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Homework Help: Rotational Kinetic Energy problem

  1. Jul 12, 2014 #1
    1. The problem statement, all variables and given/known data
    The rotating drum on a clothes washer has a rotational inertia of 1.2 kg*m^2. In the spin cycle, it rotates at 240 rev/min. (a) What length of time is required for a 0.25 hp motor to bring the drum to this rotation rate starting form rest? (b) If the angular acceleration is uniform, how many rotations will the drum make during this time?

    2. Relevant equations
    KEr = 1/2IW^2

    3. The attempt at a solution
    I am stuck on this problem because I do not know how to relate the power output of the motor to the angular acceleration or relate it to the angular velocity. I know that .25 hp means that it is putting out 186.5 j of energy per second but I am unsure of how to use that information.
  2. jcsd
  3. Jul 12, 2014 #2


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    What is the rotational energy at 240rev/min?

    If you know the energy per second being added and you can figure out the final energy, then can you use that to find the time?
  4. Jul 12, 2014 #3

    How do you find the energy added per second?
  5. Jul 12, 2014 #4


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    That is just the power

    (I'm assuming all of the energy goes towards rotational energy, and none of it is lost, because then there would not be enough information)
  6. Jul 16, 2014 #5

    So I just find the energy input in the amount of time and do E= 1/2iw^2 to find the velocity and other things?
  7. Jul 16, 2014 #6


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    What I have said only applies to part A.

    Power = Energy / Time
    The change in energy divided by power gives you the time (in quite the same way as a change in distance divided by speed gives you the time).

    For part B you'll have to use the answer from part A.

    Isn't the (final) velocity already known? The problem stated that it rotates at 240 rev/min

    but yes, you will have to use E= 1/2iw^2 to solve this problem
  8. Jul 16, 2014 #7


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    For part (a) first find the rotational energy of the washer when it is rotating at w=240 rev/min. To do this first convert 240 rev/min to rad/second (by multiplying by 2*PI and dividing by 60). Then plug that number into KE = 1/2*I*w^2 along with I (which you already have as 1.2 kg*m^2). That's your final energy of the rotating washer (379 J).

    Your initial energy is 0 J because it starts at rest.

    So your total energy put into the system by the motor has to be 379 J. If your power P is 186.5 J, then use P = E/t to solve for time (so t = (379 J)/(186.5 J) = 2.03 seconds).

    For part (b) you'll need to find the average velocity of the drum throughout the acceleration. Because the acceleration is uniform, this ends up being (w,final - w-initial)/2, which is (25.13 rad/s - 0 rad/s)/2 = 12.38 rad/s.

    Then multiply your average angular velocity by the time it's rotating (answer from (a)) to get the number of radians it rotates through in that time. So (12.38 rad/s)*(2.03 s) = 25.13 rad.

    Then just convert that back into revolutions by dividing by 2*PI. So (25.13 rad)/(2*PI) = 4 revolutions.

    Make sense?
  9. Jul 16, 2014 #8


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    Tj, it's against the forum rules to supply full solutions, let the OP solve it.

    (He will understand it best if he solves it himself)
  10. Jul 16, 2014 #9

    It does! Thanks!
  11. Jul 17, 2014 #10
    ω = 240 rpm = 25.1327 rad/sec
    i = mass moment of inertia = 1.2 kg-m²

    If the rotational acceleration is constant, then the applied torque must be constant also, which means the power must be an average value ( (min + max) / 2 ).
    (min is 0 at 0 revs)

    So maximum power (p) must be 0.25 * 2 = 0.5 hp = 372.85 Watts

    So find the torque (T) at max revs = p / ω
    = 14.835 N-m

    Rotational acceleration rate α = T / i = 14.835 / 1.2
    = 12.3627 (rad/sec)/sec

    Elapsed time = 25.1327 / 12.3627 = 2.033 seconds

    Elapsed rotations (θ) = ½ * 12.3627 * 2.033 ²
    = 25.55 radians

    Thats my take on this, comments ?
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