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Homework Help: Different values with rotational kinematic equations

  1. Jan 17, 2008 #1
    [SOLVED] Different values with rotational kinematic equations

    Hello everyone,
    I got two different values for final angular speed when I tried to use the third and fourth kinematic equations. I filled the template here with the problem with my attempts.

    1. The problem statement, all variables and given/known data
    A bicycle wheel of radius r = 1.5 m starts from rest and rolls 100 m without slipping in 30 s. Assuming that the angular acceleration of the wheel given above was constant, calculate: a) The angular acceleration, b) the final angular velocity c) the tangential velocity and tangential acceleration of a point on the rim after one revolution.


    [tex]\Delta\theta = 100/1.5 = 66.7[/tex] rad

    [tex]\omega_{i} = 0 [/tex]

    [tex]\Delta t = 30 s[/tex]

    [tex]\alpha[/tex] is constant


    [tex] \alpha = ? [/tex]

    [tex] \omega = ? [/tex]

    [tex] v_{t} = ? [/tex] and [tex] a_{t} = ? [/tex] when [tex] \Delta\theta = 2\pi[/tex]

    2. Relevant equations

    [tex]\Delta\theta = \frac{1}{2} (\omega_{f} + \omega_{i}) \Delta t[/tex]

    [tex] \omega_{f}^{2} = \omega_{i}^2 + 2 \alpha \Delta \theta [/tex]

    [tex] \omega_{f} = \omega_{i} + \alpha \Delta t [/tex]

    3. The attempt at a solution

    Part b:

    [tex]\Delta\theta = \frac{1}{2} (\omega_{f} + \omega_{i}) \Delta t[/tex]

    [tex] 66.7 = \frac{1}{2}(0 + \omega_{f})(30) [/tex]

    [tex] \omega_{f} = 4.45 [/tex] rad/s

    Part a:

    [tex] \omega_{f} = \omega_{i} + \alpha \Delta t [/tex]

    [tex] 4.45 = 0 + 30\alpha [/tex]

    [tex] \alpha = 0.15[/tex] rad/s^2

    Part c:

    My strategy is to find the final angular velocity [tex]\omega_f[/tex] using the first relevant equation. Then finding the tangential velocity by multiplying [tex]\omega_f[/tex] with the radius.

    [tex]\Delta\theta = \frac{1}{2} (\omega_{f} + \omega_{i}) \Delta t[/tex]

    [tex] 2\pi = \frac{1}{2} (\omega_{f} + 0)(30) [/tex]

    [tex] \omega_{f} = \frac{2\pi}{15} = 0.419 [/tex]

    And then:

    [tex] v_{t} = r\omega = (1.5)(0.419) = 0.6285 [/tex] m/s

    [tex] a_{t} = r\alpha = (1.5)(0.15) = 0.255 [/tex] m/s^2

    However, the solution key has a different answer. [tex]\omega_{f}[/tex] after one revolution equals 1.37 rad/s and [tex]v_{t}[/tex] = 2.06. Apparently, they used this equation:

    [tex] \omega_{f}^{2} = \omega_{i}^2 + 2 \alpha \Delta \theta [/tex]

    I'm a little confused. I double checked my calculations, and I am almost sure that I haven't made a calculation error. I appreciate it if someone could help me with this.

    Last edited: Jan 17, 2008
  2. jcsd
  3. Jan 17, 2008 #2

    D H

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    Staff Emeritus
    Science Advisor

    Your mistake is assuming that the time here is 30 seconds. It takes quite a bit less than 30 seconds for the wheel to make one revolution.
  4. Jan 17, 2008 #3
    OH! I get it now. Thanks a lot, D H
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