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Homework Help: Rotational Motion

  1. May 22, 2008 #1
    Heres the problem Im workin on

    http://img294.imageshack.us/img294/9166/rotatesm3.jpg [Broken]

    Now in order to find the magnitude of accleration Im assuming I would need to find the two components of acceleration first, Centripetal and Tangential......

    I found tangetial by the formula

    [tex] a_{t} = r\alpha [/tex]

    I converted the 2 revolutions to

    [tex] 4\pi radians [/tex]

    but im not exactly sure how to get tangential...i need angualr velocity......

    how to I get angular velocity from revolutions?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. May 22, 2008 #2

    Doc Al

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    Staff: Mentor

    You'll need to do a little kinematics. You have the angle in radians,the angular acceleration, and the initial angular speed--use kinematic formulas to find the final angular speed. Hint: Angle is the analog to distance, just like angular speed and acceleration are the analogs to linear speed and acceleration.
  4. May 22, 2008 #3
    ok how does this look

    [tex] \omega^2 = \omega_{o} + 2\alpha(\theta-\theta_{o})[/tex]

    [tex]= 0 + 2(.8)(4\pi radians)[/tex]

    [tex]\omega = 4.48 rad/s[/tex]

    [tex]a_{n} = r\omega^2 = (.6)(20.16) = 27.18 [/tex]

    [tex] a_{t} = r\theta = (.6)(.8) = .480 [/tex]

    [tex]A = \sqrt{(9.480)^2 + (27.18)^2} = 27.18 [/tex]
    Last edited: May 22, 2008
  5. May 22, 2008 #4

    Doc Al

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    Staff: Mentor

    Your method looks good, but check your arithmetic here:
  6. May 22, 2008 #5
    ohhh thanks....im having trouble reading my calculators display!

    that changes the answer to 12.09 rad/s^2
  7. May 22, 2008 #6

    Doc Al

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    Staff: Mentor

    That looks better, but be careful with units. The acceleration is in m/s^2.
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