Question

A grandfather clock has a pendulum that consists of a thin brass disk of radius r = 14.38 cm and mass 0.8261 kg that is attached to a long thin rod of negligible mass. The pendulum swings freely about an axis perpendicular to the rod and through the end of the rod opposite the disk, as shown in the figure. If the pendulum is to have a period of 1.592 s for small oscillations at a place where g = 9.822 m/s, what must be the rod length L? Rotation axis Number Units The number of significant digits is set to 4; the tolerance is +/-4%

Answer 1

Time period of pendulum is given by,

T = 2*pi*sqrt(I/(m*g*h))

here, I = moment of inertia of pendulum = 0.5*m*r^2 + m*(r+L)^2

m = mass of disk = 0.8261 kg

g = 9.81 m/s^2

h = distance of center of gravity of disk from hinge = r+L

L = length of rod = ??

T = time period = 1.592 sec.

r = radius of disk = 14.38 cm = 0.1438 m

So, 1.592 = 2*pi*sqrt[(0.5*0.8261*0.1438^2 + 0.8261*(0.1438+L)^2)/(0.8261*9.81*(0.1438 + L))]

1.592^2/(4*pi^2) = (0.5*0.8261*0.1438^2 + 0.8261*(0.1438+L)^2)/(0.8261*9.81*(0.1438 + L))

(0.5*0.8261*0.1438^2 + 0.8261*(0.1438+L)^2) = (1.592^2/(4*pi^2))*(0.8261*9.81*(0.1438 + L))

0.8261*L^2 + 2*0.1438*0.8261*L + 0.1438^2*0.8261 + 0.5*0.8261*0.1438^2 - (1.592^2/(4*pi^2))*0.8261*9.81*0.1438 -(1.592^2/(4*pi^2))*(0.8261*9.81*L = 0

0.8261*L^2 - 0.28268*L - 0.04919= 0

By solving above quadratic equatuion:

L = 0.4691 m

L = 46.91 cm