Question

The pendulum in the figure consists of a uniform disk with radius r = 13.0 cm and mass 880 g attached to a uniform rod with length L = 610 mm and mass 290 g. (a)Calculate the rotational inertia of the pendulum about the pivot point. (b) What is the distance between the pivot point and the center of mass of the pendulum? (c)Calculate the period of oscillation.

Answer 1

Given that,

Radius of disk r = 13 cm = 0.13 m

Mass of disk, M = 880 g = 0.880 kg

Length of rod, L = 610 mm = 0.610 m

Mass of rod, m = 290 g = 0.290 kg

(a)

Rotational inertia of the pendulum about the pivot point,

l = Mr^2 / 2 + mL^2 / 3 + M(r + L)^2

l = (0.88*0.13² / 2) + (0.29*0.61² / 3) + (0.88*(0.13 + 0.61)²)

l = 0.5252 kg.m^2

(b)

Let, distance between the pivot point and the center of mass of the pendulum = d

Balancing torque about pivot,

m (d - L/2) = M (L + r - d)

0.29 * (d - 0.61 / 2) = 0.88 * (0.61 + 0.13 - d)

1.17d = 0.7396

d = 0.632 m

(c)

Period of oscillation,

T = 2*pi * sqrt [ l / ((m + M)*gd)]

T = 2*pi * sqrt [0.5252 / ((0.29 + 0.88)*9.81*0.632)]

T = 1.69 s