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Homework Help: Torque and acceleration

  1. Nov 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the acceleration at the tip of the stick, ignore the effect of gravity.

    2. Relevant equations
    L = length of stick
    I = mL2/3; When the stick is swung about one end
    ω = -dθ/dt; This is the current angular velocity
    v is the current velocity of the tip

    3. The attempt at a solution
    If there is no applied torque, the tip would only have the centripetal acceleration ac, which has magnitude rω2 and points towards the hinge. With the torque, there is addition acceleration at, which points in the same direction as v.

    Since τ = I α, and at = L α, at = τL/I = 3τ/mL.

    The overall accerelation a = ac + at.
    ax = 3τ/mL cos(θ) + rω2 sin(θ)
    ay = 3τ/mL sin(θ) - rω2 cos(θ)

    Since ac was pointing towards the hinge and at is pointing at a tangent, a will be point at somewhere in between. It is obvious whether the tip is still following the circular path. What could I do to check the answer?

    - Thanks
  2. jcsd
  3. Nov 22, 2008 #2
    The tip is constrained by the stick to be a distance L from the center, so it cannot do anything but follow a circular path :smile:
  4. Nov 22, 2008 #3
    Do you mean that if the constrain L is not there, the tip would follow a path that is not circular?

    I.e.: if I were to integrate ax to get vx, and then integrate vx again to get px, I would not get a circular path (perhaps a spiral path); the tension in the stick would keep increasing under the applied torque until the stick breaks (?).
  5. Nov 22, 2008 #4
    I noticed you are using Cartesian coordinates.The problem has circular symmetry so it is going to be much nicer, intuitive etc. to work out in polar coordinates. If you still want Cartesian coordinates at the end then convert back.

    If you aren't quite sure how to do that, you should look into it. It's one of those core pieces of knowledge that show up to confuse again and again.
  6. Nov 22, 2008 #5
    The short answer to your question though is that the acceleration is:

    \frac{v^2} {r}


    \frac{r \tau} {I}
    in the angular direction

    if you want the magnitude then you just square these, add them together and then take the square root.
    Last edited: Nov 22, 2008
  7. Nov 22, 2008 #6
    Right, there is no need for it to follow a circular path in general. It would still follow a circular path if the tension along the rod was always equal to the centripetal acceleration necessary for a circular path.

    If you do the integration with the equations as you have set them up, you will get a circular path. (This integration may not be easy, and you should really work in circular co-ordinates if you want to do this, as LogicalTime suggests). But this is because you have already made the assumption that the centripetal acceleration is always something that corresponds to a circular path.

    However the second part of your statement is correct. As the angular velocity keeps increasing, the tension required to maintain the centripetal force keeps increasing. Eventually, there will be a point when the stick can no longer provide enough tension, and will break.
  8. Nov 22, 2008 #7
    In polar coordinates, the velocity vector at the tip is:

    v(t) = Lω(t)eiθ(t) ...... (1)

    When I write equation (1), I am making the assumption that the length is fixed at L. Otherwise, L would be L(t). If this assumption holds, then the acceleration is:

    a(t) = dv(t)/dt

    = L ( dω(t)/dt eiθ(t) + ω(t)eiθ(t)(idθ(t)/dt) )

    = Lαeiθ(t) - i Lω2(t)eiθ(t) ...... (2)

    Equation (2) assumes that dω(t)/dt= α, and dθ(t)/dt= -ω (my original direction of ω was in the opposite of θ). The first part of Eq(2) looks like the effect due to the applied torque, this acceleration is tangent to the motion. The second part looks like the effect due to centripetal acceleration, which is towards the hinge. How do you interpret this result? Is there an easy way to read Eq(2) so that it is obvious that the tip is moving in a circle? (I tried integrating the cartesian equations, and it was a mess. Integrating in polar coordinate doesn't seem easier.)

    Is the second part of Eq(2) still the centripetal acceleration? Such that if you only focus on an instance, the centripetal acceleration is governed by the instantaneous ω(t), independent to any tangential acceleration (which is the first part of Eq(2))? So that there is an increasing force that is pulling the tip back to the center, and a tangential force that is accelerating the tip?
  9. Nov 22, 2008 #8
    since you want the acceleration at the tip just plug in L for your radius

    if you want Cartesian coordinates we have to do that conversion thing I mentioned

    What method are you using that involves [tex] e^{i \theta} [/tex] ?
    Last edited: Nov 22, 2008
  10. Nov 22, 2008 #9
    this form might be a bit clearer

    a(t) = r \dot{ \theta }^2 \hat{r} + r \ddot{ \theta} \hat{\theta}
    by definition:
    \omega = \dot{ \theta }
    \tau = I \ddot{ \omega }
  11. Nov 22, 2008 #10
    What is the convention for defining r hat and theta hat?

