# Loop-the-loop - constant velocity assumed?

Oops! That title should read constant *speed.*

1. An object with mass m starts at rest at height h. It slides down a frictionless incline. At the bottom, it enters a vertical circular loop-the-loop of radius r. It just touches the track at the top of the loop. What is h in terms of m and r?

2. a = v^2/r, conservation of energy, etc.

3. This is of course a classic exercise. I have this question about the solution: The solutions determine the velocity at the top of the loop by using the equation a = v^2/r. My basic textbooks present this formula derived in the context of *uniform* circular motion. The speed is not constant here, is it? How is it valid to apply that formula here?

(BTW, I am not a student; I am revisiting basic physics here. I haven't taken a physics class in over thirty years.)

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Velocity is never a constant in a loop/circle when the object is moving.
Uniform circular motion means the acceleration is constant, which it is :)

I didn't say constant velocity. I said constant *speed*, which is what the textbooks assume.

For all intents and purposes, speed is velocity

Doc Al
Mentor
I have this question about the solution: The solutions determine the velocity at the top of the loop by using the equation a = v^2/r. My basic textbooks present this formula derived in the context of *uniform* circular motion. The speed is not constant here, is it? How is it valid to apply that formula here?
While it might typically be derived in the context of uniform circular motion, it applies to any circular motion. It gives the centripetal component of acceleration at any point along the path. If the speed is changing, there will also be a tangential component.

gneill
Mentor
Hello herschko. Welcome to Physics Forums.

It turns out that the formula a = v^2/r is valid at any point along a trajectory, provided that v is the speed in the direction of motion (tangent to the curve) and r is the instantaneous radius of curvature of the trajectory. How one determines r for a general curve is another matter... but it's simple for a trajectory constrained to a circle!

EDIT: And... Doc Al scoops me again! Well done that man!

Studying the text, it seems to me that in the derivation of the formula, one is using the fact that |delta v| / |v| = |delta r| / |r|, where all the letters represent vectors. This coems from the fact that the speed is constant. In the problem at hand, could it be be we are using the fact that the horizontal acceleration is 0?

BTW, I have edited the original post: The title should read constant *speed*, not constant velocity.

gneill
Mentor
Studying the text, it seems to me that in the derivation of the formula, one is using the fact that |delta v| / |v| = |delta r| / |r|, where all the letters represent vectors. This coems from the fact that the speed is constant. In the problem at hand, could it be be we are using the fact that the horizontal acceleration is 0?

Could be, I don't know if I've seen the text you're referring to, or the particulars of the derivation. But I have seen derivations regarding the decomposition of acceleration into tangential and "centripetal" components for a general trajectory. This involves ascribing an instantaneous center of curvature to the curve at any given location, and the resulting centripetal acceleration, vis v2/r, holds for every point on the trajectory.

From http://en.wikipedia.org/wiki/Mechan...coordinates_in_an_inertial_frame_of_reference , I see where we can come to this conclusion. v = rdot rhat + r theta-dot theta-hat and a = (r-double-dot - r theta-dot^2) rhat + ( other terms ) theta-hat. For motion in a circle, rdot = r double-dot = 0. There is no horizontal force, hence no theta-hat component to the acceleration at the top of the loop. This leaves v = r theta-dot theta-hat and a = -r theta-dot^2 rhat, and the relationship a = v^2/r follows.

But now I have a serious criticism of the pedagogy here. This problem is in many first-semester general physics classes. The expected solution uses the relationship a = v^2/r, which has been presented only in the context of uniform motion. (see the links at https://www.google.com/search?q=physics+loop-the-loop+problem). The student is being asked to apply a result which does not apply to the situation at hand without further considerations. Not good in my opinion.

This consideration came up when I was helping a student in a non-calculus Physics I class. The students therein have learned nothing of polar coördinates or the kinematics of motion on a curve other than uniform circular motion.

This is a uniform circular motion...

This is a uniform circular motion...

No, it is *not.* Uniform circular motion refers to constant speed. The speed varies around the loop. a does not equal v^2/r anywhere except at the top or bottom of the loop.

I've learned that uniform circular motion is when the acceleration is constant.

Doc Al
Mentor
The student is being asked to apply a result which does not apply to the situation at hand without further considerations. Not good in my opinion.
Standard derivations of the centripetal acceleration formula only involve a small segment of the circular path, not the entire circle. Over that small segment, the speed can be considered constant. So it does apply.

I agree that this should be made clear to students.

Doc Al
Mentor
I've learned that uniform circular motion is when the acceleration is constant.
In uniform circular motion, the magnitude of the acceleration is constant. As herschko has explained, this is not an example of uniform circular motion.

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