Determining k and g from a Straight Line Plot for a Bar Pendulum

[maverick280857]

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

 

[Ene Dene]

Moment of inertia around a center of mass for perfect rod of length 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?