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.

Answers

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:

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