Solving Inertia of a Record Homework

[pucr]

Homework Statement

A turntable for playing records has a moment of inertia of 5.2 kgm^2 and is rotating with an angular velocity of 36 rpm. A record, initially at rest is dropped straight down onto the rotating turntable. The record and turntable rotate together at 33 rpm. Find the moment of the inertia of the record.

Homework Equations

$$\omega_{i} = \frac{6\pi}{5} \frac{rad}{sec}$$
$$\omega_{f} = \frac{11\pi}{10} \frac{rad}{sec}$$

The Attempt at a Solution

I really don't have a clue about how to solve this. It's a problem on a practice exam, not homework. I thought I could solve it with torque, but then there is no coefficient of friction to calculate force.
The only thing I can think to do is to calculate the difference in kinetic energy but that didn't give me the right answer.
If someone could give me some direction that would be amazing. Thanks.

 

[LowlyPion]

Rotational Kinetic Energy looks to be the key.

You have something rotating with a known I*ω²/2

And you are given a modified I*ω²/2 (the added mass of the record - distributed in the same manner as the turntable)
that yields a new I*ω²/2 where they give you the new ω.

 

[thomate1]

I would approach this problem by first understanding the concept of inertia and how it relates to rotational motion. Inertia is the resistance of an object to changes in its state of motion, and in rotational motion, it is related to the object's moment of inertia.

In this problem, we are given the moment of inertia of the turntable (5.2 kgm^2) and the initial and final angular velocities (36 rpm and 33 rpm). We can use the equation for conservation of angular momentum to find the moment of inertia of the record.

Angular momentum (L) is defined as the product of moment of inertia (I) and angular velocity (\omega). Mathematically, L = I\omega. Since angular momentum is conserved, we can equate the initial and final angular momenta of the turntable and the record.

For the turntable, the initial angular momentum is given by L_i = I_{turntable}\omega_{i} and the final angular momentum is given by L_f = I_{turntable}\omega_{f}. For the record, the initial angular momentum is zero (since it is initially at rest) and the final angular momentum is given by L_f = I_{record}\omega_{f}.

Setting these two equations equal to each other and solving for I_{record}, we get I_{record} = \frac{I_{turntable}\omega_{i}}{\omega_{f}}. Plugging in the values given in the problem, we get I_{record} = \frac{5.2 kgm^2 \times \frac{6\pi}{5} \frac{rad}{sec}}{\frac{11\pi}{10} \frac{rad}{sec}} = 2.4 kgm^2.

Therefore, the moment of inertia of the record is 2.4 kgm^2. This approach uses the concept of conservation of angular momentum, which is applicable in situations where there are no external torques acting on the system.