# Another Bar Pendulum Question

## 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

$$T = 2\pi\sqrt{\frac{I}{Mgd}}$$

$$I = I_{0} + Md^2$$

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,

$$I_{0} = Mk^2$$

So,

$$I = M(k^2 + d^2)$$

Hence,

$$T = 2\pi\sqrt{\frac{k^2 + d^2}{gd}}$$

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

$$d_{1} + d_{2} = \frac{4\pi^2(d_{1} + d_{2})}{T^2}$$

and

$$d_{1}d_{2} = k^2$$

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

Related Advanced Physics Homework Help News on Phys.org
Moment of inertia around a center of mass for perfect rod of lenght 2k is:

$$I_{0}=\frac{1}{12}m(2k)^2$$

How did you got this equation:

$$d_{1} + d_{2} = \frac{4\pi^2(d_{1} + d_{2})}{T^2}$$
?

Fixed walue of T? What two walues of $d$?

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