- #1

horsegirl09

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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.

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- Thread starter horsegirl09
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- #1

horsegirl09

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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.

- #2

- 12,145

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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

Homework Helper

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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

Homework Helper

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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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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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Almost. I= (1/12)mL^2 is the moment of inertia of a thin rodI=(1/12)mL^2but there are two so it would be (1/6)mL^2

No. The center of mass will be midway between the midpoints of each piece. Figure out its distance from the axis.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)?

- #8

horsegirl09

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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

Mentor

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No. What's the moment of inertia of a rod about one end?The moment is mL^2 for one rod, 2mL^2 for both.

Yes.The center of mass of each rod 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.)therefore d is L/2.

- #10

horsegirl09

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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. So i don't understand what d is.

- #11

Doc Al

Mentor

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Good.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.

As I said before (and asSo i don't understand what d is.

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