Acceleration of a mass connected to 2 different disks

  • Thread starter Thread starter cashmoney805
  • Start date Start date
  • Tags Tags
    Acceleration Mass
AI Thread Summary
The discussion focuses on calculating the acceleration of a mass connected to two disks in a system where the mass is 0.250 kg. The equations of motion involve the sum of torques and forces, with tensions canceling out when combined. Participants highlight the importance of recognizing that the two masses have different accelerations and relate them to angular acceleration. The final equation derived incorporates gravitational forces and the moments of inertia of the disks, allowing for the calculation of angular acceleration. Clarification on the notation used for different masses is also emphasized.
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: 429
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.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top