Where a block stops on a loop-the-loop

[kinof]

Homework Statement

A small block of mass m slides along a frictionless loop-the-loop track. The block is released from rest at point P. What is the resultant force acting on it at point Q? At what height above the bottom of the loop should the block be released so that it is on the verge of losing contact with the track at the top of the loop?

Homework Equations

Wnc = ΔK + ΔU
Wc = -ΔU

The Attempt at a Solution

Point P is at the top of the loop 5R distance from the ground, and point Q is a point on the loop R above the ground and R from the center of the loop where the normal force is parallel to the surface. (Sorry if the image is a bit difficult).

I have found the force at point Q, so I am stuck on the second part.

This is how I have set it up.

From point P to point S (where N = 0) and where H is the height at which N = 0:

v^2 at point S is gR, so

0 = ΔK + ΔU = (1/2)mgR + (mgH - 5mgR)

(1/2)R + H = 5R

H = 4.5 R.

However, the correct answer is 2.5R. I don't understand, because ΔE = 0 the entire time because of the absence of friction, so can't I start at whatever point I want and so long as I have what the velocity is where N = 0, I can find the distance from the ground where that is?

Thanks.

 

[anathema]

Thank you for your question. I understand that you have found the force at point Q, but are having trouble with the second part of the problem. I would like to offer some clarification and guidance to help you find the correct answer.

Firstly, let's define some variables to make the problem easier to understand. Let h be the height at which the block is released from point P, and let H be the height at which the normal force becomes zero at point S. Additionally, let R be the radius of the loop-the-loop track.

Now, let's look at the energy conservation equation you used: Wnc = ΔK + ΔU. This equation tells us that the work done by non-conservative forces (such as friction) is equal to the change in kinetic energy plus the change in potential energy. However, in this problem, there are no non-conservative forces acting on the block, so we can simplify this equation to: ΔK + ΔU = 0.

Next, let's look at the work-energy theorem, which states that the work done by all forces acting on an object is equal to the change in kinetic energy. In this problem, the only force acting on the block is gravity, which is a conservative force. Therefore, the work done by gravity is equal to the change in kinetic energy: Wc = ΔK.

Now, let's use these equations to find the height at which the block should be released so that it is on the verge of losing contact with the track at the top of the loop. We can set up the following equation:

Wc = ΔK = (1/2)mv^2 = mgh

Where v is the velocity of the block at point S, and h is the height at which the block is released. We know that the velocity at point S is equal to the square root of the product of the acceleration due to gravity (g) and the radius of the loop (R): v = √(gR).

Therefore, our equation becomes:

mgh = (1/2)m(gR)^2

Solving for h, we get:

h = (1/2)R

Therefore, the block should be released from a height of (1/2)R above the bottom of the loop in order to be on the verge of losing contact with the track at the top of the loop.

I hope this helps you understand the problem better and find