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3 years ago

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A uniform circular disk of moment of inertia 8.0 kg m2 is rotating at 4.0 rad/s. A small lump of mass 1.0 kg is dropped on the d

isk and sticks to it at a distance of 1.0 m from the axis of rotation. What is the new rotational speed of the combined system?

1 answer:

Butoxors [25]3 years ago

7 0

The new rotational speed of the combined system is 32 rad/s

<u>Explanation:</u>

Given-

Moment of Inertia, I = 8 kg m²

speed of rotation, ω = 4 rad/s

Mass of the lump = 1 kg

distance = 1 m

New rotational speed = ?

We know,

When the lump is dropped on the rotating disc then the angular momentum remains conserved.

Angular Momentum of the disc, Lₐ = Iω

Angular momentum of the lump, L ₓ = mvR

Since the angular momentum remains conserved,

Lₐ = Lₓ

8 X 4 = 1 X v X 1

New rotational speed, v = 32 rad/s

Therefore, the new rotational speed of the combined system is 32 rad/s.

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<h3 /><h3>Potential Energy:</h3>

This is the energy due to the position of a body. The S.I unit is Joules (J)

The formula for change in potential energy.

<h3 /><h3>Formula:</h3>

ΔP.E = mg(H-h).............. Equation 1

<h3>Where:</h3>

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From the question,

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