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Angular displacement

  1. Oct 27, 2004 #1
    Alright, this problem is really starting to bug me.

    ----Starting from rest at t0 = 0s, a wheel on a mountain bike has a constant angular acceleration. When t = 5.15s, the angular velocity of the wheel is +7.66rad/s. The angular acceleration continues until t = 13.6s, after which time the angular velocity remains constant. What is the angular displacement of the wheel in the time interval from t = 0 to 28.0s?----

    I divided it into two parts, since angular acceleration is constant from t=0s to t=13.6s, and isn't for the remaining portion. So for that first part, I calculated the angular acceleration by doing change in angular velocity or change in time...7.66rad/s / 13.6s = 0.562 rad/s^2. And then I solved for Theta = 0.5 A T^2 ...and I got 52.088 rad. And then for the second part I found the acceleration over the remaining 14.4s and got an acceleration (through the same process) of 0.532 rad/s^2...and then I solved for Theta of that portion and then added my two numbers to get 107.24 rad. But it's wrong. Any guidance would be much appreciated. Thanks! :)
     
  2. jcsd
  3. Oct 27, 2004 #2

    Doc Al

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    Staff: Mentor

    You made a mistake in calculating the acceleration: 7.66 rad/s is the angular speed at t = 5.15 s, not t = 13.6 s. So the change in time is 5.15 s.

    For the second part, the acceleration is zero. They tell you that the speed is constant.
     
    Last edited: Oct 27, 2004
  4. Oct 27, 2004 #3
    O ok...so for the first part (t=0s to t=5.15s) 7.66/10.3= 0.738 rad. I had thought the acceleration would be zero because of constant velocity (now if I only followed my intuition!)...so that would mean I just multiply the speed by the time to get distance. So if I multiply 7.66 by the remaining time, 22.85s, and get 175.769 rad...and then add the 0.738 rad i found before...would that be the correct way to find the answer?
     
  5. Oct 27, 2004 #4

    Doc Al

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    Staff: Mentor

    First, let me apologize for telling you to use the average velocity to find the displacement. You can't use that method since you don't have the final speed at the end of the first part of the motion. D'oh! I must have been sleeping. (I removed that advice from the previous post.)
    Here's what you do. As you stated earlier, treat the motion in two parts (1) accelerated (from t = 0 to 13.6) and (2) constant speed (from t = 13.6 to 28.0).

    For part 1, find the acceleration. (See my previous post.) Then use that acceleration to do two things: find the displacement (using [itex]\theta = 1/2 \alpha t^2[/itex]) over the entire interval, and find the find speed at the end of the interval (using [itex]\omega_f = \alpha t[/itex]).

    For part 2, you have the speed and the time so use [itex]\theta = \omega t[/itex].

    Now just add the two displacements.
     
  6. Oct 27, 2004 #5
    For part two I did 7.66 rad/s x 14.4s = 110.304 rad

    Part one...a= deltav/deltat....7.66rad/s / 13.6s = 0.563 rad/s^2 ....
    .5 x 0.563 x 13.6^2 = 52.07 rad

    When added together, I get 162.4 rad total....but the program says I'm wrong...and unfortunately, I have only one try left. Sorry to keep bothering you, but do you see another mistake? Thanks so much for your help thus far...I don't know why I keep messing this up. :/
     
  7. Oct 28, 2004 #6

    Doc Al

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    You are still making errors in both parts.
    7.66 rad/s is only the speed at t=5.15s. It is not the speed you need for the "constant speed" part of the motion (part 2). You have to figure out that speed. See my last post for how to figure out that speed using the acceleration from part 1.
    See my first post. You are using the wrong time in calculating the acceleration: [itex]\alpha = \Delta \omega/\Delta t = (7.66)/(5.15) [{rad}/s^2][/itex].
     
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