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Compound Pendulum- Symmetric
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Aim:

 

To determine:

 

  • The acceleration g of gravity using a compound pendulum.

 

  • The radius of gyration kof the compound pendulum about an axis perpendicular to the plane of oscillation and passing through its centre of mass.

 

  • The moment of inertia Iof the compound pendulum about an axis perpendicular to the plane of oscillation and passing through its centre of mass.

 

Theory:

 

In Fig. 1, O is the point of suspension of the compound pendulum and G is its centre of mass; we consider the force of gravity to be acting at G. If h is the distance from O to G, the equation of motion of the compound pendulum is 

                  

Where I0is the moment of inertia of the compound pendulum about the point O. 

Comparing to the equation of motion for a simple pendulum

                   

We see that the two equations of motion are the same if we take

                                                          

It is convenient to define the radius of gyration kof the compound pendulum such that if all the mass M were at a distance k0 from O,

the moment of inertia about O would be I0, which we do by writing I0 = Mk02

Substituting this into (1) gives us 

                               

The point O′, a distance l from O along a line through G, is called the center of oscillation. Let h′ be the distance from G to O′, so that l=h+h'. Substituting this into (2), we have 

                                           

 

If IG is the moment of inertia of the compound pendulum about its centre of mass, we can also define the radius of gyration kG about the centre of mass by writing IG = MkG2

 

The parallel axis theorem gives us 

                                                                         

Comparing to (3), we have,

                                   

 

If we switch h with h′, equation (4) doesn’t change, so we could have derived it by suspending the pendulum from O′. In that case, the center of oscillation would be at O and the equivalent simple pendulum would have the same length l. Therefore the period would be the same as when suspended from O. Thus if we know the location of G, by measuring the period T with suspension at O and at various points along the extended line from O to G, we can find O′ and thus h′.

Then using equation (4), we can calculate kand IG = MkG2

 

Knowing h′ gives us l = h + h′, and since for small angle oscillations the period 

                                                                        

We can calculate g using

                                  

The minimum period Tmin , corresponds to the minimum value of l. Recall that l = h + h and that kG2 = hh' is a constant, depending only on the physical characteristics of the pendulum. 

Thus, l=h+kG2/h, and the minimum I occurs when, 

                                                                       

 i.e, when h2=kG2, h=h' and l=2h=2kG.

 

Thus, at Tmin, l=2kG.

 

 

 

 

 

 

 

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