Period of Small Oscillations for Rigid Body Pendulum

In summary, a pendulum with a light rigid rod of length 250mm and two identical uniform solid spheres of radius 50mm attached to its lower end has a period of small oscillations that can be calculated using the equation Period = 2*pi*sqrt(I/MgR), where I is the moment of inertia, M is the mass of the pendulum, g is the acceleration due to gravity, and R is the length of the pendulum. To find the period perpendicular to the line of centers, one must use the parallel axis theorem to calculate the moment of inertia, taking into account the distance between the pivot point and the axis of rotation. This method should give different answers for the two parts of the problem.
  • #1
willisverynic
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0

Homework Statement


A pendulum consists of a light rigid rod of length 250mm, with two identical uniform solid spheres of radius 50mm attached one on either side of its lower end. Find the period of small oscillations (a) perpendicular to the line of centers, and (b) along it.

Homework Equations


Period = 2*pi*sqrt(I/MgR)
Parallel axis theorem: I = Icm + Md^2

The Attempt at a Solution


I tried simply applying the parallel axis theorem to a ball using Icm = 2/5mr^2 and d = sqrt(.25^2+.05+2) but this yields and incorrect answer. I believe this method would be wrong because it would give the same answer for both parts, which is not the case

Thanks

ps not sure if this belongs in advanced or not
 
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  • #2
welcome to pf!

hi willisverynic! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)
willisverynic said:
I tried simply applying the parallel axis theorem to a ball using Icm = 2/5mr^2 and d = sqrt(.25^2+.05+2) but this yields and incorrect answer. I believe this method would be wrong because it would give the same answer for both parts, which is not the case

i suspect you're measuring from the pivot point rather from than the axis

the axis of rotation is through the pivot point, either North or East …

the parallel axis is through the centre of one sphere, and the distance between the two axes does depend on which way round they are :wink:
 

Related to Period of Small Oscillations for Rigid Body Pendulum

1. What is a rigid body pendulum?

A rigid body pendulum is a physical system consisting of a rigid object (such as a bar or disk) attached to a fixed point by a pivot or hinge. It can swing back and forth under the influence of gravity, similar to a simple pendulum.

2. What is the period of small oscillations for a rigid body pendulum?

The period of small oscillations for a rigid body pendulum is the amount of time it takes for the pendulum to complete one full swing. It is affected by the length, mass, and distribution of mass of the pendulum, as well as the strength of gravity.

3. How is the period of small oscillations calculated for a rigid body pendulum?

The period of small oscillations is calculated using the formula T = 2π√(I/mgd), where T is the period, I is the moment of inertia of the pendulum, m is the mass, g is the acceleration due to gravity, and d is the distance from the pivot point to the center of mass of the pendulum.

4. What is the significance of the period of small oscillations for a rigid body pendulum?

The period of small oscillations is an important characteristic of a rigid body pendulum as it determines the rate at which the pendulum swings and can be used to predict its behavior. It is also used in various applications such as timekeeping devices and seismometers.

5. How does the period of small oscillations for a rigid body pendulum compare to that of a simple pendulum?

The period of small oscillations for a rigid body pendulum is longer than that of a simple pendulum with the same length. This is because a rigid body pendulum has a larger moment of inertia due to its mass being distributed over a larger area, resulting in a slower swinging motion.

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