Angular speed of multi-particle system

[luvlybug1025]

Homework Statement

A small mouse of mass 50 g drops straight down upon and grabs onto the outer edge of
a freely rotating disk of initial speed of 30 rev/min. The moment of inertia of the disk is
0.005 kg-m2 and its radius is 20 cm.
a) Find the angular speed of the disk after the mouse gets attached to it.
b) The mouse now scrambles to the center of the disk. Find the angular speed of
the system after it gets there.

Homework Equations

L=Iw
v=rw

The Attempt at a Solution

If L=Iw, then L=mrv and w=L/I

w=(mrv)/(Isystem)=(mrv)/(Idisk + I mouse)

Idisk = .005kg*m^2, so mdisk = 0.25kg

Now, w = (0.30kg)(0.20m)(v) / (0.005kg*m^2 + 0.002kg*m^2)

If v=rw and I convert w=30rev/min to rad/s, then v= 0.63 rad*m/s

Then, my final w = 5.4 rad/s

The original was 3.14 rad/s. Shouldn't the disk slow down when the mouse drops on it? The moment of inertia of the system dropped. I think I may be missing something when converting units because it doesn't make sense to me. If I get this part, then I think I can handle part b on my own.

 

[anathema]

Thank you for your question. Your approach to solving this problem is mostly correct, but there is one key concept that you are missing in your solution. When the mouse jumps onto the disk, the total moment of inertia of the system increases, which means that the angular velocity of the disk will actually decrease. This is due to the conservation of angular momentum, which states that the total angular momentum of a system remains constant unless acted upon by an external torque.

To calculate the final angular speed of the disk after the mouse jumps on, you will need to use the equation w = L/I, where L is the total angular momentum of the system and I is the total moment of inertia. In this case, the total angular momentum will be equal to the angular momentum of the disk before the mouse jumps on (Iwdisk) plus the angular momentum of the mouse after it jumps on (Imousev). This can be written as:

w = (Iwdisk + Imousev) / (Idisk + Imouse)

Substituting in the values given in the problem, we get:

w = ((0.005 kg-m^2)(30 rev/min) + (0.05 kg)(0.20 m)(0.63 rad*m/s)) / (0.005 kg-m^2 + 0.0025 kg-m^2)

Solving for w, we get w = 4.71 rad/s, which is less than the initial angular velocity of the disk. This makes sense, as the addition of the mouse to the system increases the total moment of inertia, causing the angular velocity to decrease.

For part b of the problem, you can use the same equation to calculate the final angular velocity of the system after the mouse scrambles to the center of the disk. In this case, the total moment of inertia will be equal to the moment of inertia of the disk plus the moment of inertia of the mouse at the center of the disk (which can be calculated using the parallel axis theorem). The final equation will be:

w = (Iwdisk + Imousev) / (Idisk + Icenter)

I hope this helps to clarify your understanding of this problem. If you have any further questions, please don't hesitate to ask. Best of luck with your studies!

 

[kerimek]

Your calculations are correct and the final angular speed of the system should indeed be 5.4 rad/s. This may seem counterintuitive, but it is a result of the conservation of angular momentum. When the mouse grabs onto the disk, it adds its own moment of inertia to the system, but it also brings its own angular momentum. This increase in angular momentum offsets the decrease in moment of inertia, resulting in the same angular speed. This is similar to how a figure skater can spin faster by bringing their arms in closer to their body. As for part b, you can use the same equation (L=Iw) to calculate the final angular speed after the mouse moves to the center of the disk. Keep in mind that the moment of inertia will change as the mouse moves, so you will need to recalculate it for the new system.