The Compound Pendulum, **485 words** essay example

=Objective

1. To determine the radius of gyration of a compund pendulum about its center of gravity.

2. To investigate on the effect of fulcrum position.

3. To find the acceleration due to gravity 'g'.

4. To verify the equation of motion for a compound pendulum.

Introduction

A pendulum is a weight suspended from a pivot so that it can swing freely. When a pendulum is displaced sideways from its resting equilibrium position, it is subject to a restoring force due to gravity that will accelerate it back toward the equilibrium position. When released, the restoring force combined with the pendulum's mass causes it to oscillate about the equilibrium position, swinging back and forth. The time for one complete cycle, a left swing and a right swing, is called the period. The period depends on the length of the pendulum, and also to a slight degree on the amplitude, the width of the pendulum's swing.

Theory

In Fig. 1, O is the point of suspension of the compound pendulum and G is its centre of mass we consider the force of gravity to be acting at G. If h is the distance from O to G, the equation of motion of the compound pendulum is

Where I0is the moment of inertia of the compound pendulum about the point O.

Comparing to the equation of motion for a simple pendulum

We see that the two equations of motion are the same if we take

It is convenient to define the radius of gyration k0 of the compound pendulum such that if all the mass Mwere at a distance k0 from O,

the moment of inertia about O would be I0, which we do by writing I0 = Mk02

Substituting this into (1) gives us

The point O, a distance l from O along a line through G, is called the center of oscillation. Let h be the distance from G to O, so thatl=h+h'. Substituting this into (2), we have

If IG is the moment of inertia of the compound pendulum about its centre of mass, we can also define the radius of gyration kG about the centre of mass by writing IG = MkG2.

The parallel axis theorem gives us

Comparing to (3), we have,

If we switch h with h, equation (4) doesn't change, so we could have derived it by suspending the pendulum from O. In that case, the center of oscillation would be at O and the equivalent simple pendulum would have the same length l. Therefore the period would be the same as when suspended from O. Thus if we know the location of G, by measuring the period T with suspension at O and at various points along the extended line from O to G, we can find O and thus h.

Then using equation (4), we can calculate kG and IG = MkG2.

Knowing h gives us l = h + h, and since for small angle oscillations the period

We can calculate g using

1. To determine the radius of gyration of a compund pendulum about its center of gravity.

2. To investigate on the effect of fulcrum position.

3. To find the acceleration due to gravity 'g'.

4. To verify the equation of motion for a compound pendulum.

Introduction

A pendulum is a weight suspended from a pivot so that it can swing freely. When a pendulum is displaced sideways from its resting equilibrium position, it is subject to a restoring force due to gravity that will accelerate it back toward the equilibrium position. When released, the restoring force combined with the pendulum's mass causes it to oscillate about the equilibrium position, swinging back and forth. The time for one complete cycle, a left swing and a right swing, is called the period. The period depends on the length of the pendulum, and also to a slight degree on the amplitude, the width of the pendulum's swing.

Theory

In Fig. 1, O is the point of suspension of the compound pendulum and G is its centre of mass we consider the force of gravity to be acting at G. If h is the distance from O to G, the equation of motion of the compound pendulum is

Where I0is the moment of inertia of the compound pendulum about the point O.

Comparing to the equation of motion for a simple pendulum

We see that the two equations of motion are the same if we take

It is convenient to define the radius of gyration k0 of the compound pendulum such that if all the mass Mwere at a distance k0 from O,

the moment of inertia about O would be I0, which we do by writing I0 = Mk02

Substituting this into (1) gives us

The point O, a distance l from O along a line through G, is called the center of oscillation. Let h be the distance from G to O, so thatl=h+h'. Substituting this into (2), we have

If IG is the moment of inertia of the compound pendulum about its centre of mass, we can also define the radius of gyration kG about the centre of mass by writing IG = MkG2.

The parallel axis theorem gives us

Comparing to (3), we have,

If we switch h with h, equation (4) doesn't change, so we could have derived it by suspending the pendulum from O. In that case, the center of oscillation would be at O and the equivalent simple pendulum would have the same length l. Therefore the period would be the same as when suspended from O. Thus if we know the location of G, by measuring the period T with suspension at O and at various points along the extended line from O to G, we can find O and thus h.

Then using equation (4), we can calculate kG and IG = MkG2.

Knowing h gives us l = h + h, and since for small angle oscillations the period

We can calculate g using

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