# Solving Harmonic Pendulum Homework

• joemama69
In summary, the problem is to find the period of small oscillations of a pendulum consisting of a solid, uniform sphere attached to one end of a thin, uniform rod. Using the torque equation, we can set up a differential equation that can be simplified and solved for the period of oscillation.
joemama69

## Homework Statement

A pendulum consists of a solid, uniform sphere of mass M and radius
R attached to one end of a thin, uniform rod of mass m and length L.
The pendulum swings freely about the other end of the rod. Find the
period of small oscillations of this pendulum.

## The Attempt at a Solution

T = -mhgsinQ = Ia = Id2Q/dt2

d2Q/dt2 +mghsinQ/I = 0

does sinQ = Q for small approx

should i also be able to find the actual I

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joemama69 said:

## Homework Statement

A pendulum consists of a solid, uniform sphere of mass M and radius
R attached to one end of a thin, uniform rod of mass m and length L.
The pendulum swings freely about the other end of the rod. Find the
period of small oscillations of this pendulum.

It almost seems from the way the problem is worded that you are expected to use

$$\tau = I \alpha$$

where tau is the torque, I is the moment of inertia, and alpha is the angular acceleration. Once you have set up the differential equation that goes with this, you can pick off what you need to answer the question.

Assuming that is the case, the torque is the force (due to gravity) times the distance from the center of mass of the ball to the pivot point, plus the force acting on the center of mass of the rod times the distance between the center of mass of the rod to the pivot point:

$$Mg sin(\theta) (R+L) + mg sin(\theta) (\frac{L}{2})$$

The moment of inertia for the system is the sum of the moment of inertia of the ball about the distant pivot point (use the parallel axis theorem) plus the moment of inertia of the rod about the end point (pivot point). I'll let you figure that out. (One of those two moments of inertia you can look up.)

$$(I_{total}) \frac{d^2 \theta} {d^2 t} = \tau$$

One of your tasks will be to simplify the terms for I total and the torque into something neater. Since it's a small angle oscillation, replace the sines with just the angle. Simplify the way the equation looks and try to make it match the equation for simple harmonic motion. Check and see what terms in the D.E. for SHM determine the period, and figure out your answer.

for the system?

Hello!

It seems like you are on the right track in your attempt at solving this problem. The equation you have written is the equation of motion for a simple harmonic oscillator, where T is the period, m is the mass, g is the acceleration due to gravity, h is the distance from the pivot point, I is the moment of inertia, and Q is the angle of displacement.

Yes, for small oscillations, the approximation sinQ = Q is valid. This is known as the small angle approximation and is commonly used in solving problems involving pendulums.

To find the actual moment of inertia for the system, you can use the parallel axis theorem. This states that the moment of inertia of a system is equal to the moment of inertia of its center of mass plus the product of its mass and the square of the distance between the center of mass and the new axis of rotation. In this case, the center of mass of the pendulum is at a distance L/2 from the pivot point, and the moment of inertia of the sphere is given by 2/5MR^2. So the total moment of inertia would be 2/5MR^2 + mL^2/4.

Hope this helps! Best of luck with your homework.

## What is a harmonic pendulum?

A harmonic pendulum is a type of pendulum that swings back and forth in a regular and repeating motion, known as harmonic motion. It is often used to study oscillatory motion and has many practical applications, such as in clocks and metronomes.

## How do you solve a harmonic pendulum homework problem?

To solve a harmonic pendulum homework problem, you will need to use the equation for the period of a pendulum, which is T=2π√(L/g), where T is the period, L is the length of the pendulum, and g is the gravitational acceleration. You will also need to consider the initial conditions, such as the angle at which the pendulum is released and the initial velocity.

## What factors affect the period of a harmonic pendulum?

The period of a harmonic pendulum is affected by the length of the pendulum, the gravitational acceleration, and the initial conditions, such as the angle at which the pendulum is released and the initial velocity. Other factors, such as air resistance and friction, may also have a small effect on the period.

## What is the difference between a simple pendulum and a harmonic pendulum?

A simple pendulum is a theoretical, idealized model that assumes the pendulum has no mass, no friction, and swings in a perfect circular arc. A harmonic pendulum, on the other hand, takes into account the mass, friction, and other factors that can affect the motion of a real-life pendulum. It is a more accurate model that better represents the behavior of a real pendulum.

## Why is the harmonic pendulum important in physics?

The harmonic pendulum is important in physics because it is a simple yet effective tool for studying oscillatory motion. It is also commonly used in practical applications, such as in clocks and metronomes. Additionally, understanding the behavior of the harmonic pendulum can help us better understand other complex systems that exhibit similar types of motion.

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