Conical Pendulums and Rotating Bodies

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Picture a fair at night. In a huge wooden well, a motorcyclist rides round and round on the wall. The wall is vertical, straight up and down! Nothing is under his wheels, and gravity pulls him down all the time. So why doesn't he slide down? And a bigger surprise: if he slows down too much, he does fall! Today, conical pendulums and rotating bodies will explain it. In JEE, these horizontal circle questions come up often. So, let's begin.

0:35Conical pendulum: balance up, pull inwards

So, look at the figure on the right: a conical pendulum. The bob, of mass small m, hangs from a string of length l, this italic l, and goes round a horizontal circle at a steady speed. Theta, this Greek letter, is the angle between the string and the dashed vertical. Two forces act: the tension, capital T, the blue arrow, and the weight m g, the green arrow, with g the acceleration due to gravity. Okay? Now, the dashed arrows split the tension into two parts: T cos theta, straight up, and T sine theta, towards the centre. So tell me. Does the bob ever move up or down? No, never! It stays in one horizontal circle. So the vertical forces must balance: T cos theta equals m g. That is our first equation. Right? And T sine theta points to the centre, so it supplies the centripetal force, m omega squared r. Omega, this Greek letter, is the angular speed, and small r is the radius of the circle. Clear? Now, quickly. What is r here? It is just l, the length of the string, right? No, no, careful! The string is slanted. The radius is the horizontal distance from the axis, l sine theta, as the figure shows. Put it in, and sine theta cancels from both sides: T equals m omega squared l. No theta in it at all! With me? Now put this T into the first equation, and the m cancels: cos theta equals g over omega squared l. Spin it faster, cos theta gets smaller, and the string swings further out. And with the speed v of the bob, tan theta equals v squared over r g. Got it? Last, the period, the time for one round: two pi times the square root of l cos theta over g; pi is the Greek letter pi. The figure shows these results on the right too.

2:31Rotor and well of death: friction holds you up

Okay? Now, another ride at the fair: the rotor. You stand against the wall of a big drum of radius capital R, spinning at angular speed omega. Then, the floor is taken away! The wall pushes you towards the centre. That push, the normal reaction capital N, is your centripetal force: N equals m omega squared capital R, clear? So it is this push, N, that stops you from falling, right? No, no, wait! N is horizontal; it cannot hold anything up. Friction does, acting up along the wall. The most it can give is mu times N, where mu, this Greek letter, is the coefficient of friction. So we need mu N greater than or equal to m g; that is this sign. Got it? Put in N, and the m cancels. The double arrow means so: omega must be at least the square root of g over mu R, okay? See, no mass in it: heavy or light, the same least spin. Clear? And the well of death from our puzzle? Same physics, with the rider's speed v in place of omega: v must be at least the square root of g R over mu. Speed really keeps him up! We will prove it in example two. With me?

3:46A ball inside a spinning smooth bowl

One more case: a small ball inside a smooth bowl of radius capital R, spinning about its vertical axis at angular speed omega. The ball rides up the side and goes round with the bowl. Okay? Theta is now measured at the centre of the bowl, from the bottom. The bowl is smooth, so the bowl's only push on the ball is N, along the radius to the centre, just like a string of length R tied there. So, the same result as the pendulum: cos theta equals g over omega squared R, right? Its height h above the bottom is R minus R cos theta, and putting in cos theta, that is R minus g over omega squared. Careful: h is measured from the bottom of the bowl, not from its centre. Got it? And remember, for the ball to rise off the bottom at all, cos theta must be less than one. So this works only when omega squared is greater than g over R; that sign means greater than. Clear?

4:42Example 1: angular speed and tension of a conical pendulum

Now, example one. Here, a conical pendulum has a one metre string at sixty degrees to the vertical; the upright m is metres, and the small circle means degrees. Find its angular speed, and the tension in terms of m g. Take g as ten metres per second squared, the m s to the power minus two. Pause the video, and try it first. Done? Okay, let's check. From cos theta equals g over omega squared l, we get omega squared equals g over l cos theta. Cos sixty degrees is zero point five, so ten over, one times zero point five; the cross means times: twenty. Okay? So omega is root twenty, which is two root five, about four point five radians per second. The wavy equals sign means about, and r a d s to the power minus one is radians per second. Clear? Now the tension. The string is holding the bob up, so T is just m g, right? Oh no, careful! Only the vertical part, T cos theta, balances m g. So T equals m g over cos theta: m g over zero point five, which is two m g. Twice the weight! Got it? And a quick check with our result, T equals m omega squared l: m times twenty times one, twenty m. With g as ten, that is exactly two m g. Perfect!

6:11Example 2: least speed in the well of death

Now, let's finish our puzzle from the start. Here, a motorcyclist rides round the vertical wall of a well of radius five metres, and mu is zero point five. Find his least speed. Take g as ten again. Pause the video, and try it. Done? Okay, let's check. Many students grab the level road formula, v equals root of mu g R, from our last lesson. Root of zero point five times ten times five is root twenty five: five metres per second, the m s to the power minus one, right? No, no, careful! On a level road, friction supplies the centripetal force. Here, the wall does it: the normal reaction N points to the centre, and N equals m v squared over capital R, okay? And friction acts up the wall, against his weight. So we need mu N greater than or equal to m g. With me? Put in N, and the m cancels: v squared must be at least g R over mu. Ten times five over zero point five: one hundred. Right? So the least speed is ten metres per second, twice the wrong answer! Any slower, and friction cannot hold his weight; he slides down. That is the secret of the well of death. Puzzle solved!

7:32Quick revision: conical pendulum, rotor, well of death

So, quick revision. In a horizontal circle, the vertical forces balance, and the inward force gives m omega squared r. Conical pendulum: T equals m omega squared l, cos theta equals g over omega squared l, and the radius is l sine theta. Rotor and well of death: N is the centripetal force, and friction holds the weight. Now go and try the practice problems on this page. You can do them. All the best!