Answer: weights on a pendulum

The problem asked why weights are added or subtracted on top a pendulum, and also how much friction is needed to keep the weights from falling off.

The weights are added on top the pendulum to make it go faster when it’s too slow. Because the weights are above the pendulum’s center of mass, they raise the center of mass slightly, reducing the period.

One commenter suggested their purpose was to slow the pendulum by stretching it out, thus lowering the center of mass. The Young’s modulus of wood is roughly 10^10 N/m^2. The cross-section of that pendulum might be 5*10^-4 m^2, giving a characteristic force of 5*10^6 N. The weights are small, maybe 1N total, while the length of the pendulum is about 1m, so the stretch of the pendulum is only about 2*10^-6m.

The mass of the pendulum might be a few hundred times the mass of the weights, and they’re perhaps 5-10cm above the pendulum’s center of mass. So they raise the center of mass of the pendulum by about maybe .05cm, dwarfing the contribution from the stretching of the pendulum. The pendulum’s center of mass is moved by .05 percent, which changes the frequency by half as much (because frequency depends on length^(-1/2)). That’s a few parts in ten thousand, or around 10 seconds a day.

The weights are very unlikely to slip off, even if the friction is pretty low. The reason is that the tangential force on them (which comes from friction) is very small. They have nearly the same motion as the pendulum itself, and the non-gravitational force on the pendulum is directed purely radially.

When the pendulum is at an angle A from vertical, it experiences a tangential gravitational acceleration g sin(A). The weights feel the same gravity, but their actual tangential acceleration is only about 90-95% as large (based on being only 90-90% the distance from the pivot as the pendulum’s center of mass). So the weights experience a friction force of roughly .1 g sin(A) m.

At the peak of their oscillation, the normal force is g cos(A) m, and friction is limited to mu g cos(A) m. Thus, they will only slip if .1 sin(A) > cos(A) mu, or mu < .1 tan(A).

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