You have a bar of metal in an environment with no gravity. A force is applied on one end of it. How does it rotate?

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There is a non-zero torque on any random point selected on the bar. For example, on point A the torque is $T_A = F*y$, but on point B the torque is $T_B = F*x$. So the torque on any point on the bar except the point where the force is applied, is non-zero. If there is a non-zero torque on one point, the object must rotate around that point. But since there is a non-zero torque on every point, and the object can't rotate around every point, how does it rotate and around what point?

  • $\begingroup$ @Mostafa This isn't homework. $\endgroup$
    – dfg
    Commented Oct 19, 2013 at 16:53
  • $\begingroup$ This is better understood with a short duration force applied to a motionless bar. $\endgroup$ Commented Nov 18, 2013 at 20:21
  • $\begingroup$ See related question and answer: physics.stackexchange.com/a/43279/392 $\endgroup$ Commented Nov 18, 2013 at 20:59
  • $\begingroup$ Let me note that a pure torque can be applied anywhere on a body and it will have the same effect. Torque due to a force at a distance is different as it changes magnitude the futher away you get from the force line (axis). $\endgroup$ Commented Nov 18, 2013 at 21:51

2 Answers 2


See here for something about rotation and torque.
As is explained there, the IAOR is defined as the point about which every point is in pure rotational motion i.e. every point has a velocity perpendicular to the position vector to that point from IAOR. This point's position can vary in space and with time and is usually difficult to treat in most problems.
Theoretically, you can find the torque about any point from where you define the rotational variables ($\theta, \omega$) but a particularly useful point is the centre of mass.

Treating the rotation to be about the COM, you eliminate pseudo forces in the COM frame and that allows you to treat that frame as an inertial frame. Therefore, for any arbitrary unconstrained body, motion can be divided into two parts:-

1)The translational motion of the COM ($dv_{com}/dt=a$,$a=F_{net}/m$)
2)Rotation of the body about the COM ($d\omega/dt=\alpha$,$\alpha=\tau_{com}/I_{com}$)

Tell me if you want any more details.

  • $\begingroup$ What would the IAOR be for the bar above? $\endgroup$
    – dfg
    Commented Oct 19, 2013 at 15:38
  • $\begingroup$ @dfg Initially, at time $t=0$ it will be the end A but will keep changing. $\endgroup$ Commented Oct 19, 2013 at 16:23
  • $\begingroup$ @dfg it can also be outside the body. $\endgroup$ Commented Oct 20, 2013 at 7:50
  • $\begingroup$ The object would rotate about the IAOR. This force will also cause torque about every other point on the body(except for the points along which the force acts). What is the physical meaning of these torques? Do these torques have no part in analysing the motion of the system ? $\endgroup$
    – SNB
    Commented Jan 1, 2018 at 19:45

Consider a (motionless) bar with a short lived force $F$ applied a distance $\ell$ from the center of mass. The center of mass is on point C, while the applied load at point B. The result is an instantaneous rotation about point A. But where is A? How do you find $x$?


The total torque applied on the center of mass is equal to the mass moment of inertia times the angular acceleration

$$ \ell\,F = I_{cm} \ddot{\theta}\; \Bigr\} \; \ddot{\theta} = \frac{\ell\,F}{I_{cm}} $$

Also the total force applied equals mass times the acceleration of the center of mass. Since I want to know where A is, I assume the body will rotate about A, giving the center of mass acceleration the value of $\ddot{y}_C = x \, \ddot{\theta}$. There is only one value of $x$ that maintains this relationship.

$$ F = m x \ddot\theta = \frac{m\,\ell\,x}{I_{cm}} F \;\Bigr\}\; x = \frac{I_{cm}}{m\,\ell} $$

So the instant center of rotation A when the bar is loaded at B is given by the distance $x$. For each applied force point B there is different center of rotation A. The point B is called the sweet spot, or the axis of percussion of point A. Technically the two points consist of a pole and a polar of the rigid body.

Only the equipollent torque on the center of mass (and the distance $\ell$) is important when considering the rotational motion of a rigid bar.

Bonus Question

If the applied force is at the center of mass, with $\ell=0$ where is the center of rotation A (and $x$)?

  • 1
    $\begingroup$ Absolutely brilliant answer $\endgroup$
    – SNB
    Commented Jun 1, 2017 at 12:55

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