Centripetal Force

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You tie a stone to a string and whirl it round in a circle, at a steady speed. The speed never changes. So, by Newton's first law, no net force is needed, right? Then why must your hand keep pulling on the string? Let go, and the stone flies off! Today we settle this, with one big idea: centripetal force. Circular motion questions come up often in JEE, and they all start from here. So, let's begin.

0:34Uniform circular motion needs a net inward force

Alright, let's look carefully. Picture a body moving on a circle of radius r, at a constant speed v. This is called uniform circular motion. The speed stays the same, but look at the direction. It keeps changing, every moment, right? Now tell me. The speed is constant. So the acceleration is zero, isn't it? No, no, careful! Velocity has a direction too, and that direction is changing, so the velocity is changing. So there is an acceleration: a equals v squared over r, and it points towards the centre of the circle. The little two on v means squared. We can also write it as omega squared r. Omega is this Greek letter, the angular speed, and v equals omega times r. Okay? Now, Newton's second law, in the ground frame: an acceleration needs a net force, mass times acceleration. So the net force on the body must be F c equals m v squared over r, or m omega squared r, where m is the mass. The small c below the F stands for centripetal, which means centre seeking. And its direction? Inwards, towards the centre, always. With me?

1:47Not a new kind of force

Now, back to our stone. It needs a force towards the centre. So who supplies it? Which real force pulls the stone inwards? The tension in the string, yes, exactly! Your hand pulls the string, and the string pulls the stone towards the centre. That is why your hand must keep pulling, and that settles our puzzle. So remember: F c is not a new kind of force. It is just the name for the net inward force, and real forces supply it. Clear? Next, take a car turning on a level road. Nobody is pulling it with a string. So which force makes it turn? Friction from the road on the tyres, correct! That is the only horizontal force the road gives. So, depending on the question, the inward force may come from tension, friction, a normal force, a spring force, gravity, or a combination of them. Got it? Now, here is the method for any horizontal circle. This big Greek letter is sigma, and it means the sum. Sigma F radial, the sum of all the forces along the radius, towards the centre, equals m v squared over r. Okay? And the body is not moving up or down, so sigma F vertical equals zero. Two equations, and most problems are done. Right?

3:04The trap: never add an extra force

Now, the most common mistake in this chapter. Put our stone on a smooth table and list the forces on it, in the ground frame. Tension T along the string, inwards. The weight m g downwards, where g is the acceleration due to gravity, and the normal force N from the table, upwards. These two balance. Okay? Now many students say: it moves in a circle, so let me also draw a force, m v squared over r. Should we? No, no, never! In the ground frame, m v squared over r is not a separate force. Nobody exerts it. Draw it, and you count the same force twice. This is a classic trap, so draw only the real forces. Got it? So what is right? The tension is the only horizontal force, so the tension itself is the net inward force: T equals m v squared over r. See, m v squared over r is the result of the real forces, not one more force. Clear? One more thing. If you sit in a frame rotating with the body, the body is at rest in that frame. There, an outward centrifugal pseudo force, m omega squared r, appears, and the real forces balance it. It comes in that frame only, never in the ground frame along with m v squared over r. We will meet it in section seven point four. For now: ground frame, real forces only. Okay?

4:32Example 1: tension in a stone whirled at 2 rev/s

Alright, example one, an easy one. Read it with me: a zero point five kilogram stone, on a one metre string, is whirled in a horizontal circle on a smooth table, at two revolutions per second. Find the tension. Pause the video, and try it first. Done? Okay, let's check. The table is smooth, so there is no friction. The weight and the normal force balance, and the tension is the only horizontal force. So T equals m omega squared r. Right? Now, omega. Two revolutions per second, so omega is just two, right? Oh no, careful! Omega is in radians per second, and one revolution is two pi radians. Pi is this Greek letter. So omega is two pi times two, which is four pi radians per second; on screen, rad s to the power minus one. Got it? Now substitute. T equals zero point five, times four pi, whole squared, times one. Four pi, whole squared, is sixteen pi squared. With me? Zero point five times sixteen is eight. So T equals eight pi squared, about seventy nine newtons. This wavy equals sign means approximately, and the capital N here is the newton, the unit of force. Put omega as two, and you would get just two newtons, completely wrong! So always change revolutions into radians. Clear?

6:02Example 2: speed of a puck held by a hanging mass

Now a harder one. Read it with me: a zero point one kilogram puck sits on a smooth table. It is tied to a string that passes through a hole in the table and holds a hanging zero point five kilogram mass, at rest. The puck moves in a circle of radius zero point two metres. Find its speed. Pause the video and try it. Done? Okay, let's check. Two bodies, one string. The key is the tension. How do we find it? From the hanging mass. On screen we call its mass capital M, to keep it apart from the puck. It is at rest, so the forces on it balance: the tension upwards equals its weight, M g. Take g as ten metres per second squared: zero point five times ten, five newtons. Okay? Now, the puck moves on the circle, so we need m v squared over r. But careful with the masses. Which mass goes in it: the hanging zero point five, or the puck's zero point one? The puck's mass, zero point one, yes! The puck is the body moving on the circle, so small m is its mass. The table is smooth, so the tension is the only horizontal force on the puck, pulling it towards the hole, the centre. So T equals m v squared over r. This double arrow means implies: v squared equals T r over m. Right? Now substitute. v squared equals five times zero point two, over zero point one. Five times zero point two is one, and one over zero point one is ten. With me? So v is root ten; this is the square root sign. About three point two metres per second; on screen, m s to the power minus one. See, the hanging weight supplies the centripetal force, through the string. Very good!

7:53Quick revision: mv2/r from real forces, no extra arrow

So, quick revision. Constant speed on a circle still needs a net inward force, m v squared over r, or m omega squared r. Real forces supply it: tension, friction, a normal force, a spring, or gravity. And in the ground frame, never draw an extra m v squared over r arrow. Now go and try the practice problems on this page. All the best!