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Rotational Kinematics Help (constant angular acceleration)

  1. Nov 28, 2003 #1
    The Tub of a washer goes into a spin cycle, starting from rest and gaining angular speed steadily for 8 s, when it is turning at 5 rev/s. At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub smoothly slows to rest in 12s. Through how many revolutions does the Tub turn while it is in motion?

    Ok, I am going to do this problem in 2 stages: One for Wi =0 and Wf=5 rev/s and the second for Wi=5rev/s and W[size=.5]f[/size]=0

    First stage:

    Wf=5 rev/s

    Wf= (5 rev/s)(2(pi)rad/rev)
    Wf= 10(pi) rad or 31.4 rad

    A = (Wf-Wi)/T
    A = 31.4/8
    A = 3.9 rad/(s^2)

    Of= 0 + (10)(pi)(8) + .5(3.9)(8)^2
    Of= 80(pi) + 40(pi)
    Of= 377 rad (I am going to leave it in this form because I am going to have to plug it into stage two.

    Stage two:

    Wf= 0
    Wi= 5 rev/s or 10(pi) rad/s
    Oi= 377 rad (found in stage one)
    T=12 s


    A = (Wf-Wi)/T
    A = (0-31.4)/12
    A = -2.6 rad/s^2

    Of= 377 + 31.4(12) + .5(-2.6)(12)^2
    Of= 377 + 376.8 - 187.2
    Of=566.8 rad

    Now we need to convert to revolutions:
    566.8 rad(57.3/rad)= 32477.64 degrees

    32477.64/(360/rev) = 90. 2 rev

    Ok here is what I do not get.

    My revolutions for the washer when its acceleration is increasing in stage one is slower than the revolutions when the acceleration is decreasing in stage two. I think I am correct but it doesn't make sense.
    Last edited: Nov 28, 2003
  2. jcsd
  3. Nov 28, 2003 #2
    I didn't check your calculations, but you would expect that the second stage will undergo more revolutions than the first stage, because the deceleration is smaller in magnitude than the acceleration.

    Think about it: suppose you have a spinning flywheel, that is very gradually losing speed due to friction. It could make many revolutions over a long period of time before it finally comes to a stop. On the other hand, I can spin it up to that same speed much faster, just by applying a large torque.
  4. Nov 29, 2003 #3
    Ok, I redid this problem a couple of times and came up with a different answer. I forgot to set wi to zero in the equation Of = 0i + Wi(T)+.5(A)(T^2) in stage one. So Of=125.66 rad which changes the final answer to 50.2 revolutions
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