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Homework Help: Nonuniform Circular Motion

  1. Mar 26, 2005 #1


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    Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 20.0 s to speed up from rest to its top speed of 1 rotation every 1.10 s. The astronaut is strapped into a seat 5.60 m from the axis.

    A. What is the astronaut's tangential acceleration during the first 20.0 s?

    B. What is the astronaut's tangential acceleration during the first 20.0 s?

    Part A.

    I used [tex] a = a_r + a_t[/tex]
    To get a i find velocity, which is [tex] v = d/t , a = v/t[/tex]
    for [tex]a_r [/tex] i used [tex] a_r = v^2/r[/tex]

    Where did I go wrong ?

    Part B.

    I've no idea how to do this part, any help is appreciated.
  2. jcsd
  3. Mar 26, 2005 #2
    Part A + B are the same question.
    I dont see how this is non uniform. He's revolving an axis with r = 5.6
  4. Mar 26, 2005 #3
    Oh hes speeding up. Find the initial and final tangential velocity. If your having trouble with that find the angular acceleration, angular velocity, then multiply them by the radius. The change in velocity over the change in time will be your acceleration.

    [tex] v_i = 0, v_f = \frac{1 revolution}{1.1 sec}, \Delta T = 20s[/tex]

    For linear velocity:

    1 revolution is [tex] 2\pi r = 2\pi 5.6 = 32m[/tex]
    [tex] \Delta v = v_f - v_i = 35.2m/1.1s - 0 = 35.2m/1.1s = 32m/s [/tex]

    [tex] a = \frac{\Delta v}{\Delta T} = \frac{32m/sec}{20sec} [/tex]

    The calculation for angular is very similar.
    Last edited: Mar 26, 2005
  5. Mar 26, 2005 #4


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    Sorry for the confusion,

    Here is part B.
    How many g's of acceleration does the astronaut experience when the device is rotating at top speed? Each 9.80 m/s^2 of acceleration is 1 g.

    Do I need to find the velocity at top speed and then divide it by 9.8 ?
    How do I find the velocity at top speed. ?
  6. Mar 26, 2005 #5
    I'm not really sure how to solve that problem. Youd want to find the tangential acceleration (there should be none) and divide it by 9.8. Perhaps the centripetal acceleration is what you want.

    Your conditions are

    [tex] v_{linear} = 5.6(2pi*.9rev/sec) = 32m/s [/tex]

    [tex] a_{linear} = dv/dt = 0, a_{cent} = v^2/r = 32^2/5.6 = 182m/s^2 [/tex]

    [tex] 182m/s^2 / 9.8m/s^2 = 18.57 g's [/tex] Thats way too many g's.
    Last edited: Mar 26, 2005
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