Angular kinetic energy of a planar pendulum - Is the force at the pivot responsible for all change in the angular momentum>

Question

Is this statement below true? I cannot visualise how the pivot can cause a 'Moment of Force' on the club and be responsible for changes in its angular momentum.

Consider a golf club fixed to a pivot - like a planar pendulum - about an axis through the grip end. Two forces act on the club: A contact force at the motionless pivot point and gravity at the translating center of mass. From a work-energy perspective gravity does all the work, and more specifically, all the work done by gravity is linear. Yet, that linear work manifests itself in both linear and rotational kinetic energy in the club. The force at the fixed pivot does no work on the club. From an impulse-momentum perspective, the combined linear impulses of gravity and the force at the pivot equals the change in linear momentum, while the angular impulse – due to the moment of force from the force at the pivot – equals the change in angular momentum. To be explicit, from a work-energy perspective gravity is responsible for all change in angular kinetic energy of a pendulum, while from an impulse-momentum perspective, the force at the pivot is responsible for all change in angular momentum.

Answer 1

If only gravity is responsible for the motion, all work comes from it. It is the case of a pendulum.

But it is also possible that we want more speed at the translating mass than what is possible only by gravity. In this case the torque applied at the pivot is also responsible for the work.

In the case of pure gravity driven movement, the work done from the max. to the min. height equals the change of kinetic energy, that is equal to the loss of gravity potential energy: $E_g = mg\Delta h$.

If there is an additional torque at the pivot point: $E_a = \int T\omega dt$, where $\omega$ is the instantaneous angular velocity.

The total energy: $E = E_g + E_a$

We can think of a fan that lost all blades but one. Part of the energy for move it from up to down position comes from gravity.

Answer 2

In this case, and in many others, there is no definitive answer as to "where the angular momentum comes from."

When total linear momentum is zero, then angular momentum is equal at all observation points, near and far.

When total linear momentum is non-zero, or changing, torque $\vec{\tau}$ and angular momentum $\vec{L}$ vary, depending on your chosen observation point. Moment of inertia $I$ also varies by observation point.

If we choose the pivot point as our observation point, then the pivot provides no angular momentum. All the torque is from gravity.

If we choose the observation point that the CoM passes through at time $t_1$, then, at time $t_1$, gravity provides zero torque, and the pivot provides all the torque. Different $I$, different $\vec{L}$.

If we choose an observation point in between, or elsewhere, we get a mix.

Like a rainbow, wherever you stand, the torque is over there.

One way to view torque is as linear force multiplied by the weakness of the observation point, weakness to resist turning.

Similarly, angular momentum can be viewed as linear momentum multiplied by the weakness of of the observation point to modify it.

A fly, observed from a distant galaxy, can have more angular momentum than a freight train observed here on earth. Does the fly actually generate a huge invisible angular momentum field piercing distant galaxies?

All observation points obey the laws of physics, all produce valid conclusions that agree. So we choose an observation point, and that colors our perception of "where the angular momentum comes from."