Non ideal pulley with two masses

  • Thread starter Thread starter Voitonic
  • Start date Start date
  • Tags Tags
    Pulley Two masses
AI Thread Summary
The discussion focuses on solving a physics problem involving two masses connected by a rope over a non-ideal pulley with friction. Key parameters include the masses of the boxes, the radius and mass of the pulley, and the torque due to friction. The equations of motion for each mass and the pulley are established, emphasizing the relationship between tension, acceleration, and torque. The participant acknowledges the need for better organization in approaching such problems. Overall, the problem is deemed straightforward once the equations are properly set up.
Voitonic
Messages
2
Reaction score
0
Hi,

So I have a problem with a question upon reviewing a past physics final.

Two boxes are connected by a massless, unstretchable rope that passes over a non-ideal pulley. The radius and mass of the pulley are R = 0.100m and M = 1.75 kg. As the pulley turns, friction at the axle exerts a constant torque of magnitude of 0.480 Nm. The moment of inertia of the pulley is 1/2MR2. Box B1 has a mass m1=3.98 kg and its initial location is 0.850 m above the floor. Box B2 has a mass of m2= 2.01 kg and its initial location is at the floor. After the boxes are released from rest, B1 descends to the floor, while B2 is lifted.

I'm trying to solve for the tension between each section of rope and the pulley as well as finding the acceleration of each box.

Any help would be appreciated,
Thanks!
 
Physics news on Phys.org
It's rather straight forward.

m1, m2 - the two masses.
a1, a2 - corresponding accelerations.
T1, T2 - corresponding tensions.
α - angular acceleration of the pulley.
I, R - relevant properties of the pulley.
τf - friction torque.

Consider m1 descending, m2 rising as positive direction.

m_1 a_1 = m_1 g - T_1

m_2 a_2 = T_2 - m_2 g

I \alpha = T_1 R - T_2 R - \tau_f

a_1 = a_2 = \alpha R

Rest is just algebra.

P.S. When solving a problem like this, just write out equations of motion for every degree of freedom you have, then add any available constraints.
 
Thanks for the help.

You're right, after looking at it again the question is pretty straight forward. I guess I just need to learn how to organize my thoughts better.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
3
Views
2K
Replies
22
Views
6K
Replies
8
Views
4K
Replies
2
Views
2K
Replies
8
Views
3K
Replies
6
Views
2K
Replies
7
Views
3K
Back
Top