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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.

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- Thread starter oracleoflight
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- #1

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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

rcgldr

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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.

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rcgldr

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bottom: T-mg=(mv^2)/r

top: T+mg= (mv^2)/r

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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.

- #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]).

[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]).

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- #8

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bottom: T-mg=(mv^2)/r

top: T+mg= (mv^2)/r

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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