How can I determine the axis of rotation of a rigid when there is net external force on the Centre of mass of the body. For example consider the following semicircular disc undergoing pure rolling motion on a rough ground.How can I determine whether the body will rotate an axis passing through its COM or point O as there are torques around both the axes enter image description here

  • $\begingroup$ To work out the dynamics you have to choose the moments about the center of mass. To work out the kinematic you choose whatever point makes the equations simpler, and in this case this is the geometric center O. $\endgroup$ – ja72 Dec 6 '17 at 5:50
  • $\begingroup$ What exactly is the problem which you are trying to solve? I suspect that the question which you are asking might be unnecessary for solving that problem. See What is the XY problem? $\endgroup$ – sammy gerbil Dec 6 '17 at 21:53
  • $\begingroup$ @sammygerbil The problem asked about the angular acceleration.I spent a lot of time thinking about which point it will rotate.So I guessed a point and calculated the angular acceleration.I got the correct answer but wasn't sure why it will rotate about this point.This problem is just an example, I just needed to clarify my doubt $\endgroup$ – Gagandeep Singh Dec 7 '17 at 3:35

In the diagram shown there are external forces. The reaction from the ground is such that the velocity on the contact point is horizontal only. This means the center of rotation is somewhere along the vertical line passing through the contact point.

But where? This depends on the friction condition at the contact point. Consider the general case below:


In order to find the equations of motion we need to establish how this thing moves. We describe this with the variable $x$ for horizontal position of the center A and the angle $\theta$ the part makes with the vertical direction.

Consider sequentially the velocities of points A, B and C when $x$ and $\theta$ vary only.

$$\begin{aligned} \mathbf{v}_A & = \pmatrix{\dot{x} \\ 0} \\ \mathbf{v}_B & = \pmatrix{\dot{x} + r \dot{\theta} \\ 0} \\ \mathbf{v}_C & = \pmatrix{\dot{x} + h \,\dot{\theta} \cos \theta \\ h \, \dot{\theta} \sin \theta} \end{aligned}$$

where $\dot{x}$ and $\dot{\theta}$ are the first time derivatives.

Take the time derivative of the velocity at C to get the acceleration of the center of mass

$$ \mathbf{a}_C = \pmatrix{ \ddot{x} + h\,\ddot{\theta} \cos\theta - h\,\dot{\theta}^2 \sin\theta \\ h\,\ddot{\theta} \sin \theta + h\,\dot{\theta}^2 \cos\theta} $$

Now let's look at the equations of motion by considering all the forces acting on the body.


The equations of motion have to consider the balance of moments about the center of mass (point C) to be valid.

$$ \begin{aligned} \pmatrix{F \\ N - m g } & = m \mathbf{a}_C & & \mbox{sum of forces} \\ -(h\sin\theta) N + (r-h \cos\theta) F & = I_C \ddot{\theta} & & \mbox{sum of moments} \end{aligned} $$

There are three equations and 4 unknowns ($N$, $F$, $\ddot{x}$, $\ddot{\theta}$). To solve them you need an expression descripting the contact condition. Here are the three scenarios

  • No Slipping - Solve with $\dot{x} + r \dot{\theta}=0$, or $\ddot{x}=-r\,\ddot{\theta}$ for $$\begin{aligned} \ddot{\theta} &= - \frac{h\,m (g+r \dot{\theta}^2) \sin\theta}{I_C+m (r^2+h^2-2 h r \cos\theta)} \\ \ddot{x} & = \frac{r\,h\,m (g+r \dot{\theta}^2) \sin\theta}{I_C+m (r^2+h^2-2 h r \cos\theta)} \end{aligned}$$ The center of rotation height above the ground is $c = r + \frac{\ddot{x}}{\ddot{\theta}} = 0$ so the body is always rotating about point B. This can be confirmed by the fact that the parallel axis theorem in the denominator of $\ddot{\theta}$ contains the distance between B and C.

  • Zero Friction - Solve with $F=0$ for $$\begin{aligned} \ddot{\theta} & = - \frac{h\, m \sin\theta (g+h \dot{\theta}^2 \cos\theta)}{I_C +m h^2 \sin^2 \theta} \\ \ddot{x} & = \frac{h^2 m \sin\theta\cos\theta (g+h \dot{\theta}^2 \cos\theta)}{I_C + m h^2 \sin^2 \theta} + h \dot{\theta}^2 \sin\theta \end{aligned} $$ The center of rotation height above the ground is $c = r + \frac{\ddot{x}}{\ddot{\theta}}$ which initially (when $\dot{\theta}=0$) is equal to $c = r-h \cos\theta$. So the body rotated about a point along the contact normal with the same height as the center of mass.

