Viscosity from Terminal Velocity

Transcript

0:00Introduction

Hello everyone, here is a small puzzle. Drop a small steel ball into a tall jar of glycerine. At first it speeds up, but soon it falls at a perfectly steady speed, even though gravity is still pulling on it. Falling with no acceleration sounds impossible, right? Yet that steady speed lets us find the viscosity of the liquid, by timing the ball between two marks. Questions on this do come up in JEE, so let's begin.

0:29Why does the ball stop speeding up?

Let me ask you this: which three forces act on the ball while it falls? Its weight acts downwards. The upthrust of the liquid acts upwards, and so does the viscous force, against the motion. Okay? Stokes' law gives the viscous force. Capital F with a small v written below it equals six times pi times eta, times r, times v. Eta, the Greek letter like a small n with a long tail, is the viscosity. And r is the radius of the ball, while v is its velocity. Clear? So tell me, as the ball speeds up, what happens to this viscous force? Yes, exactly, it grows, because it is proportional to v. Soon the net force is zero, and the ball falls steadily at its terminal velocity. That is our puzzle solved. Got it?

1:23At terminal velocity: balance the forces

Now let us balance these forces. Capital W is the weight, and capital U is the upthrust of the liquid. At terminal velocity, capital W equals capital U plus F v, okay? Rho, the Greek letter like a small p, is the density of the ball. Sigma, like a small o with a line on top, is the density of the liquid. Which one goes in the weight, and which one in the upthrust? The weight uses rho: the volume, four thirds times pi times r cubed, then times rho, times g, the acceleration due to gravity. On the other side, the upthrust is the same with sigma, and then we add the viscous force. With me?

2:04Solve for v, then for η

Take the upthrust to the other side, and four thirds times pi times r cubed, times the bracket rho minus sigma, times g, equals the viscous force. Now solve for v. After cancelling, which power of r is left? It is r squared, because one r cancels, and pi cancels too. Four thirds over six gives two over nine. So v equals two times r squared, times the bracket, times g, all over nine times eta. Right? We want the viscosity, so swap eta and v. Eta equals two times r squared, times the bracket, times g, all over nine times v. Do not mix these two up: the nine sits with eta in one, and with v in the other. Clear?

2:50In the lab: a tall, wide jar of glycerine or castor oil

Here is the whole experiment in one figure. Glycerine or castor oil is in a tall, wide jar. The small circle is the ball, and the three arrows are capital W, capital U and F v, the same three forces. The text on the right repeats our balance and our formula for eta. Okay? Now the two dashed lines, capital A and capital B, are the two marks. The ball covers the distance d between them in a time t. So v is the distance between the marks, over the time, right? The note on the figure says that mark capital A, the upper mark, is well below the surface. The ball needs some distance to reach its terminal velocity, so the timing must begin only after the motion is steady. Got it? And the radius of each ball is found with a screw gauge, okay?

3:39Two precautions, and why they matter

Here are two precautions. Drop the ball gently along the axis of a wide jar, and note the temperature and keep it steady. So why must the ball stay away from the walls? That is because Stokes' law assumes a liquid of unlimited extent. Near a wall the drag is larger, and the ball falls more slowly. With me? Now repeat the experiment on a hot day. Does the ball take more time between the marks, or less? It takes less time, because viscosity falls sharply as the temperature rises, so the terminal velocity is higher. Okay?

4:17The graph of v against r²

Now, a ball of radius one millimetre falls at two centimetres per second. What about a ball of the same material with radius two millimetres? Is it four centimetres per second? No, that is the trap. The sign on screen means proportional to, and v is proportional to r squared. Double the radius, and v becomes four times, so the answer is eight centimetres per second. Clear? So for balls of the same material but different sizes, plot v against r squared, and you get a straight line through the origin. Its slope is two times the bracket, times g, all over nine times eta, so the slope gives eta. Right?

4:57Example 1: viscosity of glycerine from a falling steel ball

Let us take the easier example. Here, a steel ball of radius one millimetre and density seven thousand eight hundred kilograms per cubic metre falls through glycerine of density one thousand two hundred and sixty kilograms per cubic metre. Its terminal velocity is one point seven two centimetres per second. Take g as nine point eight metres per second squared, and find the viscosity. Pause the video and try it first. Done? Okay, let's check. Can one millimetre and one point seven two centimetres per second go straight in? No, they cannot, because the densities and g use metres. So r is ten to the power minus three metres, r squared is ten to the power minus six square metres, and v is zero point zero one seven two metres per second. Clear? The bracket rho minus sigma, seven thousand eight hundred minus one thousand two hundred and sixty, gives six thousand five hundred and forty. Put this in the formula with g as nine point eight, and you get the fraction on screen. With me? The top comes to zero point one two eight two, and the bottom to zero point one five four eight. So eta is zero point eight three pascal second, the unit you see on screen.

6:19Example 2: viscosity from an air bubble rising at 0.4 cm/s

Now let us take a harder one. An air bubble of radius zero point five millimetres rises through a liquid of density one thousand two hundred kilograms per cubic metre, at a steady zero point four centimetres per second. Neglecting the density of air, find the viscosity. Pause the video and try it first. Done? Okay, let's check. With rho as zero, the bracket is minus one thousand two hundred. So do we get a negative viscosity? Oh no, careful, we do not, because that formula was for a falling ball. The bubble rises, so the viscous force acts downwards. With no weight to count, the upthrust balances it, and eta equals two times r squared, times sigma, times g, over nine times v. Okay? In metres, r is zero point five times ten to the power minus three, so r squared is zero point two five times ten to the power minus six square metres. And v is zero point zero zero four metres per second. Clear? Now put these in, with the density of the liquid and g as nine point eight. The top comes to zero point zero zero five eight eight, and the bottom to zero point zero three six. So eta is zero point one six pascal second. If you got this on your own, very well done!

7:46Quick revision: viscosity formula, graph, precautions

So, quick revision. At terminal velocity, weight equals upthrust plus viscous force. Eta equals two times r squared, times the bracket, times g, over nine times v, and v is the distance between the marks over the time. v is proportional to r squared, so the graph of v against r squared is a straight line through the origin. Keep the upper mark well below the surface, drop along the axis, and keep the temperature steady. Now try the practice problems on this page. All the best!