Mathematics as the language of physics

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It is often said that “mathematics is the language of physics“. But what does this mean? In this post I’m going to try to explain, by using one of my favourite proofs as an example.

The Problem

The problem is this:

A ball is placed atop a sphere and released. At what angle to the vertical does the ball lose contact with the sphere?

You would be excused for thinking that the answer is 45° or 90°, but the correct answer is more complicated than that.

The Physics

For an object to move along a curved path there must be a centripetal force acting upon it, directed towards the centre of the curved path. In the case of the problem above, this centripetal force is provided by the component of the ball’s weight that acts towards the centre of the sphere.

Once the ball is released, its gravitational potential energy will be converted to kinetic energy as it falls and its speed increases. As its speed increases, the amount of centripetal force required to keep it moving along a curved path increases.

When the ball is on top of the sphere its entire weight is acting downwards, but as it rolls down the surface of the sphere the component of its weight acting towards the centre of the sphere decreases (and would be zero after it has fallen through 90°). The solution to this problem is to find the point at which the centripetal force required to keep the ball moving along the surface becomes greater than the component of the ball’s weight towards the centre of the sphere.

The Maths

The centripetal force on an object with mass m, moving at speed v along a curved path of radius r is:

F_C = \frac{mv^2}{r}

Measuring angles from the vertical, the component of the ball’s weight towards the centre of the sphere is:

F_N = mg\cos\theta

Where the subscript N indicates the force is normal to the surface of the sphere; m is the mass of the ball, and g is the strength of the gravitational field.

Therefore the ball will lose contact with the sphere when:

\frac{mv^2}{r} = mg\cos\theta

Note that the mass term appears on both sides of the equation and will therefore cancel out, which tells us that the solution to this problem is mass invariant – the answer does not depend on the mass of the ball. There are a lot of unknowns in the equation above, which we would like to get rid of, and we need to work towards an answer that has the angle between the ball and the vertical its subject.

The amount of kinetic energy gained by the ball as it falls is equal to the amount of gravitational potential energy it has lost therefore:

\frac{1}{2}mv^2 = mgh

Where h is the height through which the ball has fallen, and the other symbols have the same meanings as previously.

We now rearrange the equation above to get an expression for the speed of the ball:

v^2 = 2gh

and substitute this into our equation for centripetal force:

\frac{m\cdot 2gh}{r}=mg\cos\theta

The mass terms are again eliminated, as we would expect, as are the the gravitational field strength terms, which means that our solution is also independent of the strength of the gravitational field – if we carried out this experiment on Earth and on the Moon we would get the same answer.

\frac{2h}{r}=\cos\theta

The height h through which the ball has fallen is a function of the angle through which the ball has moved:

h = r - r\cos\theta

Which is much easier to deal with if we simplify it to:

h = r \left( 1 - \cos\theta \right)

We now substitute this value for h into our initial centripetal force equation to yield:

\frac{2r \left(1- \cos\theta \right)}{r} = \cos\theta

The radius of the sphere (r) terms cancel out, and we are finally left with an equation containing only one unknown quantity, the angle that we are trying to find:

2 - 2\cos\theta = \cos\theta

Which becomes:

\cos\theta = \frac{2}{3}

And which we can solve to yield our final answer:

\theta = \cos^{-1} \left(\frac{2}{3}\right) = 48.2^{\circ}

So there you go. The ball will lose contact with the sphere after it has fallen through 48.2°.

The sharper-eyed of you might have noticed that I have ignored the radius of the ball and treated it as a point mass. As it turns out, and as with the radius of the sphere, the radius of the ball is cancelled out, so I left it out for the sake of clarity. The answer of 48.2° therefore applies to any two balls, of any mass, in any gravitational field strength.

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6 thoughts on “Mathematics as the language of physics”

  1. Isn’t friction between the surfaces of the ball and the sphere a factor in the equation?

  2. Yes, but it would be rolling friction, which is very small, so the standard physicists’ friction-doesn’t-exist approximation applies quite well here.

  3. The conclusion that the mass of the sphere is irrelevant.

    Experiment: Place a ball on the surface of the Earth. It won’t go anywhere.

  4. Very true. I’m assuming here that gravitational effects are negligible. I don’t think that’s too unfair, given the context.

  5. Pingback: Link of the Day: A Ball Is Placed Atop a Sphere and Released. At What Angle to the Vertical Does the Ball Lose Contact With the Sphere? – IMAO

  6. This fails to take into account that some of the gravitational potential energy will be turned into rotational kinetic energy, not just kinetic energy.

    The numbers appear to be right for a ball with all of its mass at the center, but a ball of uniform density would be able to maintain contact for longer because it would have less translational velocity, and a hollow ball would be able to hold contact for even longer for the same reason.

    It would no longer be a ball, but if you added some attachments to the ball to make I >> m, you can have the new object maintain contact with the sphere for a much longer time, but never reaching 90 degrees of course.

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