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A uniform solid disk and a uniform ring are place side by side at the top of a rough incline of height h. If they are released from rest and
Question
A uniform solid disk and a uniform ring are place side by side at the top of a rough incline of height h. If they are released from rest and roll without slipping, determine the velocity vring of the ring when it reaches the bottom.
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Physics
3 years
2021-08-15T23:04:56+00:00
2021-08-15T23:04:56+00:00 1 Answers
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Answer: V = √gh
Explanation: This question can be solved by considering conservation energy.
Also we need to take the moment of inertia of the two bodies into consideration.
Total kinetic energy consists of transla-tional and rotational kinetic energies. Non slipping means v=rω and for the ring I=M R^2. The ring has kinetic energy due to transla-tional and rotational velocity.
Therefore:
mgh = (1/2 m V^2) + (1/2 . I w^2)
mgh = (1/2 . m . V^2) + (1/2 m R^2 . V^2/ R^2)
gh = V^2
the velocity vring of the ring when it reaches the bottom will be:
V = √gh