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Acceleration Of Tennis Racket (Linear And Rotational Quantites)

  1. Jun 11, 2008 #1

    During a serve, a tennis racket of mass 1.6 kg is given an angular acceleration of 157 rad/sec2. At the top of the serve, the racket has an angular speed of 13 rad/sec.
    If the distance between the top of the racket and the shoulder is 1.4 m, what is the total acceleration of the top of the racket?

    atotal = ? m/sec2

    Okay so I'm having some trouble here. I know that atotal is the combination of atranslational and acentripetal, well rather the addition of the squares of each equal to the square of atotal, but I'm not sure how to solve for atranslational.

    I know that acp = rw^2 and I solved using the values given to get 118.3m/s^2.

    However, when it comes to solving at = ra I don't know how to get the angular acceleration. I'm pretty sure the initial angular acceleration isn't what you use, as I got the wrong answer, but how else do I determine it at the top?

    Any help would be appriciated :)
  2. jcsd
  3. Jun 11, 2008 #2

    Doc Al

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    Staff: Mentor

    How did you get this result?
    Lacking information to the contrary, I would assume that the angular acceleration is constant.
  4. Jun 11, 2008 #3
    I got it by first finding the radius of the racket (1.4m / 2 = 0.7m) and then multiplying that by the square of the 13 rad/s angular speed at the top of the racket.

    0.7m * 13 rad/s = 118.3 rad/s

    Well if that were true, referring to the constant angular acceleration, then would I still be on the right track following the idea of a^2 + b^2 = c^2?
  5. Jun 11, 2008 #4

    Doc Al

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    Staff: Mentor

    1.4m is the distance between shoulder and the tip of the racket. Assume that the shoulder is the axis about which the arm and racket swings. (No need to divide by 2.)
  6. Jun 11, 2008 #5
    Oh I see! So the entire racket creates the circle, not just the top part!

    Thanks! I just plugged it in instead of 0.7 m and it came out correctly. Thanks so much :)
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