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Angular Velocity and Acceleration

  1. Apr 14, 2016 #1
    1. The problem statement, all variables and given/known data
    If a bike wheel rotates 9.4 times while slowing down to a stop from an initial angular velocity of 8.1 rad/s, what is the magnitude of the angular acceleration in rad/s/s
    2. Relevant equations
    α = at / r
    α = ω / t
    α = Θ / t^2
    ω = Θ / t
    ω = v / r
    Θ = ω t + 0.5 α t^2
    v final = v initial + at → ω final = ω initial + αt
    some of these formulas may be useless, and there possibly are some others not mentioned that i do not know :/
    3. The attempt at a solution
    Knowing the initial angular velocity is 8.1 and the final is 0 since the wheel stops, i used
    ω final = ω initial + αt
    0 = 8.1 + αt
    it spins 9.4 times in the time frame during which it slows down, but the radius of the wheel is not given. And i do not know which ω to use in the equation ω = Θ / t in order to find time. If i could get time, then i could use the equation α = ω / t or α = Θ / t^2. Am i even approaching this correctly? if not can somebody point me in the right direction?
  2. jcsd
  3. Apr 14, 2016 #2


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    There is another SUVAT equation you need to translate into uniformly decelerated angular motion: ##s = s_0 + v_0 t + {1\over 2} at^2##.

    ##\theta## takes the place of s (and one revolution is ##2\pi##) to give you a second equation. That way you have 2 eqns for ##\alpha## and ##t##, so you should be able to solve.
  4. Apr 14, 2016 #3
    i am still confused as to what two equations these are, if you could provide them for me or hint at what they would be i would greatly appreciate it. I am unsure of what s0 would be, i understand that theta substitutes for s, but it couldn't be for both sides. Is 2pi x 9.4 supposed to be s?
  5. Apr 15, 2016 #4


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    $$\theta(t) = \theta_0 + \omega_0 t + {1\over 2} \alpha t^2 $$ it's that simple (for constant angular acceleration/deceleration -- then ##\alpha < 0##).

    You don't know ##\theta(t)## and ##\theta_0## but you do know their difference (indeed, the angle you mention).
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