Angular Momentum of Disk Question

Click For Summary

Homework Help Overview

The problem involves calculating the angular momentum of a disk with a lump of clay impacting it. The disk is stationary and has a specified mass and radius, while the clay falls and sticks to the disk's edge. The scenario includes initial conditions such as the disk's angular speed and the clay's speed before impact.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between rotational and translational angular momentum, with some questioning the validity of using the equation Ltrans=mrv due to the non-perpendicular vectors involved. Others suggest using the cross product to find angular momentum.

Discussion Status

Participants are exploring the implications of the conservation of angular momentum and how to calculate the angular velocity after the impact. There is a recognition of the need to consider both the disk and the clay's contributions to the total angular momentum before and after the impact.

Contextual Notes

Some participants express confusion regarding the calculation of the cross product and the nature of momentum as a vector. There is an ongoing discussion about the assumptions related to angular momentum conservation in the context of the clay sticking to the disk.

smedearis
Messages
11
Reaction score
0

Homework Statement


A stationary uniform-density disk of radius 0.8m is mounted in the vertical plane. The axle is held up by supports that are not shown and is frictionless. The disk has a mass of 3.6 kg. A lump of clay with mass 0.3 kg falls and sticks to the outer edge at <-0.48, 0.64, 0>m. Before impact it has a speed 8 m/s and the disk is rotating clockwise with angular speed 0.4 radians/second. Just before impact, what is the magnitude of the angular momentum about the center. System=disk+clay, about the center (<0,0,0>)


Homework Equations


I know that it has something to do with:
Ltot=Lrot+Ltrans
Lrot=Iw, where I=.5*mR^2 and w is given? (the disk)
Ltrans=mrv, where the r here equals where it sticks? (the clay)


The Attempt at a Solution


I tried plugging the numbers given into this equation, but the answer isn't right.. (the answer is 0.6912 kg*m^2/s) Angular momentum is a vector, so i assumed we'd use the vector position given. What am i doing wrong?
 
Physics news on Phys.org
Ltrans=mrv is only true when the vectors r and v are perpendicular. Your vectors are not perpendicular. You need the more general vector equation involving the cross product L = r x p
 
how to do i find a cross product of something when the position is a vector, but the momentum is only magnitude?
 
smedearis said:
how to do i find a cross product of something when the position is a vector, but the momentum is only magnitude?

The momentum is also a vector. The lump of clay falls onto the disk.
 
That clears a lot up. But now, my question is after impact, what is the angular velocity (is it w - omega)?

Is there conservation of angular momentum? Do we still have translational angular momentum since the clay has been stuck to the wheel?
 
smedearis said:
That clears a lot up. But now, my question is after impact, what is the angular velocity (is it w - omega)?

Is there conservation of angular momentum? Do we still have translational angular momentum since the clay has been stuck to the wheel?

There are two ways to approach the analysis. Either way rests on the principle of conservation of angular momentum. Before impact, you have two separate angular momenta, one for the disk and one for the clay that must be added to find the total system angular momentum. After that you can treat them as separate objects with a relationship between the v of the clay and the ω of the disk, or you can find the moment of inertia of the disk-clay combination with angular velocity ω. Either way, the total angular momentum after impact must be the same as before impact.
 

Similar threads

Replies
26
Views
2K
Replies
5
Views
2K
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
335
Views
17K
Replies
30
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
3K