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Circular motion power

  1. Jun 24, 2012 #1
    1. The problem statement, all variables and given/known data[/b]

    A particle of mass m is moving in a circular path of constant radius r such
    that its centripetal acceleration a varies with time t as [itex]a = k^2rt^2[/itex], where k is a
    constant. Show that the power delivered to the particle by the forces acting on
    it is [itex]mk^4r^2t^5/3[/itex]

    2. Relevant equations



    3. The attempt at a solution

    Why isn't work done, hence power = 0 since it's moving in a circle and resultant force is centripetal?

    Edit : Assuming there's a tangential component,

    I take
    [itex]a_c = k^2rt^2 = v_{tan}^2/r[/itex]
    [itex]v_{tan} = √k^2r^2t^2 = krt[/itex]
    [itex]a_{tan} = dv/dt = kr[/itex]
    [itex]P = Fv = ma_{tan}v_{tan} = krt * kr * m ≠ mk^4r^2t^5/3[/itex]
     
    Last edited: Jun 24, 2012
  2. jcsd
  3. Jun 24, 2012 #2
    I have got the answer you have mentioned but i am not sure what could be the reason that my answer is correct. I am sitting on my computer desk and following your thread but no one is yet replying, so i think i will post my solution and see if i could receive any feedback. I know its your thread hqjb, i hope you don't mind. :smile:

    I integrated the expression for centripetal acceleration with respect to time and found [itex]v=\frac{k^2rt^3}{3}[/itex], (now i don't know which velocity is this. :confused:)

    P=Fv
    P=macv
    [itex]P=m*k^2rt^2*\frac{k^2rt^3}{3}[/itex]
    [itex]P=\frac{mk^4r^2t^5}{3}[/itex]

    But still i don't have any reasoning to why is this correct?
     
    Last edited: Jun 24, 2012
  4. Jun 24, 2012 #3
    a=k2rt2
    v=k2rt3/3

    E=1/2mv2=1/2m(k2rt3/3)2

    P=dE/dt=(6/18)m(k4r2t5)
     
    Last edited: Jun 24, 2012
  5. Jun 24, 2012 #4
    I get both of your workings, but I'm not sure why both of you used the radial component of velocity? (integrating centripetal acc = centripetal vel. right?)

    Shouldn't only tangential component count in doing work?
     
  6. Jun 24, 2012 #5
    I'm looking for the instantaneous velocity from instantaneous acceleration,irrespective of direction.
    Energy is scalar.
     
  7. Jun 24, 2012 #6
    I think that:
    v[itex]^{2}[/itex]=v[itex]^{2}_{c}[/itex]+[itex]^{2}_{t}[/itex]
    But v[itex]_{t}[/itex]=krt
    And v[itex]_{c}[/itex]=[itex]\frac{k^{2}.r.t^{3}}{3}[/itex]
    W=1/2.m.[itex]v^{2}[/itex]
    P=dW/dt=...
     
  8. Jun 24, 2012 #7
    I get my original answer + everyone else's answer in this case :

    [itex]k^2r^2tm + (k^4r^2t^5m)/3[/itex]
     
  9. Jun 24, 2012 #8
    I think this was the correct answer :-s
     
  10. Jun 24, 2012 #9
    Okay after thinking about it again, I think by definition circular motion means there is no "radial velocity". My bad for that mistake. But can anyone tell me what you get when you integrate radial acceleration? What about the v^2/r formula then(My notes says that it works for non-uniform circular motion)?
     
  11. Jun 25, 2012 #10

    ehild

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    Your result looks fine. The centripetal acceleration is v2/r, even in case of non-uniform circular motion. There is no work done by the centripetal force if the particle moves along a circle. But there should be some tangential force to ensure increase of speed. The tangential force multiplied by the velocity (which is tangential of course) gives the power.

    ehild
     
  12. Jun 25, 2012 #11
    My understanding is when a body moves at
    a=Ct2 where C=k2r
    The velocity will be
    v=Ct3/3

    KE=1/2mv2
     
    Last edited: Jun 25, 2012
  13. Jun 25, 2012 #12

    ehild

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    The velocity is tangent to the path.
    The tangent of the circle is perpendicular to its radius : there is no "radial velocity".

