1. The problem statement, all variables and given/known data The figure below shows the pendulum of a clock in your grandmother's house. The uniform rod of length L = 2 m has a mass m = 0.8 kg. Attached to the rod is a uniform disk of mass M = 1.2 kg and radius 0.15 m. The clock is constructed to keep perfect time if the period of the pendulum is exactly 3.5 s. a) What should the distance 'd' be so that the period of the pendulum is 2.5 seconds? b)Suppose that the pendulum clock loses 5.00 min/d. To make sure your grandmother will not be late for her quilting parties, you decide to adjust the clock back to its proper period. How far and in what direction should you move the disk to ensure that the clock will keep perfect time? 2. Relevant equations Parallel Axis Theorem: I = Icm + MR2 I of a rod around it's edge = (ML2)/3 I of a disk = (MR2)/2 Period of a physical pendulum = T = 2∏√(I/Lmg) 3. The attempt at a solution The I of the entire system should be I= (ML2)/3 + (M(R+d)2)/2 We are solving for d, with T = 2.5 seconds, so we plug this in for our equation for period. T = 2∏√(I/Lmg) T = 2∏√(((ML2)/3 + (M(R+d)2)/2)/Lmg) Where I am stuck, however, is what would I use for 'L' in the denominator of the equation for the period. When I assume it is 2 meters, the length of the entire stick, I get 2.8m for d, which doesn't make sense. mR is the mass of the rod, md is the mass of the disk. I got that answer by rearranging for d. I end up with T^2 *L*mg/(4∏2) = I T^2 *L*mg/(4∏2) = (mR *L2)/3 + (md (d+r)2/2) Then, 2(T^2 *L*mg/(4∏2) - (mR *L2)/3)/md = (d+r)2 I plugged in numbers for everything that is given, and end up with 8.565 = (d+.15)2, and by solving that, I get ~2.8m, which would mean the disk is off the rod. What am I doing wrong?