1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

    Thanks.
     
  2. jcsd
  3. Nov 19, 2011 #2

    dynamicsolo

    User Avatar
    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

    rock.freak667

    User Avatar
    Homework Helper

    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

    rock.freak667

    User Avatar
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rotational kinematics of a turntable
  1. Coin on a turntable (Replies: 1)

Loading...