I was trying to understand an aspect of rotational dynamics and thought of a problem to help me learn. I'm sure this problem has been considered by countless people in the past, but I'm having some trouble arriving at a solution.

Consider a disk of uniform mass distribution $\rho$ (measured in $kg / m^2$) and radius $R$ centered at the origin of an xy coordinate system at time $t=0$.

Apply a constant force $F$ in the positive $y$ direction to the disk at distance $r$ to the right of the center of the disk, where $0 \le r \le R$. What will be the rotational speed and translational velocity of the disk at time $t_1 \gt 0?$

(Note that after some time has elapsed, the disk may have rotated from its original position. If this happens, continue to apply the force at the same distance to the right of the center of the disk. Also, assume that there is no friction anywhere in the system.)

My work so far: I've figured out that if the force is applied to the center of the disk so that $r = 0$, then it's rotational speed will remain $0$ $radians/s$, and it's translational velocity can be obtained from Newton's second law $(F = ma)$. In other words,

$$ F = ma, $$ $$ F = \rho \pi R^2 \frac{dv}{dt}, $$ $$ \int_0^{t_1} F dt = \int_0^{t_1} \rho \pi R^2 \frac{dv}{dt} dt, $$ $$ F t_1 = \rho \pi R^2 \int_0^{t_1} \frac{dv}{dt} dt, $$ $$ F t_1 = \rho \pi R^2 v_1, $$ $$ v_1 = \frac{F t_1}{\rho \pi R^2}. $$

It seems to me that the same should hold true in the case where $r \gt 0,$ although intuitively, it also seems like some of the impulse from the force should cause the disk to rotate.

If the above calculation is used to obtain the translational velocity of the disk, then the same impulse applied off-center for the disk will result in the disk reaching the same amount of translational kinetic energy as in the above case along with some additional rotational kinetic energy from the applied torque, which can't be right (this is like getting something for nothing). So I'm pretty stuck at this point.

I tried considering simpler systems, like a ring with uniform mass distribution $\lambda$ $(kg/m)$, but I think I would run into the same problems.

Another related system that can be considered is when an equal and opposite force is also applied left of the center of the disk at the same distance from the center of the disk as the first force. In this case, all of the impulse from the two forces combined would go into the rotational kinetic energy of the disk, leaving its translational speed at $0$ $m/s$. We can set the total impulse imparted to the disk equal to its change in rotational momentum. So,

$$ \int_0^{t_1}2F dt = \int_0^{t_1} \int_0^R \int_0^{2\pi} \rho\;d\theta\;r^2\;dr\;\frac{d\omega}{dt}\;dt $$ $$ 2Ft_1 = \frac{2\pi\rho R^3\omega_1}{3} $$ $$ \omega_1 = \frac{3Ft_1}{\pi\rho R^3}. $$

However, I'm not sure if this helps me find how much impulse is used towards the translational versus rotational momentum of the disk in the general case of the original problem.

Is there a basic law of mechanics that I'm missing and need to apply in order to understand this system? I feel like I'm missing something conceptually...

  • $\begingroup$ The impulse isn't split, it makes the disk rotate and move at the same time. The work is increased. $\endgroup$
    – Ron Maimon
    Sep 8, 2012 at 16:21

1 Answer 1


This puzzle confused me too a long time ago, and the answer you get is absolutely right. If you apply the same impulse on the side, the disk will still translate with the same linear momentum, but it will also rotate around the center. The rotation is given by the angular momentum impulse, which is the torque $Fr$ times the increment of time.

The reason this doesn't violate conservation of energy is that the work required to apply the impulse is the force times the distance moved, and in the case that the disk rotates infinitesimally as you apply the impulse, there is an extra distance the point of the applied force moves. The extra distance means the work is larger, and by exactly the right amount.

  • $\begingroup$ Thanks Ron. I wasn't actually stuck on conservation of energy, but rather conservation of momentum. Your answer helped, though. If the same impulse is applied in both cases, the disk acquires rotational as well as translational momentum in one case, but only translational momentum in the other case. But I guess if you look at the rotational momentum of the disk as if it were translational momentum (by summing up each part of the disk), then it would actually be zero. Does that make sense? I upvoted your answer! $\endgroup$
    – Andrew
    Sep 8, 2012 at 16:34
  • $\begingroup$ @Andrew: Yes, the rotational motion doesn't give extra momentum. $\endgroup$
    – Ron Maimon
    Sep 9, 2012 at 3:38

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