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Angular velocity question

  1. Dec 8, 2016 #1
    1. The problem statement, all variables and given/known data
    The motion of a particle moving in a circle in the x-y plane is described by the equations: r(t)=3.15, Θ(t)=8.86t
    Where Θ is the polar angle measured counter-clockwise from the + x-axis in radians, and r is the distance from the origin in m.

    1. Calculate the x-coordinate of the particle at the time 2.20 s.
    2. Calculate the y-component of the velocity at the time 2.20 s?
    3. Calculate the magnitude of the acceleration of the particle at the time 1.60 s?
    4. By how much does the speed of the particle change from t=10 s to t=51 s?
    5. Calculate the x-component of the acceleration at the time 2.60s?

    2. Relevant equations

    x(t) = Rcos(ωt)
    y(t) = Rsin(ωt)

    And their derivatives for velocity and acceleration.

    3. The attempt at a solution

    I got every question except for 4.

    1. 2.52 m
    2. 22.34 m/s
    3. 247 m/s^2
    4.
    5. 124.13 m/s^2

    For 4 I tried a couple of things. The first was to multiply 247*41 but that didn't work. The next thing I did was to use the equation for velocity (The derivatives of the equations above) to find the X and Y component of the velocity of each time (x,y when t=51 and x,y t=10), calculate their magnitude, and then subtract them from each other, didn't work either.

    What am I missing?
     
  2. jcsd
  3. Dec 8, 2016 #2

    Doc Al

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    What's the speed of the particle at t = 10? (How would you calculate it in general?)
     
  4. Dec 8, 2016 #3
    Differentiate these two

    x(t) = Rcos(ωt)
    y(t) = Rsin(ωt)

    Get the speed for t=10

    And then calculate the magnitude of |x'(t), y'(t)|

    Or are you asking to demonstrate it with numbers?
     
  5. Dec 8, 2016 #4

    Doc Al

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    That will work.

    So what do you get when you do the differentiation?

    And once you have the velocity components, you can find the magnitude of the resultant velocity, which is the speed.
     
  6. Dec 9, 2016 #5
    x'(t) = -Rsin(ωt)*ω
    y'(t) = Rcos(ωt)*ω

    I know it should work but for some reason I can't get it to work...

    Edit: did a more accurate calculation, the change in speed is 0 m/s :D

    But now I'm a bit confused... I found the particle's acceleration... So how does the speed not change? Is it be because it has a constant radial speed, and the acceleration I found is the tangent acceleration? If so, I find it kind of confusing... Can anyone explain how can I know if I need to look at tangent acceleration or radial acceleration...?
     
    Last edited: Dec 9, 2016
  7. Dec 9, 2016 #6

    PeroK

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    What makes you think the particle is changing its speed? Isn't it possible to move in a circle at constant speed?
     
  8. Dec 9, 2016 #7
    It is. But I calculated it's acceleration in 3. and 5.
     
  9. Dec 9, 2016 #8

    PeroK

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    Can a particle move in a circle without acceleration?
     
  10. Dec 9, 2016 #9

    Doc Al

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    So far, so good. Now calculate the speed at time t. (You have the components of the velocity.)

    Since acceleration is a change in velocity, not necessarily speed, just going in a circle implies some acceleration.
     
  11. Dec 9, 2016 #10
    Ohhh... So velocity has a direction and a quantity (Basically a vector) while speed is just a scalar, so when moving in a circle the velocity always changes, and that's the acceleration... ?
     
  12. Dec 9, 2016 #11

    PeroK

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    Not just a circle. Any change of direction implies acceleration.

    Also, if the force/acceleration is perpendicular the the velocity then the speed does not change. That is what is special about motion in a circle at constant speed: the force/acceleration is always perpendicular to the velocity.
     
  13. Dec 9, 2016 #12
    Now I understand, thanks!
     
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