A grandfather clock has a pendulum that consists of a thin brass disk of radius

Question

A grandfather clock has a pendulum that consists of a thin brass disk of radius r = 15.90 cm and mass 0.90 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 below. If the pendulum is to have a period of 2.00 s for small oscillations at a place where g = 9.800 m/s2, what must be the rod length L to the nearest tenth of a millimeter?

http://www.webassign.net/hrw/15-54.gif

Explanation / Answer

-mg*(L+r)*sin theta = I*theta''.........where I = (1/2*mr^2) + m(L+r)^2

For small theta, we can say sin theta = theta

mg(L+r)*theta + [(1/2*mr^2) + m(L+r)^2]*theta'' = 0

Therefore, time period T = 2*pi* sqrt [((1/2*mr^2) + m(L+r)^2) / (mg(L+r)) ]

T = 2*pi* sqrt [((1/2*r^2) + (L+r)^2) / (g(L+r)) ]

2 = 2*3.14*sqrt [((1/2*0.159^2) + (L+0.159)^2) / (9.8*(L+0.159)) ]

L = 0.8221 m = 822.1 mm

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