A Spider on a rotating disk conservation of momentum

[BrainMan]

Homework Statement

A 45 rpm record in the shape of a solid disk 25 cm in diameter and mass 0.1 kg rotates about a vertical axle through its center. A 15 g spider rides along the edge of the record. Calculate the final angular speed of the record if the spider drops off without exerting a torque on the record.

Homework Equations

Conservation of momentum

The Attempt at a Solution

So I attempted to use the conservation of momentum to solve this problem. So
I first found the platforms moment of inertia
I= 1/2Mr^2
I = 1/2(.1)(.125)^2
I = 7.8125 x 10^-4
Then I found the spiders moment of inertia
I = MR^2
I = (.015)(.125)^2
I = 2.34375 x 10 ^-4
Then I found the momentum
L = IW
L = (7.8125 x 10^-4 + I = 2.34375 x 10 ^-4) (4.7129 rad/sec)
L = .0047860211
Then I compared that to the final momentum after the spider dropped off the disk
.0047860211 = (7.8125 x 10^-4)w
w = 6.126107 rad/sec or 58.50001265759 rpm. The correct answer is 45 rpm
I = (.015)(

 

[rcgldr]

Angular momentum of the disk and spider is constant. What happens to the spider's component of angular momentum if it slides off without exerting a torque onto the disk?

 

[BrainMan]

Angular momentum of the disk and spider is constant. What happens to the spider's component of angular momentum if it slides off without exerting a torque onto the disk?

The spiders component becomes zero and is made up for by the increase in angular velocity.

 

[Orodruin]

What is the angular momentum of a body in constant linear motion about a point not directly in its path?

 

[BrainMan]

What is the angular momentum of a body in constant linear motion about a point not directly in its path?

Its mass times its velocity? I don't get what you are saying.

 

[Orodruin]

Angular momentum, not linear momentum. In particular, what is the angular momentum of the spider relative to the center of the disk?

Also, if you make a free body diagram of the disk, what forces and torques are acting on it?

 

[BrainMan]

Angular momentum, not linear momentum. In particular, what is the angular momentum of the spider relative to the center of the disk?

Also, if you make a free body diagram of the disk, what forces and torques are acting on it?

Angular momentum is the angular velocity times the moment of inertia.

 

[Orodruin]

So what is the angular velocity of the spider in said situation?

Edit: Also consider ${\bf L} = {\bf r}\times {\bf p}$ for constant linear motion ...

 

[BrainMan]

So what is the angular velocity of the spider in said situation?

Edit: Also consider ${\bf L} = {\bf r}\times {\bf p}$ for constant linear motion ...

4.71239 rad/sec

 

[Orodruin]

So what makes you think the angular momentum of the spider is zero?

 

[BrainMan]

So what makes you think the angular momentum of the spider is zero?

Because it is flung off the disk.

 

[Orodruin]

This is not sufficient. Its angular momentum relative to any point on its trajectory is zero, but the angular momentum you have computed is not wrt a point on its trajectory.

 

[BrainMan]

This is not sufficient. Its angular momentum relative to any point on its trajectory is zero, but the angular momentum you have computed is not wrt a point on its trajectory.

I'm sorry but I don't understand your point.

 

[rcgldr]

I'm sorry but I don't understand your point.

The spider will move in a straight line that doesn't not intersect the center of the disk. It's angular momenum is equal to it's linear momentum (mv) times the shortest distance from the center of the disk to the linear path that the spider follows. In this case that distance is equal to the radius of the disk. You can also use the formula where the angular momentum = m v r sin(θ), where θ is the angle between the path line and a line that starts at the center of rotation, and crosses the path at some specific point. At the edge of the disk, θ would be 90°, so sin(θ) would be 1.

 

[BrainMan]

The spider will move in a straight line that doesn't not intersect the center of the disk. It's angular momenum is equal to it's linear momentum (mv) time the shortest distance from the center of the disk to the linear path that the spider follows. In this case that distance is equal to the radius of the disk.

