# Torque on puck moving on plane without friction

We have two pucks moving on a plane without friction. On one of them a force is applied on it's center of mass. On the second a force of equal magnitude is acting tangential to the puck and at a distance equal to it's radius. Which one will be faster?

Now if I try to solve the problem and take only translation in account then I would say that both forces are acting in the CM (center of mass) and both will have the same acceleration. But if I think in terms of energies, then the force acting on the CM will convert all work into translational kinetic energy while the other force would convert a part of it's work into rotational energy. So the pucks will not have the same acceleration of their center of mass.

The center of mass of a rigid body is given by: $$\vec{r}_c = \frac{1}{M} \sum_i m_i \cdot \vec{r}_i$$ with $M = \sum_i m_i$ the total mass or $$\vec{r}_c = \frac{1}{M} \int \vec{r}\ ' \cdot \varrho(\vec{r}\ ')~d^3r'$$ with $M = \int \varrho(\vec{r}\ ')~d^3r'$ for a continious mass distribution. I'll stick to the discrete case for brevity.

Therefore: \begin{align} M~\frac{d^2 \vec{r}_c}{dt^2} = M~\vec{a}_c & = \sum_i m_i~\frac{d^2 \vec{r}_i}{dt^2} \\ & = \sum_i m_i~\vec{a}_i = \sum_i \vec{F}_i \tag1 \end{align} So the center of mass moves due to every force acting on the rigid body's particles. Now, what are these forces? If we set $$\vec{F}_i = \sum \vec{F}_{\text{ext}}$$ with $\vec{F}_{\text{ext}}$ the external forces acting on a particle, this would not account for the fact, that the body is rigid. There are indeed internal forces that guarantee the body not to deform upon the action of external forces. For exmaple, if an external force is only applied to one particle and there were no internal forces, this would allow the particle to "leave" the body. This is not what one usally means if talking about a rigid body. So one rather has $$\vec{F}_i = \sum \vec{F}_{\text{ext}} + \sum \vec{F}_{\text{int}}$$. However, according to Newton's third law, for every internal force that a particle applies to its neighboors, there is a counter-force of same magnitude but opposite direction. Therefore one has $$\sum_i \left( \sum \vec{F}_{i,\text{int}} \right) = 0$$ such that the equation of motion for the center of mass becomes: $$M~\vec{a}_c = \sum_i \left( \sum \vec{F}_{i,\text{ext}} \right)$$

So concerning your problem: If the two forces, are equal in magnitude and direction, there will be no difference in the acceleration of the center of mass, no matter what the point is on which the forces are applied (e.g. at the center of mass or tangential).

However, the energies of both systems will differ, as the second system will not only move linearly but also spin. This seems odd in the first place, since the magnitude of force was the same, but think about it like this:

Take a ball of wood an lift it in the air, then take a ball of iron with equal size and lift it to the same height. The second process involves more work, since the second ball is heavier. So the extra work can be explained by the mass difference $\Delta m$ that has to be lifted in the second process. It's like you apply a force to the wooden ball plus an extra force to $\Delta m$.

This reasoning can be applied to your problem, too. If the force acts tangentially, it does not only act on the center of mass $M$ but also on the moment of inertia, because there is a torque due to this force.

edit according to: Equation of motion for the center of mass of a rigid body

We have two pucks moving on a plane without friction. On one of them a force is applied on it's center of mass. On the second a force of equal magnitude is acting tangential to the puck and at a distance equal to it's radius. Which one will be faster?...

So the pucks will not have the same acceleration of their center of mass.

Yes, your intuition is correct: the puck on which the force acts on its CM will be faster, because all energy will produce linear momentum, whereas in the other puck energy will distributed between linear and angular velocity/momentum.

• This assumes that energy is constant in the system. Apr 5, 2015 at 11:28