What is the Angular Velocity of a Disk and Rod Combination?

[Bones]

Homework Statement

A uniform disk turns at 9.5 rev/s around a frictionless spindle. A nonrotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk, see the figure. They then turn together around the spindle with their centers superposed. What is the angular velocity of the combination?

Homework Equations

The Attempt at a Solution

I1w1= 1/2mr^2*9.5rev/sec
I2w2= 1/12ml^2*w2
That's as far as I got...is it even right??

 

[tiny-tim]

A uniform disk turns at 9.5 rev/s around a frictionless spindle. A nonrotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk, see the figure. They then turn together around the spindle with their centers superposed. What is the angular velocity of the combination?
…
I1w1= 1/2mr^2*9.5rev/sec
I2w2= 1/12ml^2*w2
That's as far as I got...is it even right??

Hi Bones!

(have an omega: ω )

Sort-of … the Is are right …

but you need conservation of angular momentum,

so for the "after" side, you'll need the total I, = I₁ + I₂, so that you can find (I₁ + I₂)ω₂.

 

[thomate1]

I can provide a response to this content by further analyzing the problem and providing a more detailed solution.

First, let's define some variables:
- I1: moment of inertia of the disk
- w1: angular velocity of the disk before the rod is dropped
- m: mass of the disk and rod
- r: radius of the disk
- l: length of the rod
- I2: moment of inertia of the rod
- w2: angular velocity of the disk and rod combination after the rod is dropped

To solve this problem, we can use the principle of conservation of angular momentum, which states that the total angular momentum of a system remains constant unless an external torque acts on the system. In this case, since the spindle is frictionless, we can assume that there is no external torque acting on the system.

Before the rod is dropped, the disk is rotating at a constant angular velocity of 9.5 rev/s. This means that the angular momentum of the disk alone is given by I1*w1. When the rod is dropped, it will start rotating with the same angular velocity as the disk, since they are now attached together. This means that the total angular momentum of the system after the rod is dropped is given by the sum of the angular momenta of the disk and the rod, which can be expressed as I1*w1 + I2*w2.

Since the total angular momentum before and after the rod is dropped must be equal, we can set these two equations equal to each other:

I1*w1 = I1*w1 + I2*w2

Solving for w2, we get:

w2 = -I1*w1/I2

Substituting the values for the moment of inertia of the disk and rod, we get:

w2 = -(1/2mr^2*9.5rev/s)/(1/12ml^2)

Simplifying, we get:

w2 = -6.38*rev/s

Therefore, the angular velocity of the combination is 6.38 rev/s in the opposite direction of the original rotation of the disk. This solution assumes that the rod is dropped directly onto the center of the disk. If the rod is dropped at a different point on the disk, the calculations may be slightly different.