Arecord is dropped vertically onto a freely rotating (undriven) turntable. frictional forces act to bring the record and turntable 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?

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.

v 154.9= 0.9 length length s little over 154.9cm