Question

A horizontal vinyl record of mass 0.115 kg and radius 0.0896 m rotates freely about a vertical axis through its center with an angular speed of 5.58 rad/s and a rotational inertia of 4.84 x 10-4 kg·m2. Putty of mass 0.0484 kg drops vertically onto the record from above and sticks to the edge of the record.What is the angular speed of the record immediately afterwards?

Answer 1

Step-by-step explanation:

Angular momentum is the product of inertial and angular frequency

L = I × ω

Where

L is angular momentum

I is inertia

ω is angular frequency

So, given that

Vinyl record has a massz

M = 0.115kg

Radius R = 0.0896m

Angular velocity of vinyl record

ω(initial) = 5.58 rad/s

Rotational inertial of vinyl record.

I(initial) = 4.84 × 10^-4 kgm²

Putty drop on the record

Mass of putty M' = 0.0484kg

Angular speed after putty drop ω'

Using conversation of angular momentum

Initial angular momentum is equal to final angular momentum

I(initial) × ω(initial) = I(final) × ω(final)

So, we need to find I(final)

Inertia log putty can be determine using MR² by assuming a thin loop

I(putty) = M'R² = 0.0484 × 0.0896

I(putty) = 3.89 × 10^-4 kgm²

I(final) = I(initail) + I(putty)

I(final) = 4.84 × 10^-4 + 3.89 × 10^-4

I(final) = 8.73 × 10^-4 kgm²

Therefore,

I(initial) × ω(initial) = I(final) × ω(final)

ω(final) = I(initial) × ω(initial) / I(final)

ω(final) = 4.84 × 10^-4 × 5.58 / 8.73 × 10^-4

ω(final) = 3.1 rad/s