Circular Motion Applications

This lesson brings together the theory of circular motion with challenging application problems: the loop-the-loop, particles on the inside of bowls, and combined problems involving energy and forces.

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1 Loop-the-Loop

A particle slides down a slope and enters a circular loop of radius r. The key question: what is the minimum height h from which the particle must start to complete the loop?

Analysis

At the top of the loop (height 2r from the bottom):

For the track to maintain contact: N + mg = mv_top^2/r, where N >= 0.

Critical case: N = 0, v_top^2 = gr.

Energy conservation from start (height h) to top of loop (height 2r):

mgh = (1/2)mv_top^2 + mg(2r)

gh = (1/2)gr + 2gr = (5/2)gr

h = 5r/2

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2 Worked Example 1

A marble rolls (without sliding) from height h down a smooth track into a vertical loop of radius 0.3 m. Find the minimum h for the marble to complete the loop (treating the marble as a particle).

h = 5r/2 = 5(0.3)/2 = 0.75 m

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3 Worked Example 2

A particle enters the bottom of a loop of radius 0.5 m at speed 4 m s^(-1). Does it complete the loop?

v_top^2 = u^2 - 4gr = 16 - 4(9.8)(0.5) = 16 - 19.6 = -3.6

Since v_top^2 < 0, the particle does not reach the top. It does not complete the loop.

The maximum height reached: (1/2)(4^2) = g*h, h = 16/(2 * 9.8) = 0.816 m. Since 2r = 1 m > 0.816 m, confirmed.

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4 Particle on the Inside of a Smooth Hemisphere

A particle is placed on the inside surface of a smooth hemispherical bowl of radius r and released from the rim.

At angle theta from the vertical (measured from the centre):

Energy: (1/2)mv^2 = mgr(1 - cos theta) (taking bottom as GPE = 0, but theta is from the vertical at the centre)

Wait -- let us set this up carefully. Place the origin at the centre of the hemisphere. The rim is at the same height as the centre. The bottom is at depth r below the centre.

If the particle starts at the rim (height 0 relative to the centre, or height r above the bottom):

At angle theta below the rim: height above the bottom = r - r cos theta (if theta is measured from the vertical at the top).

Actually, let us use the standard setup: theta measured from the vertical downward from the centre.

At angle theta from the vertical: the particle is at height r cos theta above the bottom.

Starting from the rim (theta = pi/2, height = 0 above the bottom... no).

Let me use a clearer approach. Let the bottom of the bowl be the zero of GPE.

At the rim: height = r (the rim is level with the centre, which is r above the bottom).

At angle theta from the downward vertical through the centre: height = r - r cos theta above the bottom...

Actually: if theta is the angle from the downward vertical, then the particle is at a point where its height above the bottom is r(1 - cos theta). When theta = 0, the particle is at the bottom (height 0). When theta = pi/2, the height is r (the rim).

So if released from the rim (theta = pi/2, height = r):

(1/2)mv^2 = mg[r - r(1 - cos theta)] = mgr cos theta

v^2 = 2gr cos theta

Normal reaction (radially): N - mg cos theta = mv^2/r

N = mg cos theta + m(2gr cos theta)/r = 3mg cos theta

The particle leaves the surface when N = 0. But N = 3mg cos theta >= 0 for 0 <= theta <= pi/2. So the particle never leaves the bowl (it just oscillates).

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