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

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SUMMARY

The discussion centers on determining the initial height \( h \) required for a block of mass \( m \) to exert a force equal to its weight at the top of a loop with radius \( r \). The correct initial height is established as \( h = 3r \), contrary to the incorrect calculation of \( h = r/2 \). Key equations utilized include gravitational potential energy \( U = mgh \), kinetic energy \( K = \frac{1}{2} mv^2 \), and the force relationship \( F = mg = \frac{mv^2}{r} \). The solution emphasizes the need to account for the net force at the loop's top, which includes both gravitational and normal forces.

PREREQUISITES
  • Understanding of gravitational potential energy (U = mgh)
  • Knowledge of kinetic energy (K = 1/2 mv^2)
  • Familiarity with circular motion dynamics (F = mv^2/r)
  • Ability to analyze forces acting on an object in motion
NEXT STEPS
  • Study the conservation of mechanical energy in systems involving potential and kinetic energy
  • Learn about forces acting on objects in circular motion, particularly at the top of loops
  • Explore the implications of normal force in dynamic systems
  • Review problem-solving techniques for physics problems involving energy and forces
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of problem-solving in circular motion dynamics.

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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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