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Math Help - Atwood Machine Problem (Rotational Dynamics)

  1. #1
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    Atwood Machine Problem (Rotational Dynamics)

    Two masses m1 and m2 are hung over a pulley, forming an Atwood Machine.

    m1 = 0.400 kg
    m2 = 0.410 kg
    R, radius of pulley = 0.10 m
    moment of inertia I of pulley = 5E-3 kgm^2.

    a) When the distance between the two masses is 0.50 m, how fast is each moving?
    b) At that time, how fast would each mass move if the pulley was considered massless?

    Well for part a I thought that:
    (m1)a = T - (m1)g
    (m2)a = (m2)g - T

    and just solve for a. Then from a you can find the speed when the mass has moved 0.25 m.

    However I would have done the exact same thing for part b, and I didn't use moment of inertia here at all. So I'm sure this is wrong... any help?
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  2. #2
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    Quote Originally Posted by apple123 View Post
    Two masses m1 and m2 are hung over a pulley, forming an Atwood Machine.

    m1 = 0.400 kg
    m2 = 0.410 kg
    R, radius of pulley = 0.10 m
    moment of inertia I of pulley = 5E-3 kgm^2.

    a) When the distance between the two masses is 0.50 m, how fast is each moving?
    b) At that time, how fast would each mass move if the pulley was considered massless?

    Well for part a I thought that:
    (m1)a = T - (m1)g
    (m2)a = (m2)g - T

    and just solve for a. Then from a you can find the speed when the mass has moved 0.25 m.

    However I would have done the exact same thing for part b, and I didn't use moment of inertia here at all. So I'm sure this is wrong... any help?
    Are we to assume the pulley is both massles and frictionless? Mass 2 is heavier than mass 1. It is convinient in situations involving objects connected by cords to take the direction of motion as positive direction. In this case, for m_1 we take up positive, and down positive for m_2. (If we do this the acceleration will be positive for each mass).

    Writing \sum F_y = ma_y for each mass in turn we have

    F_T - (0.4)(9.81)N = (0.4)a

    And for the second mass, (0.41)(9.81)N-F_T = (0.41)a

    If we add the two equations, the unknown F_T drops out giving

    (0.41-0.4) (9.81) =0.81 a

    For which a=0.12 m/s^2

    Maybe this is how the problem needs to be approached.
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  3. #3
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    Quote Originally Posted by apple123 View Post
    Two masses m1 and m2 are hung over a pulley, forming an Atwood Machine.

    m1 = 0.400 kg
    m2 = 0.410 kg
    R, radius of pulley = 0.10 m
    moment of inertia I of pulley = 5E-3 kgm^2.

    a) When the distance between the two masses is 0.50 m, how fast is each moving?
    b) At that time, how fast would each mass move if the pulley was considered massless?

    Well for part a I thought that:
    (m1)a = T - (m1)g
    (m2)a = (m2)g - T

    and just solve for a. Then from a you can find the speed when the mass has moved 0.25 m.

    However I would have done the exact same thing for part b, and I didn't use moment of inertia here at all. So I'm sure this is wrong... any help?
    Let T_1 = tension in the cord on the m_1 side of the pulley

    T_2 = tension in the cord on the m_2 side of the pulley

    scalar equations for the net force on each side ...

    m_2g - T_2 = m_2a

    T_1 - m_1g = m_1a

    Newton's 2nd law for rotation ...

    T_2 R - T_1 R = I \alpha = I \frac{a}{R}


    solve the system for "a"
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