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Revolutions per minute, angular deceleration

  1. Oct 23, 2012 #1
    Hi all,

    1. The problem statement, all variables and given/known data

    The motor driving a large grindstone is switched off when a rotational speed of 360 rpm has been achieved. After 15s the speed has decreased to 210rpm. If the angular deceleration remains constant, how many additional revolutions does the stone make before coming to rest?

    3. The attempt at a solution

    So to start with I converted rpm to rads-1 which was easy enough (I left it in pi to make life easier)

    ω0 = 360/60 x 2∏ =12∏ rads-1
    ω15 = 210/60 x 2∏ =7∏ rads-1

    I then used v=u+at to find the acceleration

    7∏ = 12∏ + 15a
    a = -∏/3 rads-2

    This is where I'm having a mind block as to what to do next, from 15s to when the stone comes to rest - how many additional revolutions does the stone make?

    Would love some pointers (I'm sure it's a simple solution but my heads refusing to grasp it).
  2. jcsd
  3. Oct 23, 2012 #2


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    If you have the angular velocity of the motor and the angular deceleration at t=15 s,
    you should be able to easily determine how many additional revolutions occur before the motor stops.

    Imagine this: you have a vehicle going a certain velocity when the brakes are applied. How would you calculate the stopping distance from the point where the brakes were applied?
  4. Oct 23, 2012 #3
    Thanks for the reply.

    For the car thing I'd just use a kinematic equation - but surely distance doesn't relate to revolutions?

    Are there any equations that are cirular motion specific?
  5. Oct 23, 2012 #4
    Right OK just found an equation θ = ωit + 0.5αt2 which looks very useful. With that I think all I need to do is find the time it takes to completely stop, use that in the equation I've just stated then convert θ into revolutions and Bob's your uncle.

    I'll give that a go but I think I've got it from here.
  6. Oct 23, 2012 #5
    Just as a check did anybody get the answer 363 revolutions?
    Last edited: Oct 23, 2012
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