Category: 07 Gravitation

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.

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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.

127 Accelerated Tied Buoy

When stationary, the upthrust must balance the tension in the rubber band plus the weight of the ping pong ball.

ρVg = T + mg

When the bucket is accelerated upward at a, the apparent g of the water becomes g+a. The water thus exerts a larger upthrust on the ball than before. Applying N2L on the ball,

ρV(g+a) – Tmg = ma’

ρVa = ma’

a’=ρV/m a

Since water is more dense than ping pong ball, ρV > m. Hence, while the bucket accelerates at a, the ping pong ball accelerates at >a, and the rubber becomes more stretched.

On the other hand, when the bucket has a downward acceleration of a, the apparent g of the water becomes g-a. The water exert a smaller upthrust on the ball than before. Applying N2L on the ball,

T + mgρV(ga) – = ma’

ρVa = ma’

a’=ρV/m a

Since water is more dense than ping pong ball, ρV > m. Hence, while the bucket has a downward acceleration of a, the ping pong ball has a downward acceleration >a, and the rubber becomes less stretched.

The simplest case to consider is if the bucket is released from rest. As the bucket free falls, the upthrust acting on the ball disappears. While the bucket and the water free fall at g, the ball accelerates at larger than g, because it would be pulled down by gravitational force plus the tension in the rubber band.

 

126 Artificial g: Buoy in Water Pail

Whether stationary or in circular motion, the ball displaces the same amount of water. Yet, the upthrust acting on the ball varies continuously during the circular motion to provide the required centripetal force.

Consider when the bucket is at the bottom most position.

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If the bucket is stationary, then

  1. A column of water of height h would have weight hAρg. This weight must be supported by the upward pressure force exerted by the water below. Hence the water pressure at depth h must be hρg.
  2. The pressure gradient results in the ball experiencing an upthrust U = ρVg, where V is the volume of displaced water.
  3. ρVgmg = 0. Hence the ball is stationary.

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If the bucket is in circular motion, then

  1. A column of water of height h would have weight hAρg. In addition to supporting this weight, the upward pressure forces exerted by the water below must also provide for the required centripetal force of hAρ(2). Hence the water pressure at depth h must be hρ(2+g).
  2. The larger pressure gradient results in the ball experiencing a larger upthrust U = ρV((2+g).
  3. The net force experienced by the ball is thus ρV(2+g) –mg = ρV2 = mrω2. Hence the ball does the same circular motion as the bucket of water.

When the bucket is in circular motion, the water pressure becomes larger because the water molecules would be pressing harder into one another. This is equivalent to the water experiencing an artificial g that is directed in the centrifugal direction, and an “upthrust” that is always in the centripetal direction. The magnitude of this artificial g would be varying continuously (in the exact manner as the tension in the rope, and normal contact force in the bucket) during the circular motion, so as to keep the water and the ball in circular motion.