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Another Bar Pendulum Question

  • #1
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4

Homework Statement



How can one determine from a straight line plot, the value of g (acceleration due to gravity) and k (radius of gyration) of a bar pendulum?


Homework Equations



The time period of a bar pendulum is given by

[tex]T = 2\pi\sqrt{\frac{I}{Mgd}}[/tex]

[tex]I = I_{0} + Md^2[/tex]

where I is the moment of inertia about an axis passing through the pivot, d is the distance between the center of mass and the pivot, M is the mass of the rod.

In an experimental setup, we are varying d and measuring T for every chosen d.

The Attempt at a Solution



Also,

[tex]I_{0} = Mk^2[/tex]

So,

[tex]I = M(k^2 + d^2)[/tex]

Hence,

[tex]T = 2\pi\sqrt{\frac{k^2 + d^2}{gd}}[/tex]

For a fixed value of T, there are two values of d, and

[tex]d_{1} + d_{2} = \frac{4\pi^2(d_{1} + d_{2})}{T^2}[/tex]

and

[tex]d_{1}d_{2} = k^2[/tex]

But which straight line plot yields k and g? I can see that if k >> d, then we can say that T^2 is proportional to 1/d, but this would be a gross approximation, valid only for points very close to the center of mass.

Thanks
Cheers
vivek
 

Answers and Replies

  • #2
48
0
Moment of inertia around a center of mass for perfect rod of lenght 2k is:

[tex]I_{0}=\frac{1}{12}m(2k)^2[/tex]

How did you got this equation:

[tex]d_{1} + d_{2} = \frac{4\pi^2(d_{1} + d_{2})}{T^2}[/tex]
?

Fixed walue of T? What two walues of [itex]d[/itex]?
 
Last edited:

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