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Rotational Equilibrium and Dynamics Problem

  1. Dec 4, 2006 #1
    1. The problem statement, all variables and given/known data

    A hanging 4 kg mass is connected by a weightless cord on a pulley to a 3 kg mass which rests on a smooth surface. The pulley has frictionless axle and I=.5kgm^2 and r=.3m. The cord does not slip on the pulley. Find the acceleration of the two masses and the tensions of the strings (one between pulley and 3kg mass, and the other between pulley and 4 kg mass)

    2. Relevant equations

    Rotational kinematics equations...not sure which ones to use....

    3. The attempt at a solution

    I figured that downward force would be 4g. I thought about dividing that by 7+I=7.5 to get acceleration but that doesn't work. Correct acceleration is 3.12m/s^2 and tensions are 26.7 and 9.37 N.


  2. jcsd
  3. Dec 5, 2006 #2


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    Homework Helper

    The whole of the system will accelerate at the same rate - assume a.
    Start with the equations for m3 and m4. Use T3 and T4 for the tensions. What will the equations for the two masses be?
  4. Dec 5, 2006 #3
    andrevh, how does that do anything? So what, I use m3 and m4 and T3 and T4 instead of m1 and m2 and T1 and T2. I could've told you that when I was 7 for crying out loud. I really do appreciate any efforts at helping, but changing the names of variables is quite useless....or am I missing something? tell me if i am.

    In any case, I solved the problem by finding the pulleys mass to be M=I/R^2, acceleration to be 4g and F=ma.

    Can anyone show how to solve this problem by converting linear acceleration into angular and then use rotational dynamics (The above was solved linearly)? I need to know both ways for a test.


  5. Dec 5, 2006 #4

    Doc Al

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    I think andrevh was just suggesting that you name the variables something and set up your equations. (M3 for the 3kg mass; M4 for the 4kg mass, etc. But you are correct, the variable names don't matter.)

    That would be an incorrect conclusion as to the mass of the pulley, but you don't need the mass of the pulley. How did you get the acceleration to be 4g?

    Not sure what you've solved so far, but do this: Identify the forces acting on each mass and the pulley. Apply Newton's 2nd law (F = ma or Torque = I alpha, as appropriate) to each, giving you three equations. Solve those three equations together to get the acceleration.

    Be sure to use the suggestion that the cord moves with a uniform acceleration of magnitude "a" when you write your equations.
  6. Dec 5, 2006 #5
    Why does the mass of the pulley not matter? The pulley is also being accelerated at the same rate as the rest of the system, so if one wants to apply F=ma, then one needs the mass of the pulley in addition to the mass of the two masses.

    My bad about the acceleration = 4g. I meant that the downward force = 4g. I'll try to be careful.
  7. Dec 5, 2006 #6

    Doc Al

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    Unless I seriously misunderstand the problem, the center of the pulley does not accelerate--it's fixed in position (attached to the corner of a table, perhaps). Of course it is free to rotate, so its rotational inertia matters. No need to calculate the pulley's mass, since you are given its rotational inertia.

    OK, the downward force on the 4kg mass--its weight--equals 4g (using standard units). But what's the downward force on the pulley?

    Again I suggest analyzing the forces on each object separately.
  8. Dec 6, 2006 #7


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    The difference in tensions at the two masses is causes precisely by the angular acceleration of the pulley. It has the same effect as friction (due to its rotational inertia) - it will decrease the input tension and increase the output tension in the string.

    You can relate the acceleration of the system to the angular acceleration of the pulley via

    [tex]a = r \alpha[/tex]

    that is the tangential acceleration of the pulley will be the same as the acceleration of the system due to the string running over the pulley. Which means that the acceleration of the system is influenced by the angular acceleration of the pulley - that is you cannot solve for the acceleration unless you include the angular acceleration of the pulley in the maths.

    In order for me to try and help you we need to be on the same page - I have no idea how much you know and vice versa. Labelling the tensions and masses with a 3 and 4 makes it immediately clear which mass you are dealing with.

    Also to solve problems one need to work step by step - one wrong step and your answer is meaningless. So one need to build up a train of (sound mathematical and physical) thought. That is what I was trying to do.
    Last edited: Dec 6, 2006
  9. Dec 6, 2006 #8
    You can only use the simplified F=ma for a massless, non rotating pulley. For this, you have to sum up the torques and set them equal to I*alpha. Thats why they give you I and r. You dont need the mass because theyve simplified it by giving I
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