Leg oscillation question

Question

Consider a Tyrannosaurus which has a leg length of approximately 3m. Approximate the leg as a board. What would be such a boards period of oscillation if swinging freely about its end?

I keep getting the answer as 3.48 s but that's not correct and I don't know why.

Answer 1

Hi 

Arun

Is the answer 3.14 sec

If yes then get back 

Answer 2

Hi Arun!

 

A board swinging about its end is what is called a "physical pendulum."  In a simple pendulum, the mass is all concentrated at the end of the support, and the rest of the support is considered to have insignificant mass.  In a physical pendulum, though, the mass is spread all along the support (which I assume to be uniform; otherwise, we would need to have some kind of mass distribution function).  In this case, the period of the pendulum is given by:

 

T = 2π(√(I/mgL_cm))

 

where

 

I = moment of inertia of the object around the support axis

m = mass of object

L_cm = distance from the support point to the center of mass

 

For a uniform rod (or board) swinging around one end, we can look up the moment of inertia in a table (or online).  It is:

 

I = (1/3)mL²

 

where 

 

L = full length of the board

 

So we get, for the period:

 

T = 2π(√(((1/3)mL²)/(mgL_cm))

 

with, in this case:

 

L = 3 m

L_cm = 1.5 m (since the center of mas should be halfway down the board from the support point)

 

Note that the mass of the board cancels out here, so it does not matter (which is why it is not given).

 

Using g = 9.8 m/s², I get T = 2.84 s.

 

I hope this helps!  Just let me know if you have any questions at all about this.

Answer 3

For a pendulum, the period is given by:

 

T = 2pi√(L/g)

 

where L is the length of the pendulum arm from the point of oscillation to the mass and and g is the acceleration of gravity.  3.48 seconds would be correct for a pendulum of length L= 3 m where the mass is concentrated at the bottom 3 meters from the point of oscillation.  In a uniform board, however, the mass is concentrated at the center of mass only 1.5 m from the point of oscillation.  So the board is better modeled as a pendulum of length L = 1.5 m, which yields a period of 2.46 seconds.