# Homework Help: Particle, slide, forces

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1. Dec 7, 2014

### 'nuck

1. The problem statement, all variables and given/known data
A particle of mass M is released from rest at the top of a frictionless slide that is h distance from the ground. The lowest section of the slide is part of a circle with radius R. The setup looks like a candy cane. At its lowest point (bottom of circle) what is the normal force acting on the particle?

2. Relevant equations
F = ma
N = mg

These seem to be relevant...but I'm not sure.

KE = (05)Iw2
PE = mgh
L = Iw
I(particle) = mr2
T(net) = Ia

3. The attempt at a solution
KEi = (.5)m(0)^2 = 0
KEf = (.5)mvf2 + I(v2/r2)

By conservation of Energy KEf = KEi but this doesn't tell me anything about the forces on the particle.

If the particle is at the bottom then I don't see why N will be equal to other than just mg.

2. Dec 7, 2014

### Stephen Tashi

Kinetic energy isn't conserved. Conservation of energy says total energy is conserved.
Initially the particle has potential energy equal to the work needed to lift it to a height $h$.
When it is at the lowest point on the circle, all that potential energy has been converted to kinetic energy. Knowing this, you can solve for the magnitude of its velocity at that that lowest point.

Centripetal force is usually presented as an object attached to a string that is going in a circle. You should have studied the formula for the force on the string. When the string is absent the force to hold the particle in a circular path must be supplied by something else. In this problem that force is supplied by the surface of the circular slide.
.

$F = ma$ with $F$ and $a$ being vectors. If particle was sitting still at the bottom of the track there would be zero net force on it. The normal force of the track would cancel the downward force of gravity. Since the particle is moving on a circular slide, it's velocity vector is changing direction, so it has a non-zero acceleration. Hence the net force on it can't be zero. The velocity vector of the particle is horizontal and is changing so it begins to slant upwards. Hence there must be some net upward force on it. The track must exert more force than gravity.