    • Coulomb Friction - Solve with $F =\pm \mu N$ for $\ddot{x}$, $\ddot{\theta}$ and $N$. I am omitting this solution here for brevity because it is rather complex. The body rotated about a point between the two other solutions depending on the value of $\mu$.

For the particular problem, I assumed no slipping, so set $\theta = \frac{\pi}{2}$, $\dot{\theta}=0$ and $\ddot{x}-r \ddot{\theta} = 0$ to get the center of rotation at the contact point B and the angular acceleration $$\ddot{\theta} =- \frac{h m g}{I_C + m (h^2+r^2)}$$

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    $\begingroup$ This is a very complete answer to a homework question. $\endgroup$ – sammy gerbil Dec 6 '17 at 21:45
  • $\begingroup$ @sammygerbil It is a complete answer to a more general question to show the intricacies and details of the concept of center of rotation. $\endgroup$ – ja72 Dec 7 '17 at 19:00

Rotation occurs about every point - you are entirely free to choose O or C or any third point. Next, compute the torque about that point and calculate the moment of inertia about it too. The trick of course, is to choose a convenient point to calculate the torques and the moment of inertia.

Regardless of which point you choose, the angular acceleration will be the same.

  • $\begingroup$ So if I chose a point which itself is accelerating do I need to apply Pseudo force for computing torques? $\endgroup$ – Gagandeep Singh Dec 7 '17 at 7:56

As mentioned by @user1936752, you are free to choose an axis of rotation according to your wish. But if you are able to choose an axis about which the motion is purely rotational then it would simplify the problem. What is the mathematical expression that determines whether the body is in pure rotation about a given axis or not? Simple: If you take the axis to be perpendicular to the plane of your problem and passing through a point $O$ in the plane of your problem then in the frame in which point $O$ is at rest, you should be able to write $\vec{v_i}=(\vec{r_i}-\vec{r_O})\times\vec{\omega_O}$ where $\vec{v_i}$ is the velocity of the $i$th particle on the rigid body, $\vec{r_i}$ and $\vec{r_O}$ are the position vectors of points $i$ and $O$ and $\omega_O$ is the angular velocity vector aligned aling the axis passing through $O$. If such a point $O$ exists so that you can find an $\vec{\omega_{O}}$ consistent with the above relation then that point $O$ is about which the body is in the pure rotation--at an instant.

  • $\begingroup$ I disagree. The kinematics of the problem dictate a particular center of rotation such that there isn't interpenetration with the ground. Only when there is zero net force acting on the body the center of rotation is at the center of mass. $\endgroup$ – ja72 Dec 6 '17 at 5:41
  • $\begingroup$ @ja72 I never claimed the center of rotation is at the center of mass. I claimed that our of all the points, a point $O$ which satisfies the relation I stated, is the center of rotation. Probably, the confusion is due to the $O$ in my notation being a generic point whereas $O$ in the OP's notation being the center of mass. Am I correct? $\endgroup$ – Dvij Mankad Dec 6 '17 at 7:24

A rigid object's motion is divided into position, momentum, and angular momentum. So, the 'axis of rotation' is the unit vector which is the object's angular momentum vector divided by the angular momentum magnitude (and is undefined when there's no angular momentum). By convention, we'd draw that as a vector applied at the center of mass.

Force on the object will change these, but 'force on the Centre of mass' is not a contributor to angular momentum. It cannot change the angular momentum direction OR magnitude. The net force on the semicircular object consists of both gravity (a center-of-mass force) and (NOT center-of-mass) some pressure from its ground contact, and some lateral friction force from ground contact. Those other two forces change magnitude, not direction, of the rotation.

In this example, the 'axis of rotation' never changes (it is oriented into the plane of the paper) unless rotation stops.

  • $\begingroup$ I disagree. The point without angular momentum is the axis of percussion. That is the point where if you apply an impulse you stop a moving body (like a baseball bat). That point is always on the other side of the center of mass, as the center of rotation. For a bat, the center of rotation is the handle, the center of mass is somewhere in the middle of the bat and the percussion axis is towards the tip away from the pivot. $\endgroup$ – ja72 Dec 6 '17 at 5:48
  • $\begingroup$ If you want to say 'the center of rotation' of a bat is a point in the handle, that is determined not by the bat, but by the batter. A lathe would define 'the center of rotation' for any object it holds. A rigid object has its own center of mass, regardless of the holder. $\endgroup$ – Whit3rd Dec 6 '17 at 9:25

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