force and hence acc. of cm of all is same..

so to displace by same distance, all will take same time.

2.now, velocity of cm is same for all, hence tke will be same for all so, check for rke, rke = 1/2(I w^2) w = alpha t now net torque is same for all hence alpha is directly proportional to 1/I hence rke is directly proportional to I(1/I)^2 or 1/I

I is max for hollow sphere hence rke for it will be least and therefore total kinetic will also be smallest Jatin Yadav · 4 years ago

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@Jatin Yadav Thanks for a complete and concise solution. But I was wondering about the statement "friction is not sufficient to allow pure rolling". What does it have to do with the solution ? Nishant Sharma · 4 years ago

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First case mg\sin \theta - f = ma. Which is independent of I therefore all takes same time. Joe Bobby · 1 year, 8 months ago

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  1. (d)
2.(b) Jatin Yadav · 4 years ago

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