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Rotation with a Pulley and Masses

  1. Mar 16, 2005 #1
    one block has a mass M = 500 g, the other has mass m = 460 g, and the pulley, which is mounted in horizontal frictionless bearings, has a radius of 5.00 cm. When released from rest, the heavier block falls 85.5 cm in 2.47 s (without the cord slipping on the pulley).

    First I did conversions:
    M = .5kg, m = .46 kg, r = .05 m
    fall = .855m in 2.47 s

    I am unable to do part b and c right, so I am unable to go further.
    Here is my work:

    .855m = 1/2 a (2.47^2) a = .28 m/s^2

    however, It asks for 2 tensions. I never learned how to seperate them like that, especially with a pulley. I did:

    9.8 m/s^2 * .5kg = 4.9 N
    9.8 m/s^2 * .46kg = 4.508 N

    Tension of Cord = 4.9N + 4.508N. Obviously this isn't right, where do I go from here, or did I miss something?


    (a) What is the magnitude of the block's acceleration?
    .28 m/s2
    (b) What is the tension in the part of the cord that supports the heavier block?
    N
    (c) What is the tension in the part of the cord that supports the lighter block?
    N
    (d) What is the magnitude of the pulley's angular acceleration?
    rad/s2
    (e) What is its rotational inertia?
    kg · m2
     
    Last edited: Mar 16, 2005
  2. jcsd
  3. Mar 16, 2005 #2
    I got it all, thanks though
     
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