Question

Chapter 15, Problem 041 The pendulum in the figure consists of a uniform disk with radius r 16.0 cm and mass 470 g attached to a uniform rod with length L = 660 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. (a) Number Units (b) Number Units c) Number Units Click if you would like to Show Work for this question: Open Show Work

Answer 1

a) we know that

Moment of inertia of the disk is equal to the sum of the moment of inertia of the Rod and disk.

$I = 0.5 \times 470 \times 10^{-3} \times (16 \times 10^{-2})^2+ 470 \times 10^{-3} \times (66+16)^2 \times 10^{-4} +290 \times 10^{-3} (660 \times 10^{-6})^2/3$

kg-m2

b)

Again, the COM of the pendulum is equal to the sum of the COM of the disk and rod.

$COM = \frac{M(L+r) + mL/2}{M+m}$

$COM = \frac{0.47(0.66 + 0.16) + 0.29(0.66/2)}{0.47+0.29}$

$COM = \frac{0.3854 + 0.0957 }{0.47+0.29}$

c)

The time period of Oscillation can be found out by simple formula,

$T = 2 \pi \sqrt\ I /(M+m)gh$

T = 1.745 s