    This is the same problem, but now with gravity:

    Before the stick becomes vertical, the net torque is:

    τnet(t) = τ - (mg)(L/2)sin(θ(t))
    τnet(t) = I α(t)
    α(t) = 3( τ - (mg)(L/2)sin(θ(t)) )/( mL2 )

    I assume that the part of mg that is not tangential is canceled by a normal force at the hinge. v(t) is coupled to ω(t):

    v(t) = Lω(t)eiθ(t)

    a(t) is dv(t)/dt:

    a(t) = Lα(t)eiθ(t) + Lω(t)eiθ(t)(-iω(t))
    a(t) = Lα(t)eiθ(t) - i Lω2(t)eiθ(t)

    What is the force acting on the hinge?
    The radial part of mg:

    FN = mg cos(θ(t)) ei(θ(t)-π/2)
    FN = - i mg cos(θ(t)) eiθ(t)

    How do I know whether this FN is the only force acting on the hinge?
  12. Nov 22, 2008 #11
    I don't see where you get this part from :frown: It makes the rest of your algebra really messy. If you are not yet comfortable with polar coordinates, stick to Cartesian until you familiarize yourself.
  13. Nov 22, 2008 #12
    How do you write v(t) in polar coordinate?

    Like this:

    v(t) = Lω(t) r + θ(t) θ

  14. Nov 22, 2008 #13

    Look at the vector calculus part. Forget complex numbers, which is where you seem to have picked up the i.

    r is the unit vector from the origin in the direction of the radius, θ is perpendicular to r. Now can you figure out how the velocity vector is oriented?
  15. Nov 22, 2008 #14
    Then v(t) = Lω(t)θ
  16. Nov 22, 2008 #15
    Taking time derivatives with polar coordinates is pretty fundamental but a little tricky (often not taught too well either which doesn't help).

    Check the link above, it does a good job showing you how. With cartesian coordinates taking time derivatives is easy because the coordinates don't change with time. With polar, spherical, cylindrical coordinates the unit vectors do depend on each other though, but once you get this down so many things become easier like the Coriolis effect etc.
  17. Nov 22, 2008 #16

    and a(t) is pretty nasty, but much of it will go away since you are using a constant radius

  18. Nov 22, 2008 #17
    One further thing to keep in mind:θ is in the direction of increasing θ, and should be chosen so that θ is increasing in the counter-clockwise direction. You know that physically, the vector v points in the opposite direction, therefore ω(t) should be negative.
  19. Nov 22, 2008 #18
    these just represent unit vectors in the coordinate system we are using, you are representing them with bold and formally they are often represented with an e and subscript

    \hat{x} = \textbf{x} = \textbf{e}_{x}
    \hat{r} = \textbf{r} = \textbf{e}_{r}
  20. Nov 22, 2008 #19
    yeah θ is pointing in the counter-clockwise direction is the convention

    I don't understand the comment about v, if you initial conditions have the thing going CCW then v is positive and if the thing is going CW then it's negative but that's all a result of using the convention. I think that's what naresh means, am I correct naresh?
  21. Nov 22, 2008 #20
    Yes that's what I meant. The picture shows a clockwise angular velocity. Sorry if it was confusing.
  22. Nov 22, 2008 #21
    Cool, just making sure we're on the same page.
  23. Nov 22, 2008 #22
    Referring to this:


    v(t) = -Lω(t)θ

    It only has the tangential component.

    a(t) = dv(t)/dt

    To take the derivative, I treat that v(t) as a coordinate vector and bring it to the origin. Its radial component is Lω(t), it has no tangential component. (When I do this, wouldn't all vectors have 0 for θ?)

    a(t) = -Lα(t)r - Lω(t) (dθ(t)/dt) θ

    In this case, dθ(t)/dt is -ω(t), so

    a(t) = -Lα(t)r + Lω2(t) θ

    Why is the sign for θ wrong?
  24. Nov 22, 2008 #23
    The way I understand it now:

    v(t) = -Lω(t)θ

    a(t) = -L dω(t)/dt θ + -L ω(t) dθ/dt

    = -Lα(t)θ - L ω(t) ( -ω(t) (-r) )

    = -Lα(t)θ - L ω2(t) r

    This shows that the centripetal force points toward the center, and the tangetial force is still clockwise.

    dθ/dt = d ( -sinθ, cosθ )/dt = (-cosθ dθ/dt,-sinθ dθ/dt) = -ω(t)(-r) = ω(t)r in my case.
  25. Nov 22, 2008 #24
    yes, well done.
  26. Nov 23, 2008 #25
    yech, hard to read with all those ω(t) a(t) instead of having dots over the top. as long as it corresponds with the image I posted earlier you should be good. Thanks to naresh I can be lazy and not check.

    Now all you have to do is add on the gravity which from the image looks like it's going to be:

    F_{\theta} = -mg sin( \theta) \hat{ \theta }
    Last edited: Nov 23, 2008
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