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Rotational Motion and acceleration Question

  1. Jun 3, 2006 #1
    Hey guys i have only one question this time on my homework :cool:

    1) At t=0 a grinding wheel has an angular velocity of 28.0 rad/s. It has a constant angular acceleration of 29.0 rad/s^2 until a circuit breaker trips at time t=2.50s. From then on, it turns through an angle 433rad as it coasts to a stop at constant angular acceleration.

    I was able to answer part A which was: Through what total angle did the wheel turn between t=0 and the time it stopped? and figured it to be 594rad but part B is troubling me.

    Part B) At what time did it stop?
    I tried using: omega_z=omega_oz+alpha*t
    to find the speed of the wheel when it intially stops and found it to be 100.5 rad/s Then i tried using: delta_theta=(1/2)(omega_z+omega_oz)*t and moved aroudn this formula to solve for:
    t= (2*delta-theta)/(omega_z+omega_oz)
    this gave me 11.8 seconds which is wrong....anyone see my problem or have another method to share?

    Part c) What was its acceleration as it slowed down?

    I hav eno idea how to figure this out but if i had this i think i could solve b easier...any tips?
  2. jcsd
  3. Jun 3, 2006 #2

    Andrew Mason

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    What is its angular speed [itex]\omega[/itex] at t = 2.5s?
    What is its energy at t = 2.5 s (in terms of its moment of inertia, I)? How much work is done over the next 433 radians? What is that work in terms of I, the angular deceleration and angular distance? That will give you [itex]\alpha[/itex] (which is part c)). Then work out t from [itex]\alpha[/itex] and total angular distance.

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