Acceleration of a mass connected to 2 different disks

  • Thread starter Thread starter cashmoney805
  • Start date Start date
  • Tags Tags
    Acceleration Mass
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving the acceleration of a mass connected to two different disks. The original poster presents a scenario where a mass of 0.250 kg is released from rest, and the goal is to determine its acceleration. The problem involves concepts of torque, forces, and angular acceleration.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationships between the forces acting on the masses and the resulting accelerations. There is an exploration of how to express angular acceleration in terms of the different radii associated with the disks. Some participants suggest combining equations to eliminate tensions and relate linear and angular quantities.

Discussion Status

There is an ongoing exploration of the relationships between the various accelerations and forces in the system. Some participants have offered guidance on how to approach the problem, particularly regarding the summation of torques and forces. However, there is no explicit consensus on the final form of the equations or the next steps to take.

Contextual Notes

Participants note the complexity introduced by having different accelerations for the masses and the need to clarify the definitions of the variables used in the equations. There is also an acknowledgment of the potential confusion arising from using similar symbols for different masses in the equations.

cashmoney805
Messages
50
Reaction score
0

Homework Statement


The attachment gives the setup, but I can do the first part (a). The second question (not in the attachment) is "Now consider the same system as part (a) except M = .250 kg. After the system is released from rest, what will the acceleration of the .250kg mass be?


Homework Equations


sum of torques = T1*(r1) - T2*(r2) = I*angular acceleration
sum of forces on M= Mg-T1=Ma
sum of forces on m = T2-mg=ma


The Attempt at a Solution


I multiplied the two force equation by the r of the disk that each mass is attached to, then added all three equations together. The tensions cancel, giving:
RMg-rmg = RMa + rma + I*ang acc

I don't know what to do from here though. Maybe I = I big disk + I small disk. I don't know what ang acc is though. I would usually set it to a/r, but there are two different r's in this case. Any help is greatly appreciated!
 

Attachments

  • physics.jpg
    physics.jpg
    22.4 KB · Views: 439
Physics news on Phys.org
cashmoney805 said:

Homework Equations


sum of torques = T1*(r1) - T2*(r2) = I*angular acceleration
Good.
sum of forces on M= Mg-T1=Ma
sum of forces on m = T2-mg=ma
Careful! The masses have different accelerations. How do the accelerations relate?

Maybe I = I big disk + I small disk.
Yep!

I don't know what ang acc is though. I would usually set it to a/r, but there are two different r's in this case.
Good--there are two different accelerations, also. :wink:
 
Doc Al said:
Good.

Careful! The masses have different accelerations. How do the accelerations relate?
Let's see.

Is it Ma = MR(ang acc)
ma = mr(ang acc)?

If so, does the final equation, after adding all three, look like:

g(RM + rm) = ang acc (R(^2)M + r(^2)m + .5MR(^2) + .5mr(^2) ?

If so, you can calculate ang acceleration. Then do you just multiply that by R to get a of block 1?

edit: actually the final equation should be (I think)
g(RM - rm) = ang acc (R(^2)M + r(^2)m + .5MR(^2) + .5mr(^2)

oh boy, edit again. the M and m next to .5 should be masses of the disks
 
Last edited:
cashmoney805 said:
edit: actually the final equation should be (I think)
g(RM - rm) = ang acc (R(^2)M + r(^2)m + .5MR(^2) + .5mr(^2)
That looks OK. (Except for using the same letters to represent different masses!)
oh boy, edit again. the M and m next to .5 should be masses of the disks
Yep.
 

Similar threads

Replies
1
Views
1K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
8
Views
15K
Replies
11
Views
1K
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 26 ·
Replies
26
Views
3K