The physical pendulum - distance, period, and angular frequency

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SUMMARY

The discussion focuses on the analysis of a physical pendulum, specifically a 27.0 kg wedge of a circular disk with a radius of 1.87 m and an opening angle of 0.847 radians. Key calculations include finding the center of mass (Xcm = 1.21 m) and the period of oscillation (T = 2.41 s). The participants explore the use of the parallel axis theorem for determining the moment of inertia and discuss maximizing the frequency by setting the derivative of the frequency equation to zero. The maximum angular frequency is derived using the formula w = sqrt(mgd/I).

PREREQUISITES
  • Understanding of physical pendulum dynamics
  • Familiarity with the parallel axis theorem
  • Knowledge of angular frequency and simple harmonic motion
  • Ability to calculate moment of inertia for composite shapes
NEXT STEPS
  • Learn about the parallel axis theorem in detail
  • Study the derivation of the moment of inertia for a circular disk
  • Explore the relationship between frequency and period in oscillatory motion
  • Investigate the effects of varying pivot points on pendulum dynamics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for practical examples of pendulum behavior and calculations.

Hyperfluxe
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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?
 
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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.
 

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