Question

A record 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?

Answer 1

Answer:

record 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 0.41 times that of the turntable, what percentage of the initial kinetic energy is lost?

O 29%

O 38%

O 15%

O 49%

Law of Conservation Angular Momentum:

When two objects interact with one another then there are two possibilities: one is both objects will separate after the collision and the other is the two objects remain stick with each other after the collision. Therefore, in the first case, the collision is perfectly elastic and in the second case, the collision is perfectly inelastic. Mathematically,

L

=

I

ω

Where;

ω

is the angular velocity

I is a moment of inertia

According to the law of conservation angular momentum:

(

I

1

ω

1

+

I

2

ω

2

)

i

n

i

t

i

a

l

=

(

I

1

ω

1

+

I

2

ω

2

)

f

i

n

a

l

Answer 2

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