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 Object sliding (Posted on 2006-08-01)
An object is sliding from top of a ball to the ground. This object starts from rest and slides without friction. At what height will the object get apart from the ball's surface if the radius of the ball is r meters?

 No Solution Yet Submitted by atheron Rating: 4.0000 (1 votes)

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 Two methods | Comment 1 of 8

As the object slides down the side of the ball it builds up speed.  At the same time, if the gravitational force is resolved into components perpendicular to and parallel to the surface of the ball, the component perpendicular to the ball will get smaller and smaller.  As a result, the centrifugal effect will be increasing while the centripetal force will decrease. At the point where the centripetal force (the gravitational component perpendicular to the surface) is no longer sufficient to counteract the centrifugal effect, the object will go off on a tangent.  The point sought is the point where the centripetal force exactly balances the centrifugal effect.

These two balance when F = m v^2 / r, or, considering acceleration, A = v^2 / r, which could be considered F = v^2 / r for a unit mass (1 slug, in the English system).

Method 1:

The gravitational acceleration near the earth's surface is conventionally called g.  At a given stage of the object's slide, this can be resolved into two components: a1 parallel to the surface of the ball (that is, tangential), and a2, toward the center of the ball (this acceleration a2, is not actually realized as it is opposed by the solidity of the ball).

a1 = g sin(d/r)
a2 = g cos(d/r)

where d is the distance, along the surface of the ball, of the object from the top of the ball, and angular measure is in radians.

Initially the object is in equilibrium (albeit unstable) at the top of the ball.  In theory it would stay there, so any solution must consider some initial tiny offset from the top, to get the slide started.

Some differential equation could be set up and solved for v^2 / r = g cos(d/r), but that seems to be beyond me, or as they say in textbooks, left as an exercise for the reader.  The integration can be done numerically, as by the following program:

DEFDBL A-Z
pi = ATN(1) * 4
g = 32
r = 2
incr = .000001
d = r / 5000
dp = 0

DO
t = t + incr
a1 = g * SIN(d / r)
a2 = g * COS(d / r)
dpNext = dp + a1 * incr
d = d + incr * (dp + dpNext) / 2
dp = dpNext
PRINT USING "###.#### ###.#### ###.#### ###.####"; t; d; dp; 180 * d / (r * pi)
centrifAcc = dp * dp / r
IF centrifAcc >= a2 THEN EXIT DO
LOOP
PRINT USING "###.####"; 1 + COS(d / r)

At each step (time increment) in the integration, the program reports the elapsed time, the circumferential distance traveled, the circumferential speed and the angle from the top to the object, converted to degrees. The program uses English measure rather than metric, but that conversion can be taken care of later.  At the end, it reports 1+cos(d/r), as the distance from the support of the ball is r(1+cos(d/r)), so that 1+cos(d/r) is the ratio of this height to the radius of the ball.

When run as above, with r=2 and g=32, the last two lines are:

2.2630   1.6821   6.5320  48.1898
1.6667

indicating that 2.2630 seconds after the start of the slide, the object was 48.190 degrees away from the top of the ball when it lost contact with the surface, at which point it was 1.6667 r feet away from the ball's supporting surface.  That sounds like a value of 5 r / 3.

When rerun with a spherical radius of 5, the final two lines are:

3.5782   4.2053  10.3280  48.1898
1.6667

indicating this time it took 3.5782 seconds, but was still 48.190 degrees from the top, and therefore still 5 r / 3 feet above the floor or table top. (Works in meters also, when r is measured in meters, of course.)

In fact, it does not matter if g is 32 or some other value. After all, nothing in the formulae depended on the English system being used, so independence from g is to be expected. The times of course are affected by g.

Method 2:

Kinetic energy is given by m v^2 / 2.

Gravitational potential energy is given by m g h.

If x is the height above the tabletop that we are seeking, then the reduction in height (h) is given by 2r-x, so that the reduction in potential energy between the start and the point we are seeking is mg(2r-x).  This potential energy is what provides all the kinetic energy the object has at that point. Setting mv^2/2 = mg(2r-x) gives v^2 = 2g(2r-x).

Since we want v^2 / r = g cos(d/r) in order for centripetal force to match the centrifugal effect, we want

2g(2r-x)/r = g cos(arccos((x-r)/r)) = g(x-r)/r
or
2(2r-x) = x-r
4r-2x = x-r
5r = 3x

x = 5r/3

By the way, if 5/3 is considered 1+cos(d/r), then d/r is about .8410686705679302 radians, or, converted to degrees, 48.1896851042214 degrees.

This all assumes that the size of the ball is sufficiently small that:

1. The gravitational force by the ball on the object is negligible.
2. Gravitational acceleration, g, is practically the same at the top of the ball as at the bottom.

 Posted by Charlie on 2006-08-01 16:38:39

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