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Homework Help: Rotational torque problem

  1. Jan 28, 2005 #1

    A light string 4 m long is wrapped around a cylindrical spool with a radius of 0.075 m and a mass of .5 kg. A 5 kg mass is then attached to the free end of the string causing the string to unwind from the spool.

    a) What is the angular accelerataion of the spool?

    b) how fast will the spool be rotating after all of the string has unwound?

    I got an answer of 2613.333 for (a), but my teacher claims the answer is wrong. I just want to solve this problem or else it's going to kill me...so if any of you know how, please reply. He said the tension must be solved first. but i have no idea. i though the time would be .9 but seems not.
  2. jcsd
  3. Jan 28, 2005 #2
    Let's say that the string hangs on the right side of the spool, ok ?

    Suppose the y-axis is upward vertically. First we apply Newton's second law for the mass. The acceleration is negative since it moves downward:

    5*a = 5*g - T and T is the tension in the string.

    Then we apply the second law of Newton for the spool with rotational inertia denoted as I (you can look this up) and angular acceleration a' : Ia' = torque on spool

    the torque coming from the string-force (tangential to the spool) is equal to -RT (- because the spool moves along with the clock and R is the radius of the spool).

    now you know that tangential acceleration is connected to the angular acceleration via a = a' * R (a' < 0 because the spool moves clockwise and a also < 0)

    Now we have that Ia' = -RT or Ia/R = -RT or a = -R²T/I

    let's put this in the first equation for the mass

    5 (-R²T/I) = 5 * g - T

    calculate T from this equation

    once you know T, you know a...

    Last edited: Jan 28, 2005
  4. Jan 28, 2005 #3

    Thanks...i'm gonna go solve it now. And sorry for double topics...i didn't realize i was posting a highschool problem in the college section so i decided to post it again. Sorry
  5. Jan 28, 2005 #4
    What is I?

    I'm not sure what I is... is it equal to 1/2(.5).075^2?
  6. Jan 29, 2005 #5
    that is correct !!!

  7. Jan 29, 2005 #6

    If i did it correctly i got a T of 1.96. Is that the right answer?
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