How do you find the centripetal acceleration of any point on a body performing both rotational and translational motion.

For example, in pure rolling if we find centripetal acceleration of the topmost point about the IAOR, it will be $\omega ^{2}2r$ but when you find it about the COM it will be $\omega ^{2}r$

But here to find the normal force we do consider centrifugal force of the COM about the IAOR. $$mg\sin \theta + T\cos \theta - m\omega ^{2}r = N $$

ignore the T

  • $\begingroup$ $ \omega^2\,r$ is not force $\endgroup$
    – Eli
    Apr 13, 2022 at 16:02
  • 1
    $\begingroup$ ah yes my bad forgot to multiply it by mass $\endgroup$
    – Tejas
    Apr 13, 2022 at 16:11

2 Answers 2


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first put coordinate system at the center of mass

the unit vector to point A is:

$$\vec e_A=\begin{bmatrix} \cos(\theta) \\ -\sin(\theta) \\ \end{bmatrix}$$

and the components of the forces are: $$-m\,g\,\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}\quad, T\,\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}\quad, m\,\dot\theta^2\,r\,\vec e_A\quad, -N\,\vec e_A$$

from here you can "projected" the forces towards the vector $~\vec e_A~$

$$\sum F_i= -m\,g\,\vec e_z\cdot \vec e_A+T\,\vec e_x\cdot\vec e_A+m\,\dot\theta^2\,r\,\vec e_A\cdot\vec e_A-N\,\vec e_A\cdot\vec e_A=0$$

with $~\dot\theta=\omega~$


  • $\begingroup$ shouldn't the mω2r part be subtracted? Also, what about in the case of rolling, is it correct to compute centripetal acceleration about IAOR? $\endgroup$
    – Tejas
    Apr 13, 2022 at 18:29
  • $\begingroup$ Eli did the force balance in the non-inertial frame and in this frame there is a centrifugal force radially outward, and no acceleration of the center of mass. This is just for the CM. In general you want to evaluate the motion of the CM and rotation $\vec \omega$ about the CM. See if my answer helps. $\endgroup$
    – John Darby
    Apr 13, 2022 at 20:44

You can evaluate the motion of the center mass (CM) using $\vec F_{ext} = M\vec a_{CM}$ where $\vec F$ is the total external force, $M$ the total mass, and $\vec a$ is the acceleration of the CM. @Eli evaluated the acceleration of the CM in a non-inertial reference frame where there is a centrifugal force and no acceleration, then found the components of the forces in the radial direction. In the inertial frame, there is no centrifugal force and the CM accelerates; in this frame, using Eli's notation $-N + Mgsin(\theta) + T cos(\theta) = -M\omega^2r$ where $-\omega^2r$ is the centripetal acceleration.

This is not the overall motion of the body.

Here is a typical approach to evaluate the overall motion of your rolling rigid body.

Evaluate the acceleration of the center of mass (CM) in an inertial frame using $\vec F_{ext} = M\vec a_{CM}$ where $\vec F$ is the total external force, $M$ the total mass, and $\vec a$ is the acceleration of the CM. The change in angular momentum with respect to the CM is ${d\vec J_{CM} \over dt} = \vec N_{CM}$ where $\vec J_{CM}$ is the angular momentum and $\vec N_{CM}$ is the total external torque, both with respect to the CM; this is true even if the CM is accelerating. Here (rotation in a plane) $\vec J = I\vec \omega$ where $I$ is the moment of inertia, and $\vec \omega$ the angular acceleration with respect to the CM. If there is no slip of the body at the point where $\vec N$ acts, call it point $Q$, then $\vec v_{CM} = \vec \omega \times \vec d$ where $\vec d$ is the vector from $Q$ to the CM.

Here, you evaluate the above relationships using the forces $\vec T$, $m\vec g$, and $\vec N$. I would evaluate $\vec a_{CM}$ in $x, y$ coordinates where $x$ is horizontal and $y$ is vertical. Then, the rotational motion is described as $\ddot \omega$ (into the page) with respect to the accelerating CM.

  • $\begingroup$ So I understand that during pure rolling, even though the IAOR is stationary it still has some centripetal acceleration and thus it's a non inertial frame, and that's why we cannot find the absolute centripetal acceleration of any other point about the IAOR without accounting for the pseudo force. $\endgroup$
    – Tejas
    Apr 13, 2022 at 22:28
  • $\begingroup$ But I do not understand how in the curb climbing problem we can assume the COM experiences a centrifugal force about the point of contact. And in a general scenario do all bodies posses a centrifugal force about their IAOR? and could we have solved that problem from COM frame? $\endgroup$
    – Tejas
    Apr 13, 2022 at 22:32
  • $\begingroup$ There is no centrifugal force in the stationary frame. Eli evaluated the motion of the CM, then gave you the radial component with respect to the point of contact. I am not sure what you would use this result for, it is not the total motion of the CM; how would you use it? A more meaningful result is the total acceleration of the CM, and rotation about the moving CM; here every point has centripetal acceleration with respect to the CM of $\omega^2r'$ where $r'$ is the distance from the CM to the point, not necessarily at $R$. $\endgroup$
    – John Darby
    Apr 13, 2022 at 22:55

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