Two sliding masses problems

[diablo2121]

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?

 

[PhanthomJay]

One question at a time, please (you'll get better responses that way).

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.

 

[baba89]

Your approach is correct, but you have made a mistake in your calculations. The height h should be equal to v^2/2g, not v^2/2g. This is because the velocity v is not equal to the velocity at the bottom of the incline, but rather the velocity at the top of the loop. This means that the height h should be equal to v^2/2g, where v is the velocity at the top of the loop.

To find the velocity at the top of the loop, you can use the conservation of energy principle. The total energy at the top of the loop is equal to the potential energy at the top of the loop (mgh) plus the kinetic energy at the top of the loop (1/2 mv^2). Setting this equal to the potential energy at the bottom of the incline (mgh) gives you the equation mgh = mgh + 1/2 mv^2. Solving for v^2, you get v^2 = 2gh. Plugging this into the equation for h, you get h = 2gh/2g = gh/g = h. This means that the initial height h should be equal to the radius of the loop, which is 3r.