I thought I included that in my calculation. I did the sum of all the moment of inertia's and then used that to find the total angular momentum. Then I removed the spiders moment of inertia and calculated what the speed had to be for the two momentum's to be equal.

 

[Orodruin]

Angular momentum is always computed relative to a paricular point in space. In your case, the angular momentum (that is conserved) is computed around the center of the disk. You must therefore also express the angular momentum of the spider with respect to this. Hint: angular momentum for constant linear motion is the linear momentum multiplied by the shortest distance between the trajectory and the reference point.

The alternative to using conservation of angular momentum is to draw the free body diagram for the disk. If the spider exerts no torque on the disk, what is the net torque on the disk?

 

[BrainMan]

Angular momentum is always computed relative to a paricular point in space. In your case, the angular momentum (that is conserved) is computed around the center of the disk. You must therefore also express the angular momentum of the spider with respect to this. Hint: angular momentum for constant linear motion is the linear momentum multiplied by the shortest distance between the trajectory and the reference point.

The alternative to using conservation of angular momentum is to draw the free body diagram for the disk. If the spider exerts no torque on the disk, what is the net torque on the disk?

I'm very confused how the spider still be effecting the momentum of the disk when it is not on the disk?

 

[rcgldr]

Then I removed the spiders moment of inertia.

That's part of the issue. Since the spider slips off the disk without exerting any torque on the disk, then the spiders component of angular momentum doesn't change.

 

[rcgldr]

I'm very confused how the spider still be effecting the momentum of the disk when it is not on the disk?

In this case it never effects the angular momentum of the disk. It doesn't generate any torque on the disk when it slides off. It continues to move in a straight line that is tangent to the outer edge of the disk.

 

[Orodruin]

I'm very confused how the spider still be effecting the momentum of the disk when it is not on the disk?

It is not doing anyhing with the disk momentum by definition (no torque), but it is the total momentum that is conserved (that of the spider+disk system).

 

[BrainMan]

That's part of the issue. Since the spider slips off the disk without exerting any torque on the disk, then the spiders component of angular momentum doesn't change.

So how do I solve this problem?

 

[rcgldr]

So how do I solve this problem?

Well if the spider's component of angular momentum doesn't change when the spider slides off the disk, and knowing that the total angular momentum of the disk + spider system doesn't change, and also knowing that the spider did not generate any torque on the disk when it slid off, then is there any effect on the spinning disk?

 

[Orodruin]

One of two ways:

1) find the angular momentum of the spider *after* falling off with respect to the disk center
2) the free body diagram - what torques act on the disk?

 

[BrainMan]

Well if the spider's component of angular momentum doesn't change when the spider slides off the disk, and knowing that the total angular momentum of the disk + spider system doesn't change, and also knowing that the spider did not generate any torque on the disk when it slid off, then is there any effect on the spinning disk?

No. Should the speed of the disk remain the same?

 

[Orodruin]

Yes.

 

[BrainMan]

Yes.

That's so cool! It's expected that the speed of the disk should increase. So if I jumped off a merry go round while it was moving the speed of the mery go round should stay the same?

 

[Orodruin]

Unless you push off the merry-go-round in the tangential direction (ie, generate torque), yes.

 

[BrainMan]

Unless you push off the merry-go-round in the tangential direction (ie, generate torque), yes.

cool

 

[rcgldr]

Note that the angular momentum of the spider and disk would be the same if the spider didn't start off on the disk, but instead was always traveling in a straight line tangent to the edge of the disk, initially approaching the edge of the disk, then reaching the edge of the disk where it's velocity and the velocity of the point on the disk touched by the spider would be the same, and then the spider continuing in a straight line away from the disk.

 

[BrainMan]

Note that the angular momentum of the spider and disk would be the same if the spider didn't start off on the disk, but instead was always traveling in a straight line tangent to the edge of the disk, initially approaching the edge of the disk, then reaching the edge of the disk where it's velocity and the velocity of the point on the disk touched by the spider would be the same, and then the spider continuing in a straight line away from the disk.

That's very interesting