The physical pendulum shown on your paper is a 27.0 kg wedge of a circular disk of uniform density with radius, R=1.87 m and opening angle β=0.847 radians. The pivot point of the pendulum can be moved along the center line of the wedge as shown on your paper.
a) Find the distance of the center of mass from the tip of the wedge.
b) If this physical pendulum is supended by a pivot at its tip, and oscillates with a small amplitude, find the period of oscilation.
c) Find the distance from the tip at which the pivot point should be placed to maximize the frequency for the pendulum.
d) Find the maximum angular frequency for this pendulum
HINT: The `angular frequency' is the name often used for the parameter `w' (omega) in the general equation for Simple Harmonic Motion: i.e., Acos(`w't+phi). Note that it is also the angular velocity for the circular motion whose projection is SHM. Since you have already calculated the distance from the pivot to the CM and also Icm, you can now calculate the maximum value of `w'.
The Attempt at a Solution
a) Xcm=(4/3)(R)(sin(0.5A)/A)=1.21m (correct)
b) T=2pisqrt(I/mgXcm)=2.41s (correct)
A friend told me to use the parallel axis theorem but I didn't use it and still got the answer right, do I have to use it here?
c) I have no clue...I'm guessing I use D=Acos(wt+phi), and to maximize the whole cos term will be equal to 1, so D=A? It's clearly wrong but I don't know where to go from here.
d) Once I find the distance in part c, w=sqrt(mgd/I) and that's it?