Oscillations of a Balanced Object

  • #1
horsegirl09
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Two identical thin rods, each of mass m and length L, are joined at right angles to form an L-shaped object. This object is balanced on top of a sharp edge. If the object is displaced slightly, it oscillates. Assume that the magnitude of the acceleration due to gravity is g.

Find omega, the frequency of oscillation of the object. Your answer for the frequency may contain the given variables m and L as well as g.

I got that w= (g/L)^.5 but it keeps saying I am off by a multiplicative factor.
 

Answers and Replies

  • #2
Redbelly98
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In the formula (g/L)^0.5, "L" is the distance between the pivot point and the center-of-mass.
 
  • #3
alphysicist
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Hi horsegirl09,

The formula

[tex]
\omega = \sqrt{\frac{g}{L}}
[/tex]

applies to simple pendulums in which the mass is concentrated at a single point at the end of a massless string. In this problem that is not true; when the mass is extended out, it is (usually) called a physical pendulum. Do you have the formula for the frequency of a physical pendulum?
 
  • #4
horsegirl09
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w= sqrt(Lmg/I_cm + mL^2)? with the parallel axis theorem?
 
  • #5
alphysicist
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w= sqrt(Lmg/I_cm + mL^2)? with the parallel axis theorem?


It might be tedious to calculate directly the moment of inertia (rotational inertia) I_cm for the L shape using an integration. However, the L is just two rods, each rotating about a pivot point at one of their ends. What is the moment of inertia of each of the rods about this pivot? Once you have the separate values of I for each rod, what is the combined value for the entire L shape?

The formula for the physical pendulum is

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

where I is the moment of inertia of the entire object about the pivot point, and d is the distance from the pivot point to the center of mass. So you still need to find where the center of mass is.
 
  • #6
horsegirl09
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I= (1/12)mL^2 but there are two so it would be (1/6)mL^2
Center of mass of the object is the moment of the L shape divided by the mass of the object. ((1/6)mL^2)/2m)?
 
  • #7
Doc Al
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I= (1/12)mL^2 but there are two so it would be (1/6)mL^2
Almost. I= (1/12)mL^2 is the moment of inertia of a thin rod about its center; you need it about one end.
Center of mass of the object is the moment of the L shape divided by the mass of the object. ((1/6)mL^2)/2m)?
No. The center of mass will be midway between the midpoints of each piece. Figure out its distance from the axis.
 
  • #8
horsegirl09
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Oscillation of a Balanced Object

The moment is mL^2 for one rod, 2mL^2 for both. The center of mass of each rod is L/2. therefore d is L/2.
w= sqrt((2m*g*L/2)/(2mL^2))
 
  • #9
Doc Al
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The moment is mL^2 for one rod, 2mL^2 for both.
No. What's the moment of inertia of a rod about one end?
The center of mass of each rod is L/2.
Yes.
therefore d is L/2.
No. "d" is the distance from the pivot point to the center of mass of the object. Where's the center of mass of the the object? (Draw yourself a diagram.)
 
  • #10
horsegirl09
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Oscillation of a balanced object

moment of inertia about the end of a rod is (1/3)mL^2. The pivot point is where the ends of the two rods meet. So i don't understand what d is.
 
  • #11
Doc Al
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moment of inertia about the end of a rod is (1/3)mL^2. The pivot point is where the ends of the two rods meet.
Good.
So i don't understand what d is.
As I said before (and as alphysicist explained in post #5) "d" is the distance from the pivot point to the center of mass of the object. To help you find the center of mass of the object, draw a diagram.
 

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