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Homework Help: Find angular velocity from constant linear velocity

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A rotating disk of 2.50m in diameter serves tu connect a counter weight to a mass through a massless rope. The rope does not slip on the disk so, there is no friction on the rim. What angular velocity in rpm must the disk turn to raise the elevator at 25.0 m/s?

    2. Relevant equations

    3. The attempt at a solution
    I am confused here.
    since the linear velocity of the rope is constant, its linear acceleration should be zero, and every point on the rope should have the same velocity. Becasue the radius of the rim is constant, the angular velocity should be zero and angular acceleration as well.
    because the tangential acceleration is zero, then the acceleration vector should have a radial component only: a(rad)=omega^2*r
    I could factor out omega from the equation and obtain:
    omega=square root of a(rad)/r.. but I dont have the radial component of the acceleration....
    should I just use the equation v=omega*r, and factor out omega from it?

    Just wondering, if I were given the tension on the rope, caused by the weight of the mass connected to it, or a given displacement with this constant pulling force by the rope on the mass, would these factors just be ignored by the fact that the bofy is moving with a constant velocity, thus, a=0, and the net force on the object is zero as well?
    please help
  2. jcsd
  3. Nov 14, 2009 #2
    I believe that you should find the circumference of the wheel [ PI * diameter ].
    Divide the circumference by the speed the lift is expected to travel (25 m/s); this will give you the period (time for one full rotation of the wheel) in seconds.

    Work out the angular frequency by finding the reciprocal of the period [ 1 / period ], but to get your answer in RPM convert the period into minutes rather than seconds by dividing it by 60.

    This should give you the answer expected in the question; the required revolutions per minute to raise/lower the lift at 25 m/s.

    Please do correct me if I'm wrong, anyone.
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