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?

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

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .