Physical pendulum frequency

In summary: No, the masses are just attached at different ends of a rigid rod. So, in this case, the third law is still obeyed.Also, if you know Lagrangian mechanics, deriving the correct answer is really straightforward: you just sum over two oscillators with different lengths. So, the correct answer is (6/5)^1/2.
  • #1
ehrenfest
2,020
1

Homework Statement


A long, straight, and massless rod pivots about one end in a vertical plane. In configuration I, two small identical masses are attached to the free end; in configuration II, one mass is moved to the center of the rod. What is the ratio of the frequency of small oscillations of configuration II to that of configuration I?
(A) (6/5)^1/2 (the correct answer)
(B) (3/2)^1/2
(C) 6/5
(D) 3/2
(E) 5/3

Homework Equations



Angular Frequency of a physical pendulum of length L with a mass at its endpoint: sqrt(g/L)

"The center of mass of a system moves as if it were a single particle of mass equal to the total mass of the system, acted on by the total external force, and independent of the nature of the internal forces" (Marion and Thornton page 333)

The Attempt at a Solution



Why is it wrong to apply the quote from Marion and Thornton and say that this situation is equivalent to the situation of having a physical pendulum of length 3/4 L with both of the masses at the endpoint?
 
Physics news on Phys.org
  • #2
The mass switch affects potential and kinetic energies in a different way. I guess M&T are describing a system that is in uniform motion (every particle moving with the same velocity).
 
  • #3
Also, if you know Lagrangian mechanics, deriving the correct answer is really straightforward: you just sum over two oscillators with different lengths.
 
  • #4
ehrenfest said:
1.


Homework Equations



Angular Frequency of a physical pendulum of length L with a mass at its endpoint: sqrt(g/L)

"The center of mass of a system moves as if it were a single particle of mass equal to the total mass of the system, acted on by the total external force, and independent of the nature of the internal forces" (Marion and Thornton page 333)



The formula you cited refers to the "mathematical pendulum". For a physical pendulum with mass m and moment of inertia I with respect to the axis of rotation, the angular frequency is

[tex] \omega = \sqrt{\frac{mgs} {I}} [/tex]

where s is the distance of CM from the axis.

ehild
 
  • #5
clamtrox said:
The mass switch affects potential and kinetic energies in a different way. I guess M&T are describing a system that is in uniform motion (every particle moving with the same velocity).

Marion and Thornton derive that principle for any system of particles that obeys the weak form of Newton's third law (that forces are equal and opposite). Is the weak form of Newton's third law violated here?
 

What is a physical pendulum?

A physical pendulum is a type of pendulum that consists of a rigid body suspended from a fixed point and can swing back and forth under the influence of gravity.

What is the frequency of a physical pendulum?

The frequency of a physical pendulum is the number of complete swings or oscillations it makes in a given amount of time. It is dependent on the length of the pendulum, the mass of the object, and the acceleration due to gravity.

How is the frequency of a physical pendulum calculated?

The frequency of a physical pendulum can be calculated using the formula f = 1/2π√(g/L), where f is the frequency, g is the acceleration due to gravity, and L is the length of the pendulum.

What factors affect the frequency of a physical pendulum?

The frequency of a physical pendulum is affected by the length of the pendulum, the mass of the object, and the acceleration due to gravity. Changing any of these factors will result in a change in the frequency of the pendulum.

How is the frequency of a physical pendulum different from a simple pendulum?

A simple pendulum consists of a mass hanging from a massless string, while a physical pendulum consists of a rigid body suspended from a fixed point. The frequency of a physical pendulum is affected by its mass distribution and the length of the body, while the frequency of a simple pendulum is only affected by its length.

Similar threads

  • Mechanical Engineering
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
974
  • Advanced Physics Homework Help
Replies
9
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
636
  • Advanced Physics Homework Help
Replies
9
Views
3K
Replies
10
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Mechanical Engineering
Replies
1
Views
1K
Back
Top