Question

4) Extra credit if correct. The pendulum of my grandfather clock is a rigid 1kg 1 meter that dangles from one end and a .1 kg point mass that is attached to the rod. The frequency of oscillation can be adjusted by changing what part of the rod the .1 mass is attached to. How far bellow the top of the rod should the .1 kg mass be attached to give the a) minimum and b) maximum period?

Answer 1

For a physical pendulum, the period is giving by: T = 211 Mgd Where M is the total mass (M = mrod + m) d is the distance between the pivot point and the center of mass I is the moment of inertia about the pivot point. Let be x the distance between the pivot point and the point mass: The moment of inertia of the system is: 1 = Irod + Imass = mrod + mx? The position of the center of mass is: d-mrod () + mx d=- mrod + m Then, the period is: z mrodL? + mx2 T = 20 (mrod + m)g + mx mrod + m Using the giving data: (0.3229, 1.6129) The minimum period is (1)12 +0.1.x2 T = 210 9.8(1. (1+ 0.1x) 1 + 0.3x2 T = 21 9.8(1.5 +0.3x) Let's plot T vs x For x = 0.3229 m we get the minimum period and for x= 1m we get the maximum period