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3 years ago

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A flat, rigid object oscillates as a physical pendulum in simple harmonic motion with a frequency f. The mass of the pendulum is

m, and the pivot point is a distance d from the center of mass. What is the moment of inertia of the pendulum about its pivot point

1 answer:

lyudmila [28]3 years ago

7 0

Answer:

I = mgd/4π²f²

Explanation:

The period of the physical pendulum T = 1/f where f = frequency,

T = 1/f = 2π√(I/mgd) where I = moment of inertia of the physical pendulum, m =mass of pendulum, g = acceleration due to gravity and d = distance of center of mass of pendulum from pivot point.

1/f = 2π√(I/mgd)

dividing both sides by 2π, we have

1/2πf = √(I/mgd)

squaring both sides, we have

(1/2πf) = [√(I/mgd)]²

1/4π²f² = I/mgd

multiplying both sides by mgd, we have

I = mgd/4π²f²

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1. 218.55 N

2. $30.96^{o}$

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Part 1;

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Part 2.

Using the relationship that

Frictional force $F_{s} = \mu_{s} mg cos \theta$

$mg sin \theta =\mu_{s} mg cos \theta$

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$\mu_{s}= tan \theta$

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$\theta= 30.96^{o}$

Part 3:

Net force on the object is given by

$ma = mg sin 38 - \mu_{k} mg cos 38$ where $\mu_{k}$ is the coefficient of kinetic friction

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