Problem : A disk of mass 2 kg and radius .5 m is hung from a wire, then rotated a small angle such that it engages in torsional oscillation. The period of oscillation is measured at 2 seconds. Given that the moment of inertia of a disk is given by I = , find the torsional constant, κ , of the wire.

Solution for Problem 1 >>

To solve this problem we use the equation for the period of a torsional oscillator:

T = 2Π

Solving for κ ,

κ =

We are given T , and must simply compute I . Given the dimensions of the disk, we can simply plug into the formula we are given for the moment of inertia: I = = = .25 . Thus:

κ = = = 2.47

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Problem : The disk from problem 1 is replaced with an object of unknown mass and shape, and rotated such that it engages in torsional oscillation. The period of oscillation is observed to be 4 seconds. Find the moment of inertia of the object.

Solution for Problem 2 >>

To find the moment of inertia we use the same equation:

T = 2Π

Solving for I,

I =

>From last problem we know that κ = , and we are given the period (4 seconds). Thus:

I = = 1

In the last two problems we have established a method for determining the moment of inertia of any object.

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Problem : A pendulum of length L is displaced an angle θ , and is observed to have a period of 4 seconds. The string is then cut in half, and displaced to the same angle θ . How does this effect the period of oscillation?

Solution for Problem 3 >>

We turn to our equation for the period of the pendulum:

T = 2Π

Clearly if we reduce the length of the pendulum by a factor of 2 we reduce the period of the oscillation by a factor of 4.

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Problem : A pendulum is commonly used to calculate the acceleration due to gravity at various points around the earth. Often areas with low acceleration indicate a cavity in the earth in the area, many times filled with petroleum. An oil prospector uses a pendulum of length 1 meter, and observes it to oscillate with a period of 2 seconds. What is the acceleration due to gravity at this point?

Solution for Problem 4 >>

We use the familiar equation:

T = 2Π

Solving for g:

g	=
	=	= 9.87 m/s2

This value indicates a region of high density near the point of measurement- probably not a good place to drill for oil.

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Problem : What is the angular velocity of a particle moving in uniform circular motion that has the same period as a mass of 2 kg on a spring with constant 8 N/m?

Solution for Problem 5 >>

Recall from our comparison of circular and oscillatory motion that the angular velocity of a particle in circular motion corresponds with the angular frequency of a particle in oscillatory motion. We know the angular frequency of the mass- spring system:

σ = = = 2

Thus we can infer that the particle moves in a circle with constant angular velocity of 2 rad/s.

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