# Angular Momentum; rod & disk inelastic collision

• tennisgirl92
In summary, the conversation discusses a problem involving a plastic rod attached to a table with a disk sliding towards the other end with an initial velocity, and then sticking to the rod upon collision. The questions focus on finding the angular velocity and kinetic energy before and after the collision. The conversation also includes a discussion on conserving angular momentum and using the moment of inertia of a rod about its end. Finally, the conversation concludes with the correct set up for finding the kinetic energy after the collision.
tennisgirl92

## Homework Statement

The figure shows an overhead view of a 2.50-kg plastic rod of length 1.20 m on a table. One end of the rod is attached to the table, and the rod is free to pivot about this point without friction. A disk of mass 39.0 g slides toward the opposite end of the rod with an initial velocity of 33.0 m/s. The disk strikes the rod and sticks to it. After the collision, the rod rotates about the pivot point.

(a) What is the angular velocity of the two after the collision?

What is the angular momentum of an object moving with a linear velocity, about a given point? rad/s

(b) What is the kinetic energy before and after the collision?
KEi
= J
KEf
= J

## Homework Equations

L=Iw
(Iw+Iw)I=(iw+iw)f
w=v/r
KErot=1/2Iw^2
Irod=ML^2/3
Idisk=MR^2/2

## The Attempt at a Solution

I believe the set up would look something like this:
(Irodxwrod)i+(Idisk x wdisk)I=(Idisk+Irod)wf
The initial speed of the rod is zero, canceling that out.
MR^2/2 x V/R=(MR^2/2 + ML^2/3)wf
The problem I am having is we are not given the radius of the disk. Is it possible I am missing that the radius cancels out?

And again, with the KE, we need the radius in order to solve for angular momentum in the equation.

Thank you for any help!

Assume that the disk is a point mass. It appears you are conserving angular momentum. About what point is angular momentum conserved?

Ok, so if this is a point mass I=MR^2, where r would be the distance from mass to the axis. So in this case, 1.2? Would this be applied on both the initial and final sides of the equation? I can see how it would be a point mass when attached to the pole, but why when it is solo?

And to answer what point angular momentum conserved, would this be at the pivot point?

I apologize but I do not know what LaTex is!

Thank you so much for your help.

tennisgirl92 said:
Ok, so if this is a point mass I=MR^2, where r would be the distance from mass to the axis. So in this case, 1.2?
Yes, but call the length of the rod L and put in numbers at the very end. Also, the mass of the disk is not equal to the mass of the rod, so use different symbols for the two.
tennisgirl92 said:
I can see how it would be a point mass when attached to the pole, but why when it is solo?
Because, as you correctly pointed out, you are not given the radius of the disk, therefore you are to assume it's a point mass. If it were an extended mass, you would have to worry about exactly where on the rod its center of mass is stuck.
tennisgirl92 said:
And to answer what point angular momentum conserved, would this be at the pivot point?
Yes. That's also the point about which you calculate all moments of inertia.
tennisgirl92 said:
apologize but I do not know what LaTex is!
Look all the way to the left of the POST REPLY button. Click LaTeX to the right of the question mark. It is a powerful, easy-to-learn equation writer that you should invest some time in.

So here is my final set up.

MdiskR^2 x V/R=(MdiskR^2 + MrodL^2)w
which can simplify to
Mdisk xR x V=(Mdisk R^2 + Mrod L^2) w

I tried to make it clear-I will check out the LaTex when I have a little more time.

would this be an accurate set up? I obtained .4212 rad/sec, which seems small.

What is the moment of inertia of a rod about its end?

ah I forgot! ML^2/3

Mdisk R^2 x V/R=(MdiskR^2 +Mrod L^2/3) w

I now obtained 1.226 rad/sec, which seems more acceptable.

I trust you used R = L when you put in the numbers.

Yes! And this was the correct answer. THANK YOU for your help!

And one more question if you don't mind. For the KE after the collision, I understand this is now rotational kinetic energy which is 1/2Iw^2. We would use the w found in part 1, but for the moment of inertia would it be the combined values for the disk and rod? So:

KEf=1/2 (Mdisk R^2 + [Mrod L^2/3]) w^2

Yes.

Thank you! You are most patient and helpful. I really appreciate your time.

## 1. What is angular momentum?

Angular momentum is a measure of the rotational motion of an object. It is defined as the product of an object's moment of inertia and its angular velocity.

## 2. How is angular momentum conserved in a collision?

In a collision between a rod and a disk, angular momentum is conserved as long as there are no external torques acting on the system. This means that the total angular momentum of the rod and disk before the collision is equal to the total angular momentum after the collision.

## 3. What is an inelastic collision?

An inelastic collision is a type of collision in which kinetic energy is not conserved. This means that some of the initial kinetic energy is converted into other forms, such as heat or sound.

## 4. How does a rod and disk inelastic collision differ from an elastic collision?

In an elastic collision, kinetic energy is conserved, meaning that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. In an inelastic collision, some of the initial kinetic energy is lost, resulting in a decrease in the total kinetic energy of the system.

## 5. What factors can affect the outcome of a rod and disk inelastic collision?

The outcome of a rod and disk inelastic collision can be affected by factors such as the mass and velocity of the objects, the angle and point of impact, and the coefficient of restitution, which is a measure of the objects' elasticity. Other external forces, such as friction or air resistance, can also influence the collision.

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