Pseudo Forces

Transcript

0:00Introduction

Hello everyone! Okay, picture this. You are sitting in a car, and a small toy hangs from the mirror on a string. The car speeds up, and the toy swings back. Then it settles there, tilted, at rest relative to you. Now, on the toy there are only two forces: its weight, straight down, and the pull of the string, along the string. Can two forces at an angle ever balance? No, right? And yet the toy sits still. So, has Newton failed inside the car? Today one new idea fixes this: the pseudo force. And look, JEE problems on accelerating frames come up often, so let's begin.

0:42The fix: a pseudo force

So, what went wrong? The car is accelerating. Let's call its acceleration a nought: the a with an arrow on top, a vector, with a small zero below. And the car is not rotating. Now remember, a frame like this is not inertial, so Newton's laws do not hold in it. But we want to work from inside the car. So, suppose we add one extra force on the toy, to make it balance. The toy hangs tilted back, so the string pulls it up and a little forwards. Which way must this extra force point? Yes, exactly! Backwards, opposite to the car's acceleration. So, we give every body of mass m an extra force: minus m a nought vector. The minus sign simply means opposite to a nought. This is the pseudo force. Pseudo means false: in the ground frame, there is no such force at all. Right? And with it, Newton's second law works again, inside that frame. The big Greek letter sigma means sum, so sigma F real vector, the F with the little arrow, is the sum of all the real forces. Add the pseudo force, minus m a nought, and the total equals m times a rel vector, where a rel, with the small letters r e l below, is the acceleration measured relative to that frame. Clear?

2:04Effective gravity

Now look at the figure, on the left. There is a car with acceleration a, the dark arrow just to its right, and a bob hanging inside, just like our toy. On the bob, its weight m g, the green arrow, acts down, and the pseudo force m a, the red arrow, acts backwards. Together they act like one gravity: the effective gravity. g eff vector, the g with the small letters e f f below, equals g vector minus a nought vector. Okay? Normally, a plumb line, a string with a weight at the end, hangs along gravity, right? So inside this car, which way will it hang? Very good! It hangs along g eff. In the car, g eff is g down, combined with a, backwards. So g eff tilts back, and the string tilts back with it. That is exactly our toy! With me? The angle theta, the Greek letter, measured behind the vertical, is tan inverse of a over g. The minus one up on tan means inverse tan, not one over tan, okay? Puzzle solved: in the car's frame, the string's pull balances weight plus pseudo force, so the toy rests at an angle. Newton is safe! Same for anything dropped inside the car. Relative to the car, it falls along g eff, with acceleration g eff, not g. Here g eff without the arrow means just its size. Keep this in mind for Example two. Got it? And a pendulum in that frame? Its period T, the time of one full swing, is two pi times the square root of l over g eff, where l is the length of the string. Just g eff in place of g. Right? Now a topper's trick, on the right of the figure: a block on a smooth wedge with angle theta, and the wedge accelerates with a, to the left. In the wedge's frame, the block feels m g down, the pseudo force m a, the red arrow, to the right, opposite to the wedge's acceleration, and the normal reaction N, the blue arrow. Smooth, so no friction. The block stays at rest when g eff is perpendicular to the face, and that happens when a equals g tan theta. Think of it as tilted gravity, and it is one line!

4:18Trap: which frame are you in?

Now, the trap that catches many students. Suppose you solve a problem from the ground frame. Should you add the pseudo force there too? No, no! In the ground frame there is no pseudo force. There, you use only the real forces, and the true acceleration. Okay? Use the pseudo force only when you describe motion relative to the accelerating frame. Then it is real forces plus pseudo force, and the acceleration relative to that frame. Right? And never use both in one equation. Pick one frame, and stick to it. Clear?

4:55Example 1: friction on a passenger in an accelerating train

First example. So, a passenger of mass sixty kilograms, the k g on screen, stands in a train accelerating at two metres per second squared, the m s to the minus two. What horizontal force does the floor exert on him? And describe it in the frame of the train. Pause the video, and try it first. Done? Okay, let's check. First, from the ground. The friction from the floor must give him the train's acceleration, forwards. With me? Call this force capital F, the force from the floor. So, F equals m times a: sixty times two, one hundred and twenty newtons, forwards. Capital N is the newton, the unit of force. Only real forces, and the true acceleration. Simple! Now, the same man from inside the train. In the train's frame, he is at rest. But friction still pushes him forwards. So what balances it? Yes, exactly! The pseudo force, m a, acting backwards, opposite to the train's acceleration. Right? So friction balances it: F equals m a, again one hundred and twenty newtons, forwards. Same answer from both frames, and we never mixed them in one equation. Perfect!

6:13Example 2: where a ball falls in an accelerating train

Now a slightly harder one. So, a ball is dropped from one point two five metres, the m on screen, above the floor of a train. The train is accelerating at five metres per second squared. Where does the ball land, relative to the point directly below its release? Pause and try. