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Lilit [14]
4 years ago
12

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

ble to a common angular speed. If the rotational inertia of the record is 0.54 times that of the turntable, what percentage of the initial kinetic energy is lost?
Physics
1 answer:
BartSMP [9]4 years ago
8 0

Answer:

35%

Explanation:

Assuming no external torques present during the collision between the record and the turntable, total angular momentum must be conserved.

For a rotating body with some angular velocity and moment of inertia, the angular momentum can be expressed as follows;

L = I* ω

So, as initial angular momentum and final angular momentum must be the same, we have:

Li = Lf

⇒ I₁ * ω₁ = I₂ * ω₂ (1)

where I₁ is the rotational inertia of the turntable, and I₂, is the combined rotational inertia of the turntable and the record:

I₂ = I₁ + 0.54 I₁ = 1.54 I₁

We can solve (1) for the new common angular speed, as follows:

ω₂ = ω₁ / 1.54 (2)

The initial rotational kinetic energy is given by definition for the following equation:

Kroti = 1/2 * I₁ * ω₁² (3)

The final rotational kinetic  energy takes into account the new rotational inertia and the common final angular speed:

Krotf = 1/2* I₂ * ω₂² = 1/2* 1.54 I₁* (ω₁/1.54)² (4)

Dividing both sides in (3) and (4), we get:

Krotf/Kroti = 1/1.54 = 0.65

This means that the final rotational kinetic energy, has reduced to 0.65 of the initial value, or that has lost 35% of the initial kinetic energy.

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3 years ago
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3 years ago
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Refer to the diagram shown below.

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For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
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T₃ = 1.2259T₂             (1)

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3 years ago
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