Torque/acceleration with masses on pulley

  • Thread starter Thread starter vtech
  • Start date Start date
  • Tags Tags
    Pulley
AI Thread Summary
The discussion revolves around a physics problem involving two masses connected by a cord over a pulley, requiring the calculation of acceleration and tensions in the cord. The user has derived three equations based on the system's dynamics but is confused about the signs of the acceleration and the resulting calculations. They initially obtained a negative acceleration value, which led to uncertainty about the direction of motion. A suggestion is made to establish a consistent sign convention to clarify the relationship between the masses and their respective accelerations. This approach will help in correctly solving for the acceleration and tensions in the system.
vtech
Messages
12
Reaction score
0
Wer are given two objects of mass m1 = .20 kg and m2 = .67 kg, which are connected by a massless cord that is wrapped around a uniform disk of mass M = 0.50 kg and radius R = 0.12 m. The disk can rotate without friction about a fixed horizontal axis through its center; the cord cannot slip on the disk. The system is released from rest.

(a) Find the magnitude of the acceleration of the blocks____m/s2
(b) Find the tension T1 in the cord at the left____ N
(c) Find the tension T2 in the cord at the right____N
--------------------------------------------------------------------------

well so far i have three equations - clockwise motion
1) T2*R - T1*R = I*alpha = net torque where alpha = a/R for pulley
2) T2 - m2*g = m2*a for mass on pulley
3) -T1 + m1*g = m1*a for mass on pulley

Then I solve 2) and 3) for respective tensions T2, T1 and substitute into 1) and solve for "a" I mean, am I close here or no? Cuz I don't want to write all of my solving out if I'm at the wrong starting point anyway...kinda messy... but I'm getting an answer of

a = - 7.429m/s2 but I get it as wrong. Don't have actual answer. Negative? So I'm guessing goes in opposite direction...?

Without correct a can't solve for T1 & T2 ... for answer a) since magnitude is being asked, even if acceleration is negative answer is +ve value.

I'd really appreciate some input on this questions! Thanks in advance!
 

Attachments

  • hrw7_10-53.gif
    hrw7_10-53.gif
    3.8 KB · Views: 694
Physics news on Phys.org
can anyone show me how to include a figure/graph in my question area instead of putting in as an attachment??
 
can anyone help on this one?

surely there is someone that can help out with this problem? i understand the concept of the torque acting with the forces, but i am unsure as to how i can relate this with the tensions! help anyone? i would really appreciate it!
 
sign problems

vtech said:
well so far i have three equations - clockwise motion
1) T2*R - T1*R = I*alpha = net torque where alpha = a/R for pulley
2) T2 - m2*g = m2*a for mass on pulley
3) -T1 + m1*g = m1*a for mass on pulley
You are mixing up the signs for the acceleration. You need to use a consistent sign convention. If you take up as positive, then the acceleration of m1 is up (+a), but the acceleration of m2 is down (-a). Rewrite equations 2 and 3.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top