Question

Question 3 of challenging questions from module 22 (Torque and SHO) ...

part a) why is the moment of inertia about pt P the same as that about the disc's CM?

part b) Why can't I use (1/2)mv^2 but the rotational one as my K.E. equation? The disc is fixed to the pendulum rod anyway...can't I treat it as a particle?

Thanks

Answer 1

Usually you can only treat teh dis a a particle if its dimensions are very small compared to teh distance from the rotation axis. The KE is pure rotational KE, so you must use \[K _{rot}=1/2I _{net} \omega^{2}\] (maybe that's what you're saying) but you have to use the parallel axis theorem to come up with I for the disk

Answer 2

sory about the spelling - typing too fast

Answer 3

as for part a, they are not saying it is the same as about the CM, they are saying use the parallel axis theorem\[I _{P}=I _{CM} + md^{2}\]where d is the distance from the CM of the disk to point P

Answer 4

So ... if I have a point particle instead of a disk swinging down, should I use the translational or rotational KE equation? (Or do both apply?)

As for the moment of inertia...the solutions say \[I _{p} = 0.5m_{1}R _{}^{2}\]
Yet, if according to the //-axis theorem, there should an additional ml^2 term, where l is the length of the massless rod attached to the disk.

Answer 5

Illian
Sorry - I did not read the question carefully enough. The whole point of the question is that the disk is attached by a frictionless bearing, meaning that it is NOT fixed to the pendulum rod in the sense that it does not rotate about its CM as it falls thus all the potential energy gained by the disk goes into translation rather than rotation + translation - thus it does act like a point mass