    There is radial acceleration: it is called "centripetal acceleration". There can be also tangential acceleration.
    The centripetal force does not do any work.
    The integral of the centripetal acceleration is neither velocity nor speed.
    Think of the uniform circular motion. The centripetal acceleration is constant. Does the speed increase linearly with time?

    ehild
     
  14. Jun 25, 2012 #13
    ehild,

    I don't think so. Let ω be constant and vector R = cosωt*i + sinωt*j. Then velocity vector V = dR/dt = -ω*sinωt*i + ω*cosωt*j , where V and R are at right angles to each other. So the acceleration vector is A = dV/dt = -ω^2(cosω*t*i + sinωt*j) = -ω^2*R, which is a vector in the opposite direction of R and whose magnitude is |V|^2/|R|. This derivation requires that ω be constant (uniform). Besides, the problem statement says that the centripetal acceleration is k^2*r*t^2.

    Ratch
     
    Last edited: Jun 25, 2012
  15. Jun 25, 2012 #14
    Last edited by a moderator: May 6, 2017
  16. Jun 25, 2012 #15

    ehild

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  17. Jun 25, 2012 #16
    @ehild: Can i get some feedback on my solution? That would be very helpful. The solution is posted on page 1, post #2. :smile:
     
  18. Jun 25, 2012 #17

    ehild

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    Pranav, think of the uniform circular motion. The centripetal acceleration is constant. Do you get anything reasonable if you integrate it with respect time? Is it the velocity or speed or anything? Is the speed of the uniform circular motion a linear function of time instead of being constant?

    Read my post #13.

    ehild
     
  19. Jun 25, 2012 #18

    ehild

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    We totally agree then. The acceleration is the vector sum of Ac and At. Ac is the centripetal acceleration. Its magnitude is v2/r.

    ehild
     
  20. Jun 25, 2012 #19
    The speed is constant in uniform circular motion. The centripetal acceleration is constant in uniform circular motion but here the centripetal acceleration is varying with time.
    I really have no idea what we get if we integrate the centripetal acceleration.
     
  21. Jun 25, 2012 #20
    Hello hqjb and others,
    I think the solution pasted in the attempt is correct.Circular motion uniform or non uniform “indeed requires the centripetal acceleration to be equal to(V^2/R)”.Let us prove it through calculus and polars.(2Dimensional avoiding the usage of axis of rotation)
    We know that vector r of any point on the circle with O as origin at an elevation from say the horizontal (the diagram can be drawn easily) and with unit vectors as i and j is given as
    R=R(cos(θ)i+sin(θ)j)
    We define two other unit vectors e(r) and e(t) ,with the first pointing away from the center towards the radius and the second in the direction of the advancing tangent.
    e( r )= cos(θ )i+sin(θ )j
    e( t )= -sin(θ )i+cos(θ )j
    e’( r )= e(t)*θ’(Differentiation w.r.t any variable)
    Similarly
    e’( t)=-e( r )* θ’
    As such r=r*e( r ) ...........$
    V=rω*e(t)+r’e(r) ………………$$
    /* Unless radius is changing w.r.t time velocity is always tangential as r’=0 */so
    V=rω*e(t) ………………………$$$
    (ω is angular velocity)
    So that
    a=-r(ω^2)e(r) +rαe(t) ………………$$$$ (radius taken constant once again)(alternatively use complex numbers for the same result)
    In short only given that radius is constant the centripetal acceleration is ((v^2)/r) no matter what (we used v=rω in the radial portion of the last equation which arrives from our velocity vector expression)For other curvilinear trajectories the centripetal acceleration is defined by the general term “normal component of acceleration”The same equation works with only the r in the denominator being radius of curvature of the curve at that point .
    So
    (1) Integrating centripetal acceleration in any case of circular motion is not an act that brings forth any velocity.
    (2) One need not worry about radial component of velocity here as it reduces to zero given the fact that radius is constant.
    (3) Something is wrong with the question, it yields fine with the acceleration being defined plainly but is incorrect as regards circular motion and centripetal acceleration.
    Apologies in advance for straying the discussion off topic if I did so.Correct me if I am wrong.
    Regards
    Yukoel
     
    Last edited: Jun 25, 2012
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