# What is the speciality of the velocities of the constituent particles of a body in rolling motion?

Do we need to apply more energy to move a wheel (ie. Make the wheel roll) than a solid object (both with same masses) so that both of them have the same velocity assuming friction is negligible? (Since we need to make the particles of the wheel undergo both translational motion and rotational motion).

Also I have learnt that rolling motion is a combination of both translation and rotation. Does that mean that all the particles will have the same velocity as that of the centre of mass while they undergo translation? Does that mean all the particles will have more velocity than the centre of mass (due to the contribution from rotation )?

Am I correct in assuming that the excess energy we apply (assuming my first question was correct) simply is utilised to make an entirely independent rotational motion and that it, sort of, gets incorporated into a body with pure translational motion so as to create rolling motion?

A wheel has its mass distributed far away from its axis of rotation (i.e. its centre), whereas a solid cylinder has this mass distributed at essentially equal distances from this point.

Consider each object as being made up of lots of small parts of mass. Let's let these objects rotate at the same number of revolutions per second.

You know that the parts of a rotating body further from the axis of rotation need to move faster than closer parts. It is then clear to see that in the case of the wheel there are lots of parts moving quickly, and few moving slowly. Compare this to the solid cylinder, and we see that there are more parts moving slowly than in the case of the wheel. Remembering that each object has the same mass, we can then say that more mass is moving at higher speeds in the wheel, than in the cylinder.

Of course, kinetic energy is given by $$\frac{1}{2}mv^2$$, so clearly, if both the objects rotate at the same number of revolutions per second, the wheel has a greater kinetic energy.

We actually have a tool for doing this mathematically, called the moment of inertia. This is defined as:

$$I = \sum_i \delta m_i r_i^2$$

i.e. the sum of all the little parts of mass that make up a body multiplied by the square of their perpendicular distance $$r$$ from the axis of rotation.

We can also write this as an integral, if you are familiar with calculus:

$$I = \int_{M} r^2 dm$$

Anyway, you can actually quite easily use the first of these two formulae to see that the moment of inertia of a 'wheel' of mass $$M$$ and radius $$R$$ is given by $$I = MR^2$$. I will give you that for a cylinder, however, $$I = \frac{1}{2}MR^2$$.

Ok, so we've found the moment of inertia, but what use is it to us?

Well, kinetic energy can be expressed in terms of the moment of inertia. We know we can work out the total KE by summing over the KEs of all the little bits of mass:

$$E_K = \frac{1}{2} \int_{M} v^2 dm$$

but $$v = \omega r$$ in a rotating body, where $$\omega$$ is angular frequency, thus

\begin{align} E_K &= \frac{1}{2} \int_{M} \omega^2 r^2 dm \\ &= \frac{1}{2} \omega^2 \int_{M} r^2 dm \\ &= \frac{1}{2} \omega^2 I \end{align}

That is, the kinetic energy of a rotating body is directly proportional to its moment of inertia. Using this, and the momenta of inertia for our two bodies, we can say that the kinetic energy of a rotating wheel is twice that of the solid cylinder rotating at the same speed.

In your question you also mention translational energy. The key trick to working with translation kinetic energy is to understand that you need only consider the motion of the centre of mass.

The centre of mass of both the example objects above is, well, at the centre, and we have been considering that they both move at the same velocity. This is implies that the centres of mass also move at the same velocity and thus that the translation KE of the bodies are exactly the same. They differ only in rotational KE.

• @ jumpboat what if we considered a body undergoing pure translation and a body undergoing rolling motion? the wheel has both rotational and translational motion. So, does it not mean that the body undergoing pure translation would reach a particular postion faster if there was a 'race' between them? Isn't it counter-intuitive? Commented May 31, 2021 at 15:58
• OK, so you're asking a different question there. In a race between a rolling wheel and a sliding wheel, a sliding wheel would make it to the bottom of an incline first. You can calculate this, and you find that the sliding wheel actually takes $\sqrt{2}/2$ the time that the rolling wheel takes. The intuitive explanation for this, I suppose, is that the sliding wheel has no rotational kinetic energy and thus all of the GPE can be converted to translational KE, whereas a rolling wheel is constrained to have a certain amount of rotational KE as well as translational. Hope that helps. Commented May 31, 2021 at 16:08
• It was really helpful. Could you tell me whether my assumption in the last para of the question is correct or not? Commented May 31, 2021 at 16:38
• It is just that I want to convince myself that rotational motion and translational motion are entirely seperate entities and that they don't affect each other's properties Commented May 31, 2021 at 16:41
• @Proxima you are familiar with the concept of momentum, surely? Well, there are two types of momenta, linear momentum (i.e. translational) and angular momentum (rotational). Both are conserved in a closed system, and you cannot convert one to the other. This is pretty much the basis for why we can consider rotational and translational motion separately. The only way they affect each other is when we impose constraints - for example, when we say that the wheel must roll. Mathematically, this states that the angular velocity $\omega = Rv$ where $v$ is the translational velocity. Commented May 31, 2021 at 17:16