Fluids in Accelerating Containers

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. You are standing in a lift, holding a bucket of water, and a wooden block is floating in it. Suddenly the lift shoots upward. You feel heavier, right? Your legs feel it. So, does the block feel heavier too, and sink deeper into the water? And one more: what if the lift falls freely? Today we learn how a liquid behaves when its container accelerates, and one simple idea will answer both questions. This idea appears in JEE problems too, so let's begin.

0:36One idea: effective gravity

So, take any container, and give it an acceleration a. See the small arrow on top of the a? It means a vector: it has a size and a direction. And the liquid inside has settled down, moving along with the container. Okay? Now think of yourself sitting in a car. The car speeds up, forward. Which way do you get pushed? Backward, into your seat, exactly! It feels like an extra gravity, pointing opposite to the acceleration. The liquid feels the very same thing. So it behaves as if gravity were g effective, equal to g vector minus a vector. The small e f f written below is a subscript, short for effective. This one line is the heart of today's lesson. Clear? Now recall, for a liquid at rest: P equals P naught plus rho g h. P naught, with the small zero, is the pressure at the free surface; rho, the Greek letter, is the density of the liquid; and h is the depth. With me? In an accelerating container, the rule is really simple: wherever you see g, use g effective instead. Pressure, upthrust, the shape of the surface, everything follows from this. Got it?

1:48Lift accelerating upward: gravity feels stronger

Now, back to the lift from our puzzle. Look at the left half of the figure: a block floating in water. Its weight, m g, mass times g, pulls it down. The upthrust, rho l V sub g, pushes it up. Rho with a small curly l is the density of the liquid, and V sub is the volume under water. When it floats, the two are equal. Now the lift accelerates upward with a. Gravity points down, the acceleration points up. So what is g effective? g plus a, yes! Minus a vector, when a points up, means an extra a pointing down. So it adds to g. That is exactly why you feel heavier in a lift going up. Right? So the pressure at depth h becomes P naught plus rho, times g plus a, times h; the round brackets keep g plus a together. More pressure at the same depth. Okay? And the upthrust? Rho l V sub, times g plus a: bigger! But wait. Inside the lift, the block's effective weight, its real weight plus the extra downward push it feels in the lift, also becomes bigger: m times g plus a. Clear? So now, our puzzle. Does the block sink deeper, or does it rise? Neither! Both forces grew by the same factor, g plus a, so the balance does not change. The submerged fraction stays exactly the same. Many students say it sinks deeper. Oh no, careful! That is the trap. Got it? And the second question: free fall. The lift accelerates downward with a equal to g, so g effective is g minus g: zero! No pressure difference with depth, so no upthrust at all. The block and the water simply float freely together. With me?

3:32Horizontal acceleration: the surface tilts

Now the right half of the figure. The tank accelerates to the right; see the arrow marked a. And the surface is no longer flat. It tilts by an angle theta, the Greek letter, measured from the dashed horizontal line. So, gravity points down, and a points sideways. Which way does g effective point? Remember, it is g minus a. So g effective has g pointing down, and a pointing backward, opposite to the motion. It leans back, just like you in that car. Okay? Now, a liquid at rest relative to its container keeps its free surface at right angles to the gravity it feels. In a still tank, gravity is straight down, so the surface is flat. Here g effective leans back, so the surface tilts. Right? And the angle? The sideways part is a, and the downward part is g. So tan theta equals a over g, where tan is the tangent of the tilt angle. Bigger acceleration, steeper surface. Clear?

4:33Rotating container: the surface becomes a bowl

One more case. Spin a cylinder of water about its vertical axis, with angular speed omega, the Greek letter that looks like a curly w. The water spins along with it. Okay? Now tell me. Where does the water climb higher: at the centre, or at the rim, near the walls? At the rim, correct! The water climbs the walls and dips in the middle, like a bowl. This bowl shape is called a paraboloid, and its lowest point is at the centre, on the axis. With me? And how high? At a distance r from the axis, the surface is above the lowest point by y equals omega squared r squared over two g. The small two on top means squared. Careful with two things: r is measured from the axis, not from the wall, and do not forget the two in the bottom. Got it?

5:24Example 1: gauge pressure in a bucket accelerating up at g/2

Alright, example one, an easy one. Here is a bucket of water, accelerating upward at g over two. Find the gauge pressure at depth h. Pause the video, and try it first. Done? Okay, let's check. Upward acceleration, so gravity feels stronger: g effective is g plus g over two. Right? Now, careful. The question asks for the gauge pressure, not the total pressure. So, do we add P naught, the pressure at the open surface? No, no, careful! Gauge pressure is P minus P naught: the pressure above the atmosphere. So the P naught term simply drops out. Clear? So the gauge pressure is rho, times g plus g over two, times h: one point five rho g h. One and a half times its value at rest. Simple, isn't it?

6:21Example 2: level difference in a tank accelerating at 5 m/s2

Now a harder one. Here is a tank, two metres long, partly filled with water; the small m on screen stands for metres. It accelerates horizontally at five metres per second squared; the minus two on the s means per second squared. Take g as ten. Find the difference in water level between its front and back. Pause the video, and try it. Done? Okay, let's check. Tan theta equals a over g: five over ten, which is one half. Okay? So the level difference is just tan theta: half a metre, right? No, no, wait! Tan theta is a slope: rise over run. For every metre of length, the surface drops by tan theta. So over the whole two metres, the drop is two times tan theta. With me? Two times one half: one metre. One whole metre of difference! Now, which end is higher: the front or the back? The back, exactly! Opposite to the acceleration, just as in the figure. The liquid piles up at the back, like the tea in your cup when the car speeds up. Got it?

7:30Quick revision: g eff, lift, tilted and rotating surfaces

So, quick revision. In an accelerating container, g effective equals g vector minus a vector; use it in place of g. Lift going up: P equals P naught plus rho, times g plus a, times h; the floating fraction does not change; and in free fall, there is no upthrust. Sideways: tan theta equals a over g, higher at the back. And spinning: y equals omega squared r squared over two g. Now go and try the practice problems on this page. One idea, and every case falls into place. All the best!