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Centripetal Motion

  1. Nov 12, 2011 #1
    I have a question about the difference between tangential acceleration and centripetal acceleration. I understand that in centripetal motion, centripetal acceleration is toward the center of a circle. However perpendicular to that is tangential acceleration which depends on tangential velocity. So as one is going up the circle the tangential velocity decreases and is a minimum at the top of the circle. Now this tangential velocity is decreasing in speed due to the force of mgsin(theda) countering it on the way up. Since acceleration is the derivative of velocity, isn't tangential acceleration a min at the top of the circle as well? If not, then where would it be a min?

    Since centripetal acceleration is the acceleration toward the center of the circle, is it ever a minimum or is it constant? What about the component of centripetal velocity? Is it a minimum ever?

    Thanks and sorry for the questions.
     
  2. jcsd
  3. Nov 13, 2011 #2

    rcgldr

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    m g sin(θ) is the component of gravitational force parallel to a plane with an angle of θ from horizontal. In your question about circular motion, you've mentioned a gravitational component of force, but did not explain what is restricting the moving object into a circular path. Perhaps you're thinking of an idealized case of a block sliding on a frictionless circular surface, with no external forces (like aerodynamic drag) other than gravity. For this situation, the minimal speed would occur at the top of the circle and the maximum velocity at the bottom of the circle. Gravity would always exert a downwards force of m g on the object. The frictionless circular surface would exert a centripetal force = v2 / r on the object, which would vary with the speed2 of the object. The assumption is the object is moving fast enough that it stays in contact with the frictionless circular surface at the top of it's circular path. Is this the situation you're trying to describe?
     
  4. Nov 13, 2011 #3
    Exactly. I am sorry that I did not clarify.

    Would you please be able to explain the interplay between tangential acceleration and velocity in conjunction with centripetal acceleration and velocity? In the idealized case that I discussed, when are each of them a minima and maxima and why. Thank you.
     
  5. Nov 13, 2011 #4

    rcgldr

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    As I mentioned before tangental acceleration is at a maximum on the left and right of the circle, and at a minimum (zero) at the top and bottom. Velocity and centripetal acceleration are at a minimum at the top, and a maximum at the bottom.
     
  6. Nov 13, 2011 #5
    I am a little confused why centripetal acceleration is a max at the bottom if you have:

    bottom: T-mg=(mv^2)/r
    top: T+mg= (mv^2)/r
     
  7. Nov 13, 2011 #6
    If we are talking about a rollercoaster going around a circular loop, then I believe you have those equations backwards . In the rollercoaster case, the restraining normal force aways points radially outward and gravity always points downwards.

    So it would be

    N - mg at the top
    N + mg at the bottom

    Most variations of this problem have been examined in detail and not just at the top and bottom points either.
     
  8. Nov 13, 2011 #7

    BruceW

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    If we have an object, whose position is given by [itex]r(cos(\theta) \hat{i} +sin(\theta) \hat{j})[/itex] then the acceleration of that object is: [itex](\ddot{r} - r \dot{\theta}^2) \hat{r} + (r \ddot{\theta} + 2 \dot{r} \dot{\theta}) \hat{\theta}[/itex] Now if we are interested in a cart on a circular track, we can specify that its radius is constant, which makes the acceleration equal to:
    [tex] -r \dot{\theta}^2 \hat{r} + r \ddot{\theta} \hat{\theta}[/tex]
    And we also know there are two forces acting on the cart: first is the normal force on the cart, which acts radially inward on the cart [itex]-N_{(t)} \hat{r}[/itex] (where N is the magnitude of the normal force, which is time-dependent generally). And we also have the gravitational force on the cart, which written in polar coordinates is equal to: [itex]-mg(sin(\theta) \hat{r} + cos(\theta) \hat{\theta})[/itex] Also we know that the total force on an object is equal to the acceleration of that object times its mass, we can combine the equation for the acceleration of the object with the equation for the forces on the object, to get two equations:
    [tex]mr \dot{\theta}^2 - mgsin(\theta) = N[/tex]
    [tex]r \ddot{\theta} = -g cos(\theta)[/tex]

    This is the general solution of the problem. The other equation would be energy conservation, I guess. Note that [itex]\theta= \frac{\pi}{2}[/itex] when the cart is at the top of the track. (Just to make sure it is clear what I mean by the angle [itex]\theta[/itex]).
     
    Last edited: Nov 13, 2011
  9. Nov 13, 2011 #8
    It seems that you are talking about motion on the inside of a vertical loop.
    In the equations above T and v are different for the two positions.
    They are actualy:

    bottom: T1-mg=(mv_1^2)/r
    top: T2+mg= (mv_2^2)/r

    Conservation of energy will provide a relationship between v1 and v2.
    v_1^2=v_2^2+2Rg and so v_1>v_2
     
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