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Rotational Kinematics of bicycle breaks

  1. Feb 29, 2008 #1
    [SOLVED] Rotational Kinematics

    1. The problem statement, all variables and given/known data
    A person is riding a bicycle, and its wheels have an angular velocity of +18.0 rad/s. Then, the brakes are applied and the bike is brought to a uniform stop. During braking, the angular displacement of each wheel is +13.5 revolutions.

    (a) How much time does it take for the bike to come to rest?

    (b) What is the angular acceleration of each wheel?

    2. Relevant equations
    (a) θ = (ω0 + ω)t

    (b) ω = ω0 + αt

    3. The attempt at a solution
    (a) +13.5 rev = (18.0 rad/s - 0 rad/s)t
    t = (+13.5 rev)/(18.0 rad/s) = 0.75 - wrong answer

    (b) (ω - ω0)/t = α
    (0 rad/s - 18.0 rad/s)/0.75 = α - wrong answer


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    1. The problem statement, all variables and given/known data
    After 10.0 s, a spinning roulette wheel at a casino has slowed down to an angular velocity of +1.70 rad/s. During this time, the wheel has an angular acceleration of -5.05 rad/s^2. Determine the angular displacement of the wheel.

    2. Relevant equations
    ω^2 = ω0^2 + 2α(θ − θ0)

    3. The attempt at a solution
    (+1.70 rad/s)^2 = 0 + 2(-5.05 rad/s^2)θ
    θ = [(1.70 rad/s)^2]/2(-5.05 rad/s^2) = -0.286 - wrong answer

    ------------------------------------------------------------------

    Please help me, I don't know what I'm doing wrong.
     
  2. jcsd
  3. Feb 29, 2008 #2

    Dick

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    Start with 1). Your equation a) is wrong. This is accelerated motion. The equation should be:
    theta=omega_0*t+(1/2)*alpha*t^2. Solve your equation b) for alpha and substitute that into a). Now the only variable is t.
     
  4. Feb 29, 2008 #3
    The equation you have listed as (a) is only valid for uniform angular velocity. Since the wheel is slowing down, velocity is not constant, so you have to use an equation for uniform acceleration (since the problem states that the wheel comes to a uniform stop).Since the acceleration is uniform, you can assume that [itex]\alpha = \frac{\Delta\omega}{\Delta t}[/itex].
     
    Last edited: Feb 29, 2008
  5. Feb 29, 2008 #4
    equation 1: θ = ω0t + 1/2αt^2

    equation 2: ω = ω0 + αt
    (ω - ω0)/t = α

    solve for α from equation 2 & plug into equation 1:
    θ = ω0t + 1/2[(ω - ω0)/t]t^2

    θ = ω0t + 1/2[(ω - ω0)]t
    13.5 = 18.0t + 1/2(0 - 18.0)t
    13.5 = 18.0t - 9t
    13.5 = 9.0t
    t = 1.5 seconds - but it's still wrong

    Did I simplified it wrong?
     
  6. Feb 29, 2008 #5

    Dick

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    A 'rev' is not a 'rad'. How many radians is 13.5 revolutions? Other than that, well done.
     
  7. Feb 29, 2008 #6
    (a) 13.5 x 2Pi = 9.0t
    t = 9.42 s

    (b) (0-18.0 rad/s)/9.42s = -1.91 rad/s^2

    Thank you so much.

    I'm still confuse about the second problem though. I tried several ways, but I've gotten some weird answers.
     
  8. Feb 29, 2008 #7

    Dick

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    omega_0 in that problem isn't zero. The wheel starts at velocity omega_0, slows down at a rate of -5.05rad/sec^2 for 10 sec. Then it's speed is 1.7rad/sec. Can you figure out the correct value for omega_0. Think about it...
     
  9. Feb 29, 2008 #8
    ω = ω0 + αt
    ω - αt = ω0
    (1.70 rad/s) - (-5.05 rad/s^2)(10.0 s) = ω0
    ω0 = 52.5 rad/s

    ω^2 = ω0^2 + 2αθ
    (1.70^2) = (52.2^2) + 2(-5.05)θ
    θ = [(1.70^2) - (52.2^2)]/(2 x -5.05) = 269.5 rad

    GOT IT!
    Thank you so much for helping me. ^-^
     
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