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Rotational kinematics of a turntable

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data

    An electric turntable 0.710 m in diameter is rotating about a fixed axis with an initial angular velocity of 0.230 rev/s. The angular acceleration is 0.887 rev/s^2.

    A. Compute the angular velocity after a time of 0.192 s.

    B. Through how many revolutions has the blade turned in this time interval?

    C. What is the tangential speed of a point on the tip of the blade at time t = 0.192 s?

    D. What is the magnitude of the resultant acceleration of a point on the tip of the blade at time t = 0.192 s?

    2. Relevant equations

    Any rotational kinematic.
    Conversions to radians/revolutions

    3. The attempt at a solution

    A.) I took the angular velocity 0.230 and multiplied it by 2pi which equals 1.445 rad/s.

    I then plugged it into a kinematic equation, 1.445+(0.887*0.230)^2 = 2.56 rad/s = .400 rad/s.

    This is right.

    B.) The I used the distance rotational kinematic equation, 1.445 * 0.230 +.5(0.887)(0.230)^2 = 0.3558 rad. = 0.57 rev.

    This is wrong. Need help.

    C) To to find the tangential speed, I did 0.710 m * 2.56 rad/s = 1.82.

    This is wrong. Need help.

    D) Need help.

  2. jcsd
  3. Nov 19, 2011 #2


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    Homework Helper

    On part (b), have you perhaps "plugged in" the initial angular velocity where the time interval 0.192 seconds belongs?

    I think you have misled yourself because you have omitted a factor (and wrote another incorrectly) in the typing of your equation in part (A):

    I believe that should be 1.445 + [(0.887*2*pi)^2] (0.192) = 2.56 rad/s . I agree with your result, but you didn't get it using what you've typed.
    Last edited: Nov 19, 2011
  4. Nov 19, 2011 #3


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    Conversion is correct.

    From what I gather, your formula is Ω = ω + (αω)2 ?

    I am not following this line as that computation gives me 1.48 or so.
  5. Nov 19, 2011 #4
    The angular velocity w = w0 + a*t where w0=0.23*2*Pi rad./s
    also where 'a' = 0.887*2*Pi rad./s^2
    At t=0.192, w(0.192) = 0.23*2*Pi + 0.887*2*Pi*0.192
    =2.5 rad/s

    On part B, I need to plug in time(0.192 s) where I plugged in initial angular velocity?

    For, C, should i be using the radius of .355 m? Then multiply it by 2.56?
  6. Nov 19, 2011 #5


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    The formula is θ = ωt + ½αt2

    You plugged in t as 1.445 and ω as 0.230, you need to plug in ω as 1.445 and t=0.192. Just makes sure α is in rad/s2

    Yes that would be correct since you know that v = rω
  7. Nov 20, 2011 #6
    Ok, thank you guys for all the help! much appreciated.
  8. Nov 20, 2011 #7
    For part D.

    I received the acceleration of the radius by multiplying 0.887 rad/s^2 by 2pi.
    =5.73 rad/s

    I received the tangential acceleration by the change in speed of (2.56 rad/s)^2 multiplied by the radius of 0.355 m.
    = 2.195 rad/s.

    (5.57)^2+(2.19)^2 and the square root = 5.99 rad/s which is wrong.

    The I tried (2.515)^2(.355) = 2.25 = radial acceleration
    with (5.573)(0.355) = 1.98

    The use Pythagorean theorem for 2.99 m/s^2

    Is this right?
    Last edited: Nov 20, 2011
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