Compound Pendulum

In the previous section we dealt with a simple pendulum which consisted of a point mass connected to a fixed axis by a massless rod.  How do we deal with a real object that has mass distributed over a finite volume?  We call a rigid body suspended and free to swing about an axis a compound pendulum.

We assume the axis does NOT pass through the center of mass.  In the drawing, O is a point on the axis;  G is the point at the center of mass of the object; and O’ is a point at a distance (to be defined in a little bit) that is

at a distance, l, from the axis.  Note that we specify point O, and we can determine point G by simply seeing where the object balances.

As with the simple pendulum, we start with Newton’s 2nd law:  St = Iθ’’  .

Recall that we could write the moment of inertia (about an axis through point O) as:
IO = MkO2  where the kO is called the radius of gyration and M is the total mass of the object.  Also the total gravitational torque acting about point O is the same as if all of the mass were concentrated at the center of gravity.  Thus the gravitational torque is:
t = -Mgh sin(θ).  Putting all three of these together, we get:

-Mgh sin(θ) = MkO2θ’’  .

We see that this is equivalent to the simple pendulum equation (-Mgl sin(θ) = Ml2θ’’) if we define l as

l = kO2/h  ,

and so the compound pendulum will oscillate like a simple pendulum with a frequency (small angle approximation) of  wo = [g/l]½ .  Thus, by experimentally measuring wo, we could get an experimental value for l.

We then define h’ as the difference between l and h:  l = h + h’.  Combining the two expressions for l we have:

h + h’  =  kO2/h   ,   or  h2 + hh’  =  kO2  .

Using the parallel axis theorem (which states that the moment of inertia about a point is equal to the moment of inertia about the center mass plus the moment of inertia of a point of mass at the center of mass about the original point), we have

MkO2 = MkG2 + Mh2   ,   or    kO2 = kG2 + h2   .

Combining the two equations for kO2, we get:    h2 + hh’  =  kO2 = kG2 + h2   ,  or

kG2 = hh’  .

Symmetry:  Since kG2 is symmetric with respect to h and h’, we can conclude that if the object were suspended about an axis through O’, it would oscillate at the same frequency as if it were suspended about an axis through O.

Sweet spot on a bat (or racket, or club, or ...)

Consider the same figure as above: a rigid object pivoted to rotate about an axis through the point O; the object has a center of mass (center of gravity) at point G.

We will consider an impulse (caused by a force of short duration – such as an impact of a ball on the bat) J’ at the point O’ where the force, F’, is perpendicular to the line through O and G and O’.  We will also allow for the existence of an impulse at the axis, J at point O, where the force, F, is also perpendicular to the line through O and G.

J’ =   ∫ F’ dt    at O’;         J  =  ∫ F dt    at O  .

Recall that Newton’s 2nd Law says:

SFexternal  =  F’ + F  =   dP/dt      where P = MV = momentum of center of mass

where the magnitude of V is:  V = hθ’    (since G is a distance h from O).  Since all of the vectors are in the same direction at the instant of collision, we have:

F’ + F = d(Mhθ’)/dt,    or      D(Mhθ’)  =  J’ + J   .

Here we are assuming θ’ = 0 initially, so that we have  Mhθ’ = J’ + J due to the impulses.

In terms of torque and angular momentum:

St = dL/dt,    where t = F’l     and   L = Iθ’  =  MkO2θ’

since l is the distance from the axis through O to the location of the force, F’, at O’.  Note that F does not cause a torque since it is located at the axis of rotation (through point O).  This gives:

J’l = D(M kO2θ’)    ,  or  with θ’ being initially zero,      J’l = M kO2θ’  .

If we want J to be zero (so that there is no force at the axis), the above two red equations give:

J’ = Mhθ’ = MkO2θ’/l   ,   or  hl = kO2 .

We note that

l = h + h’,   so that     hl = h2 + hh’;   and so   kO2 = h2 + hh’,      and that

kO2 = kG2 + h2     (from parallel axis theorem);

so that we finally get the condition for zero force on the axis as:

h2 + hh’  =   kG2 + h2 ,   or   kG2  =  hh’   .

Note that this is the same condition we had for the oscillation of the compound pendulum.  Thus the sweet spot on the bat is located at the same distance from the axis that the effective length is for the bat to behave like a simple pendulum.

To locate the sweet spot, we could fix the bat to rotate about point O; measure the period of small angle oscillations; and then use w = [g/l]½  = 2p/T  to calculate l.  If the impact of the ball on the bat is not at this distance, there will be a force on the axis.  This is what causes the bat to jar the hands when you hit a ball at a location other than the sweet spot.