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2 years ago

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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 f

rom rest and roll without slipping, determine the velocity vring of the ring when it reaches the bottom.

1 answer:

Reil [10]2 years ago

3 0

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

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here * stand for multiplication

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radius of cylinder = 1.1 cm

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$E = \frac{[a+b(L/2)^2]I }{\pi r^2}$

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$E = 1.7 *10^{-4} V/m$

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c ).

75 cm means length will be half

that is x = L/2

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$R_{2} = R- R_{1}$

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$R_{2} = 1.16 *10^{-4}$ Ω

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