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General motion with constant speed

  1. Aug 25, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle moves along any path in 3d space with constant speed. Show that its velocity and accelerations vectors must always be perpendicular to each other.[hint. Differentiate the formula v dot v = v^2 with respect to t.


    2. Relevant equations

    possible equations: v=dr/dt=dr/ds*ds/dt, a=dv/dt=(dv/dt)t +(v^2/rho)n,
    v=v*t where t is the unit tangent vector and n is the normal unit vector.
    3. The attempt at a solution

    If a particle is moving in 3d space with constant speed, then the type of motion could be uniform circular motion and the acceleration is therefore equal to zero.

    Probably, v dot v= v^2(x) + v^2(y)+ v^2(z) , x , y and z making up the dimensions of the 3-D plane; since the particle moves along a 3D plane. If the particle is in uniform motion, the position vector r probably is in polar coordinates: r= b*cos(theta)i + b*sin(theta)j and theta is equal to ut/b where u is the speed of the particle , b is the arc length and t is the time its traveled. Would I differentiate r and then plugged in dr/dt into the v dot v equation? I expect I also have to show that x hat times y hat or x hat times z hat or y hat times z is equal to zero, since the problem asks me to show that velocity and acceleration vectors are always perpendicular to each other.

    Maybe the simplest solution is to show that v cross a = 0 since the acceleration of the particle is zero and therefore I would be done with the problem.
     
    Last edited: Aug 25, 2008
  2. jcsd
  3. Aug 25, 2008 #2

    Doc Al

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    Just follow the hint! Take the derivative of both sides of:
    [tex]\vec{v}\cdot\vec{v} = v^2[/tex]

    Hint: If two vectors are perpendicular, what must their dot product be?
     
  4. Aug 25, 2008 #3

    Kurdt

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    In uniform circular motion the acceleration is not zero. The acceleration is however orthogonal to velocity in uniform circular motion.

    Try doing what the hint says first. You're making this way more complicated.
     
  5. Aug 25, 2008 #4
    I'm having trouble determining what must be the derivative of v dot v. Would the derivative of v dot v be dv/dt*v+v*dv/dt=0 => 2*(dv/dt *v)=0?, since v is constant , therefore dv/dt=0 and thus 2*(dv/dt*v) must be zero, right?
     
  6. Aug 25, 2008 #5
    The problem doesn't say anything about the particle being in uniform circular motion, problem only states that v is constant, and therefore if their is no change in velocity, the acceleration o the particle must be zero.
     
  7. Aug 25, 2008 #6

    Kurdt

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    I was responding to the first paragraph in your attempt at the answer.

    Anyway if the derivative of [itex] \mathbf{v} \cdot \mathbf{v} = v^2 [/itex] is almost as you have it.

    [tex] \mathbf{v} \cdot \frac{d\mathbf{v}}{dt} + \frac{d\mathbf{v}}{dt} \cdot \mathbf{v} = 2v \frac{dv}{dt} [/tex]

    As you have said the derivative of the speed must be zero if it is constant and if that is true how will the left hand side be made zero? Use Doc Al's hint above.

    EDIT: oops made the same speed/velocity mistake :uhh:
     
    Last edited: Aug 25, 2008
  8. Aug 25, 2008 #7

    Doc Al

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    Yes, but it's essential to distinguish the velocity (a vector) from the speed (a scalar). In this problem you are told that the speed is constant, not the velocity. All the v's on the left side of your equation stand for velocity, not speed. (See Kurdt's post.)
     
  9. Aug 25, 2008 #8

    HallsofIvy

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    No, the problem said the speed was constant, not the velocity!
     
  10. Aug 25, 2008 #9
    since the acceleration is zero, then v*dv/dt+dv/dt*v=2*v* dv/dt => v*0+0*v=2*v*0=> 0=0 which implies acceleration vector is perpendicular to velocity vector
     
  11. Aug 25, 2008 #10

    Doc Al

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    No. The acceleration is NOT zero.

    [tex]\vec{a} = \frac{d\vec{v}}{dt} \ne 0[/tex]

    But:

    [tex]\frac{dv}{dt} = 0[/tex]
     
  12. Aug 25, 2008 #11
    so if dv/dt ,without the vector, is not did acceleration, is dv/dt the magnitude of the acceleration?
     
  13. Aug 25, 2008 #12

    Kurdt

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    Its the rate of change of speed. The question asks about the velocity and acceleration vector though. So how can the differentiated dot product side equal zero? Consider Doc Al's first hint.
     
  14. Aug 25, 2008 #13

    HallsofIvy

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    Not exactly [itex]|d\vec{v}/dt|[/itex] would be the magnitude of the acceleration. If the acceleration is not 0 then its magnitude also cannot be 0. Doc Al is using v to mean [itex]|\vec{v}|[/itex] so dv/dt is the "derivative of the magnitude of the velocity" which is NOT the same as the "magnitude of the derivative of the velocity". The order is important!
     
  15. Aug 25, 2008 #14

    Doc Al

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    No. v, without the vector sign, stands for speed. dv/dt is the rate of change of the speed.
     
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