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Net angular acceleration of a two wheel system

  1. Jun 15, 2015 #1
    1. The problem statement, all variables and given/known data

    The problem

    QCc0zj6.jpg

    Direct imgur link to problem: http://i.imgur.com/QCc0zj6.jpg

    2. Relevant equations

    [tex]\alpha = \frac{\tau_{net}}{I}[/tex]

    [tex] I = \frac{1}{2}MR^2[/tex]

    [tex]\tau = rF = rmg[/tex]

    3. The attempt at a solution

    I have figured that the moment of inertia of the entire system must be used for both torques pulling on the system, so the one I is

    [tex]I = \frac{1}{2}(5.0\times (0.05)^2 + 20\times (0.10)^2)[/tex]

    So shouldn't the net angular acceleration just be the net torque of the two competing weights divided by the moment of inertia of the entire system?

    [tex]\frac{0.10\times 2.5\times 9.8 + (- 0.05 \times 4.0 \times 9.8)}{\frac{1}{2}(5.0\times (0.05)^2 + 20\times (0.10)^2)} \approx 4.6 \frac{rad}{s^2} [/tex]

    [The second torque trying to turn CW, so therefore is negative torque.]

    However, the answer from the back of the text says [itex]3.5 \frac{rad}{s^2}[/itex]

    What am I doing wrong? Do I need to consider this is terms of the tensions in the ropes instead? When considering the torque applied to the system by one hanging weight, does the other hanging weight count towards the moment of the inertia of the system?
     
  2. jcsd
  3. Jun 15, 2015 #2

    Doc Al

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    Yes. It is the rope tension that acts on the cylinders, not the weight of the hanging masses.

    Hint: Apply Newton's 2nd law three times.
     
  4. Jun 15, 2015 #3
    Are the tensions in the ropes equal? The masses have different accelerations, though one is twice the other.
     
  5. Jun 15, 2015 #4
    Since the angular acceleration for the entire system is the same throughout

    [tex]\frac{\tau_1}{I} = \frac{\tau_2}{I} ?[/tex]

    [tex]\frac{r_1(m_1a_1+m_1g)}{I} = \frac{r_2(-m_2 \frac{1}{2}a_1+m_2g)}{I}[/tex]

    And now I just solve for a, and then solve for the angular acceleration knowing the radii...?
     
    Last edited: Jun 15, 2015
  6. Jun 15, 2015 #5
    I am really struggling with this one.

    I can see that the angular acceleration will be the same for the whole system, and so the tangential acceleration of the weight on the inner cylinder will be -1/2 the tangential acceleration of the weight on the outer cylinder.

    I know that the net torque divided by the moment of inertia of the entire system will get me the angular acceleration. I am just struggling to resolve how the forces and torque are interacting to produce the net torque.

    So, net torque will be +ve torque from 2.5 kg weight - 0.50 Nm from friction - torque from 4.0 kg weight. I am assuming that the CCW is the positive direction.

    Ok, so the force creating the torque will be the tension in the rope minus the weight on the end of the rope.

    I get [itex]T_1 - 24.5 = 2.5a_1[/itex] and [itex]T_2 - 39.2 = 4.0a_2[/itex], but [itex]a_2 = -\frac{1}{2}a_1[/itex] so [itex]T_2 - 39.2 = -2.0a_1[/itex]

    But I don't think the two tensions in the ropes are the same, they are different ropes, and the masses that they are supporting are undergoing differing linear accelerations.

    I've tried [itex]T_2 = \frac{\tau_{net}}{r_2}[/itex] is this correct? Is the second tension raising the other weight the result of the overall torque divided by the radius of that particular cylinder. But subbing back in to other equations gives an acceleration for the first dropping weight greater than gravity.

    I AM DYING HERE, someone please help me!!!
     
  7. Jun 15, 2015 #6

    SammyS

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    The tensions are not equal.

    Yes, the acceleration of one mass is twice the other. Which has the greater acceleration ?

    True, there is only one angular acceleration for the combined cylinders. However, it's the difference (or the sum) of the torques that produces the angular acceleration. Those quantities are not equal.
     
  8. Jun 16, 2015 #7
    For the total moment of inertia (I), add the two seperate cylinder moment of inertias together:
    I = ( ½ * 5 * 0.05 ² ) + ( ½ * 20 * 0.1 ² )

    Calculate the net torque (T) provided by the hanging masses.
    Calculate the angular acceleration (in rad/sec) from ( as you said ) α = T / I
     
  9. Jun 16, 2015 #8

    Doc Al

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    No.

    Exactly. You'll need to relate the accelerations of the masses with the angular acceleration of the cylinder.
     
  10. Jun 16, 2015 #9

    Doc Al

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    To make things easier, I recommend that you write equations for each mass and the cylinder, expressing Newton's 2nd law. That will give you three equations.

    Using the relationship between the linear accelerations of the masses and the angular acceleration of the cylinder, you should be able to solve for the angular acceleration. (And you can solve for the other unknowns, if you like, such as the tension in the rope and the accelerations of the masses.)

    Express everything symbolically. Don't plug in numbers until the very end.
     
  11. Jun 16, 2015 #10
    I did that it got the wrong answer...maybe I'm putting the friction factor in in the wrong way.

    edit: Wait, there is no friction factor...great, that was a part of a different problem that got mixed in...
     
  12. Jun 16, 2015 #11

    haruspex

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    How are you proposing to find the net torque? The masses accelerate too.
    You can handle that by including the masses in the MoI as well.
     
  13. Jun 17, 2015 #12
    My previous post was wrong, it did not not account for the two hanging masses in the acceleration.
    Apologies.
    This problem can be solved by moving the cylinder(s) at an arbitrary constant rate (say 10 rad/sec) and calculating the KE of each part at that rate, then you can give each part a relative moi value by dividing its KE value by the KE value of the large cylinder and multiplying by the large cylinders moi.
     
  14. Jun 17, 2015 #13

    haruspex

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    As I wrote, your first method can be rescued by representing the hanging masses as moments of inertia. Can you see how to do that?
     
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