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.

Question

Physics

Killian Burch

9 May, 12:00

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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Answers (1)

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Jaslene Oconnor · Answer 1

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