Physical Pendulums and Other Oscillators

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Imagine a straight, perfectly smooth tunnel through the Earth, between two cities far apart. You drop a ball in. It falls, reaches the far end, and comes back to you. Now dig a second tunnel, much shorter, between two nearby towns. Which ball comes back first? The short tunnel, surely, right? Hold on to your guess. This is a classic JEE question, so let's begin.

0:29The physical pendulum

Look at part c of the figure; parts a and b are the springs from earlier in this chapter. Here a rigid body hangs from a pivot, the point O, and its centre of mass, the point C, is a distance d from the pivot. The red arrow is its weight, m times g. This is a physical, or compound, pendulum, and the formula under it is where we are heading today. Okay? Now turn it by a small angle theta, the Greek letter between the dashed vertical line and the body. Let me ask you this: what pulls it back? Its own weight, exactly! About the pivot, the weight gives a torque, written with the Greek letter tau, equal to minus m times g times d, times sine theta; the minus sign means it turns the body back. For small angles, sine theta is about theta; the wavy equals sign means approximately. So the torque is proportional to the angle, and that is angular SHM, right? So capital T is two pi times the square root of capital I, over m times g times d; the tick-shaped sign is the square root. Careful: capital I is the moment of inertia about the pivot, not about the centre of mass. Got it?

1:45Three standard bodies

Let's use it. For a uniform rod of length capital L, about one end, capital I is one third of m times capital L squared, and d is capital L over two. The mass cancels. The double arrow on screen means 'which gives', so capital T is two pi times the square root of two times capital L, over three times g. Okay? Next, a ring of radius capital R hanging on a peg. Quick question: about the peg, is capital I just m times capital R squared? No, and that is exactly the trap, so be careful! That value is about the centre. The peg is a distance capital R away, so the parallel axis theorem adds another m times capital R squared. So capital I is two times m times capital R squared, d equals capital R, and capital T is two pi times the square root of two times capital R, over g. With me? And a disc pivoted at its rim gives two pi times the square root of three times capital R, over two times g. Proving this one is the practice problem on this page, with the same steps as the ring.

2:54Twisting oscillators: torsional pendulum and dipole

Right? Now twist a body hanging from a wire by an angle theta. The wire twists it back: tau equals minus capital C times theta, where this capital C is the torsional constant of the wire. This torsional pendulum has period two pi times the square root of capital I over capital C, okay? Next, an electric dipole in a uniform electric field; p is its dipole moment, and capital E is the field. For small angles, its period is two pi times the square root of capital I, over p times capital E, clear? Now a topper's trick: see the pattern. For the physical pendulum, m times g times d plays the role of capital C, and for the dipole, p times capital E does. So one formula covers all three, got it?

3:43Floating bodies and liquid in a U-tube

Now, two oscillators with no pivot. First, a body of uniform cross-section floating upright, like a wooden cylinder; the letter h stands for its submerged depth at rest. Push it down a little and let go, and it bobs with capital T equal to two pi times the square root of h over g. Okay? Second, liquid in a U-tube. Push it down by y on one side, and it rises by y on the other. So how far apart are the two levels now? They differ by two y, exactly! That extra liquid pushes it back, and that is where the two in the formula comes from. So capital T is two pi times the square root of capital L, over two times g, where capital L is the total length of the liquid column, not one arm. Clear?

4:34Example 1: period of a 1.5 m rod swinging about one end

Alright, time for examples. Example one: find the period of a uniform rod, one point five metres long, swinging about one end, with g equal to nine point eight metres per second squared. Pause the video and try it first. Done? Okay, let's check. Rod about one end, so we use the formula with two times capital L, over three times g, and put in one point five and nine point eight. Right? Two times one point five gives three, and three times nine point eight gives twenty nine point four. Three over twenty nine point four is one over nine point eight, so capital T is two pi times the square root of one over nine point eight, about two point zero one seconds. Okay? Look at that answer again. It is the period of a one metre simple pendulum, from our last lesson! Two thirds of one point five gives one, so this rod swings exactly like a simple pendulum one metre long. That length, small l, is called the length of the equivalent simple pendulum. With me?

5:40Example 2: SHM in a smooth tunnel along a chord of Earth

Now, remember our puzzle? Here it is. Example two: a particle is dropped into a straight, smooth tunnel along a chord of the Earth. Show that it executes SHM, and find the period. Here is a hint: think about how g changes inside the Earth. Pause the video and try it first. Done? Okay, let's check. Inside the Earth, g falls in proportion to the distance from the centre: at a distance r, g of r equals g times r, over capital R, the radius of the Earth. Clear? But this pull points to the centre, and the particle can only move along the tunnel. So which part of the pull moves it? Only its component along the tunnel, exactly! Let x stand for the particle's distance from the tunnel's mid-point. Then x over r is the cosine of the angle between the pull and the tunnel, so that component is g of r, times x over r. The r cancels, leaving g times x, over capital R, towards the mid-point. Okay? So the acceleration along the tunnel, written a, is minus g over capital R, times x: it always points back to the mid-point, and it is proportional to x. That is SHM, with omega squared equal to g over capital R, where omega, the Greek letter, is the angular frequency. Right? So capital T is two pi over omega, which is two pi times the square root of capital R over g. With capital R equal to six thousand four hundred kilometres and g equal to nine point eight, that is about eighty four point six minutes. Got it? Now look closely: the length of the chord appears nowhere in this answer. So every smooth tunnel, long or short, takes the same time, and both balls in our puzzle come back together! If you said the short tunnel, that was the trap: it is shorter, but the pull along it is weaker too. Clear?

7:40Quick revision: physical, torsional, U-tube, tunnel periods

So, quick revision. Physical pendulum: two pi times the square root of capital I, over m times g times d, with capital I about the pivot, for small angles. The torsional pendulum and the dipole follow the same pattern, and the floating body and U-tube results are on screen too. And for a tunnel along any chord, the period is about eighty four point six minutes. Now go and try the practice problem on this page. All the best!