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Rotational motion of a washing machine

  1. Oct 13, 2003 #1
    Hi everyone….well I have just started to learn about rotational motion, therefore, I am really confused so I have two different problems that I need help on.

    1.) Q: A washing machine goes into the “spin cycle”. From rest the washer starts rotating, gaining 1 rev/sec of rotational velocity each second. How long before it is spinning at 300 RPM? What it its angular acceleration? How many revolutions does it go through in getting up to 300 RPM?

    When the spin cycle ends the washer comes to rest from 300 RPM in 4 seconds. What it its angular acceleration? How many revolutions does it go through?

    A: I think this is totally wrong but here it is:
    300rev/1min * 1min/60sec = 5 rev/sec, so would that be 5 sec?
    alpha = change omega/ time
    omega = 5 rev/ 1sec * 2pi = 31.4rad/sec
    (31.4-0)/5 = 6.26rads, which is not even in the right units, so I know it is wrong. I don’t even know how to broach this problem?

    2.) Q:Two 5 kg masses are connected by an arm that measures 5 ft from end to end. In the middle of the arm is a low-friction pivot point. A lightweight pulley 2 ft in diameter is mounted at the pivot point, with a cord wrapped around the pulley. The cord is pulled with a steady 20 N tension.

    How long before the masses achieve a tangential speed of 40 mph?

    A: torque = m * r^2 * alpha
    20kgm/s^2 = 10kg * 1.83^2 * alpha
    alpha = .597m/s^2

    40mph = 17.88m/s
    acceleration = change speed/ time
    .597m/s^2=(17.88m/s-0m/s)/t = 30.8s
  2. jcsd
  3. Oct 13, 2003 #2
    For #1 you should have quit while you were ahead. I agree with 5 s.
    If you want to do it in radians, well, 1 rev. = 2Π rad. so
    acceleration of 1 rev./s2 = 2Π rad./s2
    300 rev./min = 5 rev./s = 10Π rad./s
    so we have (10Π rad./s)/(2Π rad./s2) = 5 s.

    For the next one, I don't know what your 1.83 represents, but let's try it this way:

    First, the moment of inertia is ∑(m*r) and in this case r=2.5 ft. =.762 m. so
    I = 10*.7622 kg*m2
    For the torque, the moment arm r=2 ft.=.6096m so
    T=20*.6096 N*m
    Next, T= Iα so
    α = I/T = (20*.6096)/(10*.7622)

    You want 40 mi/hr tangential but you have to convert that into a radial velocity ω to relate it to the radial acceleration α. ω=v/r so you need
    40 mi/hr * 1609 m/mi / 3600 s/hr / .6096 m (the radius) = 29.33 rad./s

    Now, you can relate angular acceleration to angular velocity using
    ωf = ωi + α*t (Looks familiar?)
    Of course, in this case, ωi=0.

    You finish it. (I get 14 sec.; hopefully I didn't screw up someplace.)
  4. Oct 15, 2003 #3
    thank you for your help gnome
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