Month: June 2016

129 Accelerated Buoys

If we think of the water as one body, then normal contact force (of bucket on water) must be larger when the bucket is accelerating upward, and smaller when the bucket is accelerating downward.

Likewise, if we look within the water itself, then the water pressure must also be larger when the bucket is accelerating upward, and smaller when the bucket is accelerating downward.

By thinking about the pressure forces required to accelerate a column of water of height h, we can deduce that the water pressure at depth h must be hρ(g+a).

Since upthrust is the net resultant force from all the pressure forces, the upthrust that the water exerts on the ball must also be affected the same way. Thus the upthrust is now ρV(g+a). We can actually think of (g+a) as the apparent g of the water.

So when the bucket is stationary, the ball experiences zero net force because ρVgmg = 0.

When the bucket is undergoing acceleration a, the ball experiences a net force of ma because ρV(g+a) – mg = ma. This implies the ball must also be accelerating at a.

Amazingly, when the water becomes more or less pressurized in order to accelerate itself at a, it also pushes exerts a larger or smaller upthrust on the ball that ensures the ball also accelerates at a, for the same amount of water displaced. That’s why the ball does not become more or less submerged.

Actually, this conclusion is obvious if we remember that the floating ball and the displaced water have the same mass. If the displaced water accelerates up and down at the same rate as the rest of the water (which it obviously must), then the floating ball must also accelerate up and down at the same rate as the water.

128 Tied Buoy Accelerator

You can think of this contraption as an acceleration meter. The length of the rubber is an indicator of the acceleration. When it is at the usual extension, acceleration is zero. When it stretches beyond this length, acceleration is upward. When it shrinks, acceleration is downward.

As in any oscillation, the acceleration is always opposite to the displacement. So in this vertical oscillation, the acceleration is upward (shown by the stretched rubber band) whenever it is below the equilibrium position. It does not matter whether it is going downward or upward.

Likewise, regardless of whether the velocity is upward or downward, whenever it is above the equilibrium position, the acceleration is downward (as shown by the shrunk rubber band)