Part B: Harder Problems

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. The radius of a pendulum bob is not known, and you cannot measure it. You have only some thread, a metre scale and a stopwatch. Can you still find g, the acceleration due to gravity, and the radius of the bob too? We will solve it in the second example. This is part B of the unit problem bank, the harder problems, and these are JEE pattern questions, so let's begin.

0:28The falling ball: time and radius

First, the falling ball. In the figure, a ball falls through a tall jar of liquid; the arrows are its weight, capital W, the upthrust, capital U, and the viscous force, capital F with a small v below it, and the text on the right repeats their balance. At its steady terminal velocity, v equals two times r squared, times the bracket rho minus sigma, times g, all over nine times eta. Here r is the radius, rho, the Greek letter like a small p, is the density of the ball, sigma is the density of the liquid, and eta, like a small n with a long tail, is the viscosity. Okay? In our problem the balls are of the same metal, in the same liquid, so rho, sigma, g and eta are all fixed. Then what can still change v? Yes, exactly, only the radius can change it. So v is proportional to r squared; the curly sign on screen means proportional to. Right? But in the lab we time the ball between two marks, the dashed lines capital A and capital B in the figure. It covers a distance d in a time t, so v equals d over t. And, as the note in the figure says, mark A sits well below the surface, so the ball is already at terminal velocity when we start timing. With me? Now let me ask you this. Take a bigger ball of the same metal, with the same marks. Does it take more time between the marks, or less? It takes less time, because it falls faster. For the same marks, t is proportional to one over v, and so to one over r squared. Clear?

2:09Pendulum: the radius of the bob is unknown

Now the simple pendulum, on the left of the picture; the graphs beside it are about damping, and we do not need them today. Capital L is the length marked, and capital A is the amplitude. The period, capital T, equals two pi times the square root of L over g; pi is the Greek letter. Square both sides, and the double arrow means so: capital T squared equals four pi squared over g, times L, okay? But L is the effective length, from the point of suspension to the centre of the bob: thread, plus hook, plus radius of the bob. Our problem has no hook, so, as the note under the drawing says, L is just thread plus radius of the bob. So L equals small l, the thread, plus r, the radius. Right? Now here is the hard part. We know neither r nor g. So how can we get rid of r? Yes, we take two readings, with threads of lengths l one and l two; the small one and two below are just labels. So capital T one squared equals four pi squared over g, times the bracket l one plus r, and capital T two squared is the same with l two. With me? Now subtract the first equation from the second. The r cancels, and g equals four pi squared times the bracket l two minus l one, over capital T two squared minus capital T one squared. So g comes out without the radius. Got it? Once g is known, capital L one equals g times capital T one squared, over four pi squared, and r is capital L one minus small l one. Here is a topper's trick: when the same unknown sits in every reading, take two readings and subtract, like the end correction in the resonance tube. Clear?

3:52Example 1: fall time of a 1.5 mm ball if 1.0 mm takes 12 s

Now, example one, the easier one. Here, a ball of radius one point zero millimetres takes twelve point zero seconds to fall between two marks in a jar of oil. How long does a ball of the same metal, of radius one point five millimetres, take? On screen, m m means millimetres, and s means seconds. Pause the video and try it first. Done? Okay, let's check. The radius is one and a half times as big, so the time is twelve point zero over one point five, which gives eight point zero seconds, right? No, it is not, and this is the trap. The time goes as one over r squared, not one over r. So t two over t one equals r one over r two, the whole bracket squared; the ones and twos label the two balls. Okay? Now the numbers. One point five squared is two point two five, so one point zero over one point five, squared, is one over two point two five. Right? So t two equals twelve point zero over two point two five, which gives five point three seconds. The bigger ball takes less time, just as we said. If you avoided the eight seconds answer, very good!

5:07Example 2: g and bob radius from two thread lengths

And now example two, the harder one, our puzzle from the start. The radius of a pendulum bob is not known. With threads of length sixty point zero centimetres and one hundred point zero centimetres, the periods are one point five seven four seconds and two point zero two two seconds; c m means centimetres. Find g and the radius of the bob. Pause the video and try it first. Done? Okay, let's check. Many students put L as sixty point zero centimetres straight into g equals four pi squared times L, over capital T squared; the slash on screen means over. Is that correct? No, that is the trap. Sixty centimetres is only the thread; the real L also includes the radius. So we use our two thread method. First, l two minus l one is one hundred point zero minus sixty point zero, which gives forty point zero centimetres, or zero point four zero zero metres. Okay? Now the periods. One point five seven four squared is about two point four seven seven, and two point zero two two squared is about four point zero eight eight. Their difference is one point six one one, in seconds squared. Right? Now into our formula. Four pi squared is about thirty nine point four eight, so g equals thirty nine point four eight times zero point four zero zero, over one point six one one. That gives nine point eight zero, in m s to the power minus two, meaning metres per second squared. With me? So g is done. Now, how do we get the radius itself? Yes, exactly, we go back to one thread. Capital L one equals g times capital T one squared, over four pi squared. Nine point eight zero times two point four seven seven, over thirty nine point four eight, gives zero point six one five metres, or sixty one point five centimetres. Clear? So the radius is sixty one point five minus sixty point zero, which gives one point five centimetres. Our puzzle is solved: with two threads and a stopwatch, we found both g and the radius. If you got both answers, really well done!

7:18Quick revision: t as 1 over r squared, two-thread method

So, quick revision. For the same metal in the same liquid, v is proportional to r squared, so the time between the marks goes as one over r squared. For a pendulum, L is thread, plus hook, plus radius of the bob. And with an unknown radius, use two threads: g equals four pi squared times the bracket l two minus l one, over capital T two squared minus capital T one squared. Now try the rest of part B on this page before reading the solutions. All the best!