I am trying to understand why you can use $F=ma$ for a simple pendulum, yet need the rotational equivalent for a physical pendulum. I understand it is because the rigid body can rotate too, whereas you don't consider that for a simple pendulum and consider just the mass under translational motion. I just don't understand why you don't have to consider translational motion with the physical pendulum? My understanding was kinetic energy is the sum of the translational kinetic energy of the centre of mass and rotational kinetic energy where it is rotating about the centre of mass? I'm not sure how to use this information though for a physical pendulum when it is rotating about a different axis?

  • $\begingroup$ The formula $\tau= I_{support}\alpha$ (which uses an angular coordinate system) can also be used on a simple pendulum. You will have to assume a point mass "rotating" about the hinge. $\endgroup$ – harshit54 Aug 2 at 21:48

With any extended object such as a physical pendulum, you are considering not the force on a single particle but on all of the particle (atoms, molecules, whatever) that make up the object: right off the bat, you're considering the positions of $10^\text{something}$ particles.

But that's not the whole story, because these particles exert forces on each other, and you need to keep track of that too. These forces are not fixed beforehand like gravity: they are whatever is needed to keep the body rigid, which adds more complication since we have to calculate them too. So if you want to analyze the physical pendulum the same way you would the point-mass one, you can.

The whole point of the rotational equations, however, is that they take care of all of this for us. They are nothing more than a convenient way to simplify the problem using that we don't actually need to write down one equation per particle, since the fact that the body is rigid makes everything a whole lot simpler: we actually only care about a single number, the angle of the pendulum.

So to answer your question: we do have to consider the translational motion, since that's the only kind of motion there is in reality. But doing it directly would be too complicated, which is why we have the equations of rotational motion, that help us simplify the whole thing down to a single variable.

  • $\begingroup$ That makes sense that it is just a simplification. My only other question is when you can treat a rigid body as a point mass at the centre of mass, and when you can't? I understand you can't when the body can rotate about it's centre of mass. But I don't understand why if something rotates about a point other than the centre of mass you can't treat it as a point mass in this case (as in the physical pendulum), as then surely you could just use F=ma etc like you do for a simple pendulum? As you can treat a rigid body as a point mass in other situations? $\endgroup$ – May J Aug 3 at 20:03

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