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Tangential Speed and Acceleration

  1. Dec 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Hello, I am a relatively new member of this website and I wanted to get some help or feedback in helping me understand how to answer these questions. If my format is poor, I am sorry because I just don't know how to use this site that well yet. Anyhow thanks for looking.

    A disk with a radius of 0.2 m is initially spinning counterclockwise at a rate of 45 revolutions per minute. At t = 0, a constant angular acceleration of −1.5 rad/s^2 is imparted onto the disk.
    a) What is the tangential speed of a point on the outer edge of the disk at t = 2 s?
    b) What is the tangential acceleration of a point on the outer edge of the disk at t = 2 s?


    2. Relevant equations
    ωr = v
    a = αr

    3. The attempt at a solution
    Okay so I started with:

    45 rpm = 45rpm* (2pi/1) = 282.743rad/m
    then 282.743rad/m * (1/60 seconds) = 4.71 rad/s
    Since I know the radius and angular velocity, I did
    4.71rad/s * .2m = .94 m/s
    According to the answer my professor gave me, the answer is .34m/s. Can someone please help me explain where I went wrong? Am I suppose to associate t =2 somewhere in my answer?

    For the second part, I did something similar, except I used constant acceleration (-1.5/s^2) so it went
    -1.5rad/s^2 * .2 = -0.3m/s^2 but the answer was .30m/s^2. Did I do this right?
     
  2. jcsd
  3. Dec 5, 2015 #2

    TSny

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    Welcome to PF! You did a great job in formatting your question.

    Did you find the tangential speed at the instant of time that was asked?

    Your answer for the tangential acceleration is correct. Perhaps the answer given is for the magnitude of the acceleration, so they did not specify it as negative.
     
  4. Dec 5, 2015 #3
    Thanks for taking my question and I probably didn't do that. Am I suppose to use the time "t =2 " into the equation? I assumed I didn't use the right formula to find it at t= 2 which is where I got lost at.
    By the way, thanks for checking my second answer.
     
  5. Dec 5, 2015 #4

    TSny

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    The angular acceleration indicates the rate at which the angular velocity is changing. So, in this problem the angular velocity is changing at a rate of -1.5 rad/s2. That is, the angular velocity is changing by -1.5 rad/s every second.
     
  6. Dec 5, 2015 #5
    Hm, I got an answer that seems close, but I don't know if I am correct.
    Since the disc decelerates over time, I did
    vf = .94 (from the trangential speed) + (-.3 from the acceleration from (b)*2s from the time
    .94-.6 = .34 m/s
    Is this right?
     
  7. Dec 5, 2015 #6

    TSny

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    Yes, that is correct. However, if you want to work part (a) without first working part (b), then try to find the angular velocity at t = 2 s by considering the fact that the angular velocity is decreasing at a rate of -1.5 rad/s each second.
     
  8. Dec 6, 2015 #7
    Thanks! I appreciate the help! I hope you enjoy the holidays!
     
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