Question

A grandfather clock has a pendulum that consists of a thin brass disk of radius r = 13.62 cm and mass 1.199 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 Fig. 15-54. If the pendulum is to have a period of 1.583 s for small oscillations at a place where g = 9.846 m/s2, what must be the rod length L (in m)?

Answer 1

Answer:

Explanation:

Expression for time period of a pendulum is as follows

T = $2\pi\sqrt{\frac{l}{g} }$

l is length of pendulum from centre of bob and g is acceleration due to gravity

Given

Time period T = 1.583

g = 9.846

Substituting the values

1.583 = $2\pi\sqrt{\frac{l}{9.846} }$

l = $\frac{(1.583)^2\times9.846}{4\times(\frac{22}{7})^2 }$

l = .6244 m

= 62.44 cm

Length of rod  = length of pendulum - radius of bob

= 62.44 - 13.62

= 48.82 cm

= .488 m