What Initial Height Ensures a Block's Force Equals Its Weight at the Loop's Top?

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To ensure a block of mass m pushes against the top of a loop with a force equal to its weight, the initial height h must be calculated correctly. The equations of potential energy (U = mgh) and kinetic energy (K = 1/2 mv^2) are essential for this analysis. The net force at the top of the loop includes both the gravitational force and the normal force, which must be accounted for in the calculations. The correct relationship shows that the speed at the bottom of the loop differs from that at the top, leading to the conclusion that the initial height h should be 3r, not r/2. Understanding these dynamics is crucial for solving the problem accurately.
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Homework Statement


A block of mass m starts from rest from a height h and slides down along a loop with radius r. What should the initial height h be so that m pushes against the top of the track with a force equal to its weight? (ignore friction)

Homework Equations


U = mgh
K = 1/2 mv^2
F = mg = mv^2/r

The Attempt at a Solution


So the force should be equal to its weight (mg) which I set equal to mv^2/r. So at rest, m has a potential energy of mgh. At the bottom on the incline before entering the loop, m has a kinetic energy of 1/2 mv^2. Since there is no external forces, U = K. Solving the force equation, I got v^2 = gr. The height h = v^2/2g. My answer ends up being h = r/2, but I have the correct answer, 3r. What am I missing here?
 
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One question at a time, please (you'll get better responses that way).
diablo2121 said:

Homework Statement


A block of mass m starts from rest from a height h and slides down along a loop with radius r. What should the initial height h be so that m pushes against the top of the track with a force equal to its weight? (ignore friction)

Homework Equations


U = mgh
K = 1/2 mv^2
F = mg = mv^2/r
What force is this?

The Attempt at a Solution


So the force should be equal to its weight (mg) which I set equal to mv^2/r.
The net force acting at the top of the loop is equal to mv^2/r. One of the forces acting is the normal pushing force on the block , given as mg. There is another force acting on the block also. You must include it in determining the net force.
So at rest, m has a potential energy of mgh. At the bottom on the incline before entering the loop, m has a kinetic energy of 1/2 mv^2. Since there is no external forces, U = K. Solving the force equation, I got v^2 = gr. The height h = v^2/2g. My answer ends up being h = r/2, but I have the correct answer, 3r. What am I missing here?
You should note that the net force at the top of the circle is more than just mg; and that the speed of the block at the bottom of the loop (bottom of incline) is not the same as its speed at the top of the loop. You might want to consider writing your U=K equation at the top of the loop, after drawing a sketch and identifying all the forces acting on the block at the top of the loop.
 
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