Question

A uniform rod of mass M and length L is free to swing back and forth by pivoting a distance x from its center. It undergoes harmonic oscillations by swinging back and forth under the influence of gravity. Randomized Variables M=2.2 kg x= 0.49 m 33% Part (a) In terms of M, L, and x, what is the rod' s moment of inertia / about the pivot point?

Answer 1

Answer:

The moment of inertia is 0.7500 kg-m².

Explanation:

Given that,

Mass = 2.2 kg

Distance = 0.49 m

If the length is 1.1 m

We need to calculate the moment of inertia

Using formula of moment of inertia

$I=\dfrac{1}{12}ml^2+mx^2$

Where, m = mass of rod

l = length of rod

x = distance from its center

Put the value into the formula

$I=\dfrac{1}{12}\times2.2\times(1.1)^2+2.2\times(0.49)^2$

$I=0.7500\ kg-m^2$

Hence, The moment of inertia is 0.7500 kg-m².