# The physical pendulum - distance, period, and angular frequency

• Hyperfluxe
In summary: In this case, the moment of inertia is (4/3)(1.87)(0.847)=2.03kgm. Therefore, the distance from the pivot to the CM should be 2.03m to maximize the frequency.

## Homework Statement

The physical pendulum shown on your paper is a 27.0 kg wedge of a circular disk of uniform density with radius, R=1.87 m and opening angle β=0.847 radians. The pivot point of the pendulum can be moved along the center line of the wedge as shown on your paper.

http://i.imgur.com/TqMkX.gif

a) Find the distance of the center of mass from the tip of the wedge.

b) If this physical pendulum is supended by a pivot at its tip, and oscillates with a small amplitude, find the period of oscilation.

c) Find the distance from the tip at which the pivot point should be placed to maximize the frequency for the pendulum.

d) Find the maximum angular frequency for this pendulum
HINT: The angular frequency' is the name often used for the parameter w' (omega) in the general equation for Simple Harmonic Motion: i.e., Acos(w't+phi). Note that it is also the angular velocity for the circular motion whose projection is SHM. Since you have already calculated the distance from the pivot to the CM and also Icm, you can now calculate the maximum value of w'.

Variables:
m=27.0kg
R=1.87m
A=0.847m

## Homework Equations

Xcm=(4/3)(R)(sin(0.5A)/A)

T=2pisqrt(I/mgXcm)

D=Acos(wt+phi)

## The Attempt at a Solution

a) Xcm=(4/3)(R)(sin(0.5A)/A)=1.21m (correct)

b) T=2pisqrt(I/mgXcm)=2.41s (correct)
A friend told me to use the parallel axis theorem but I didn't use it and still got the answer right, do I have to use it here?

c) I have no clue...I'm guessing I use D=Acos(wt+phi), and to maximize the whole cos term will be equal to 1, so D=A? It's clearly wrong but I don't know where to go from here.

d) Once I find the distance in part c, w=sqrt(mgd/I) and that's it?

Anyone? For part c) I realize that we have to use the equation of frequency (1/T), then find the derivative and set it equal to zero to find the distance which gives the maximum frequency, but what is the moment of inertia (Icm) in this case?

Hyperfluxe said:
Anyone? For part c) I realize that we have to use the equation of frequency (1/T), then find the derivative and set it equal to zero to find the distance which gives the maximum frequency, but what is the moment of inertia (Icm) in this case?

The moment of inertia of a single particle (which we can assume in this case, as the mass of the string is negligible) is equal to the mass of the particle multipled by the square of the moment arm, or the distance from the particle to the rotating axis.

## 1. How does the distance affect the period of a physical pendulum?

The period of a physical pendulum is directly proportional to the distance between the pivot point and the center of mass of the pendulum. This means that as the distance increases, the period also increases. This relationship can be described by the equation T = 2π√(I/mgd), where T is the period, I is the moment of inertia, m is the mass of the pendulum, g is the acceleration due to gravity, and d is the distance.

## 2. What factors influence the period of a physical pendulum?

The period of a physical pendulum is influenced by several factors, including the length of the pendulum, the mass of the pendulum, the distance between the pivot point and the center of mass, and the gravitational acceleration. The shape and density of the pendulum can also affect its period.

## 3. How is the angular frequency related to the period of a physical pendulum?

The angular frequency of a physical pendulum is equal to 2π divided by the period. This means that as the period increases, the angular frequency decreases, and vice versa. The angular frequency is a measure of how quickly the pendulum oscillates back and forth.

## 4. Can the period of a physical pendulum be calculated if the distance is unknown?

Yes, the period of a physical pendulum can be calculated even if the distance is unknown. This can be done by measuring the period of the pendulum at two different distances and using the equation T = 2π√(I/mgd) to solve for the unknown distance. Alternatively, the period can be measured at one distance and then the pendulum can be moved to another distance, allowing for the calculation of the period at that distance.

## 5. How does the shape of a physical pendulum affect its period?

The shape of a physical pendulum can affect its period by changing its moment of inertia. A pendulum with a larger moment of inertia will have a longer period compared to a pendulum with a smaller moment of inertia, given all other factors are the same. This is because a larger moment of inertia requires more energy to rotate, resulting in a longer period.