(a) If thependulum is modeled as a simple pendulum, what would be theperiod? (b)

Question

(a) If thependulum is modeled as a simple pendulum, what would be theperiod?
(b) Christyobserves the actual period of the clock, and finds that itis 0.80% faster than that for asimple pendulum that is 0.800 m long. If Christy models the pendulumas two objects, a 0.800 muniform thin rod and a point mass located 0.800 m from the axis of rotation, whatpercentage of the total mass of the pendulum is in the uniform thinrod?


I have no idea how toapproach part b. Anything to send me in the rightdirection?

Explanation / Answer

f' = 1.008 f   so P' = .99206 P P = 2 (L / g) P' = 2 (I / (M g l))   where l is thedistance to the center of mass and M the total mass of thependulum P' / P = (I / M l) / L l = [L / 2 * (M - m) + m L] / M = L / 2 * (1 + m /M)       distance to C.M. where M is the total mass and m the mass at theend I = (M - m) L2 / 9 + m L2 =L2 / 9 * (M + 8 m)    Moment of inertia I / l = 2 L M / 9 * (1 + 8 m / M) / (1 + m / M) P' / P = [2 / 9 * (1 + 8 m / M) / (1 + m / M)] =.99206 (1 + 8 m / M) / (1 + m / M) = 4.42885 m / M = .9601 So the mass of the rod is 1 - .9601 = .0399 or3.99% the mass of the pendulum

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