# 17.40 Rolling

Before you read this, I suggest you read posts 17.38 and 17.39.

We are all familiar with rolling objects (a ball rolling along the ground or the wheel of a moving car); as a result, we tend not to think much about rolling and may not appreciate how complicated it is.

The picture above shows the point P on a wheel whose centre is at C. Initially P is on the ground but some time, t, later the wheel has rolled and P has moved to P’. This rolling motion consists of two parts. One: C has moved to C’ along a line of length vt, where v is the speed of the wheel (post 17.4). Two: P has rotated around C by an angle ωt where ω is the angular speed of the wheel (post 17.12). We can see that rolling is a combination of translation and rotational motion (post 17.39). Note that v and ω are not independent because v = ωr (post 17.12), where r is the radius of the wheel.

The picture above shows how the position of P (a green dot on each position shown for the red wheel) depends on time; the blue curve passes through all the positions of P and is called a cycloid. The appendix shows how this curve can be calculated.

What is the kinetic energy of an object of mass m rolling with a constant speed v? You might expect it to be ½mv2, for the reasons explained in post 16.21. But it isn’t. To explain why, we’ll think about a ball of radius r rolling with a constant speed v. The ball has translational kinetic energy ½mv2 but it also has rotational kinetic energy ½Iω2 (posts 17.38 and 17.39), where I is its rotational inertia (post 17.38) and ω is its angular speed (post 17.12). So its total kinetic energy is

K = ½mv2 + ½Iω2.

But, since v = ωr (post 17.12) and, for a sphere I = (2/5)mr2,

K = ½mv2 + ½(2/5)mr2(v/r)2 = ½mv2 + (1/5)mv2 = (5/10)mv2 + (2/10)mv2 = (7/10)mv2

which is higher than the kinetic energy of a sphere sliding with the same speed.

So rolling objects have more kinetic energy than objects undergoing translational motion with the same speed.

Related posts

Appendix

The picture above shows that the coordinates of P’, relative to P, are given by

x = vtrsin(ωt) and y = rrcos(ωt).

Since v = ωr (post 17.12),

x = ωrtrsin(ωt) = r(ϑ – sinϑ) and y= r(1 – cosϑ)

where I have defined ϑ by ϑ= ωt. (See post 16.50 appendix 3 for definitions of sin and cos). These two equations define x and y in terms of a third variable ϑ. If we assign a value to ϑ, the corresponding values of x and y can be calculated. A pair of equations like this is called the parametric equations of the cycloid.

I first learnt about the parametric equations of a cycloid when I was 17 years-old. I was taught that the way to deal with these parametric equations was to eliminate ϑ and so find y in terms of x as the only variable. After a lot of algebra, you get the result that

x = racos(1 – y/r) – {y(2ry)}1/2

where arcos(u) is the angle whose cosine has the value u.

But now we have personal computers and life is a lot simpler. To plot a cycloid, I note that it must repeat itself every 360o (one revolution of the wheel) and so assigned ϑ the values in the spreadsheet above. These values were used to calculate corresponding values of x and y. The results are scaled by r, the radius of the wheel, that I assigned a value of 1. Plotting y against x gives the cycloid shown below. Increasing the range of ϑ values would have given more cycles of the cycloid.

If you want to investigate parametric equations further, I suggest you look at the pair

x = rcos ϑ and y = rsinϑ.

You might be able to recognise what shape these equations represent straight away. But you can assign values to ϑ and so plot y against x. If you want to eliminate ϑ from the two equations, try squaring them and adding the results (you might want to look at post 16.50 appendix 4). If you do this and want to plot the curve, remember that 1 has two square roots – +1 and -1.