Centripetal acceleration in rotational and translational motion

Question

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. $$mg\sin \theta + T\cos \theta - m\omega ^{2}r = N $$

Answer 1

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 $~\dot\theta=\omega~$

$$m\,g\,\sin(\theta)+T\cos(\theta)+m\omega^2\,r=N$$

Answer 2

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.

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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.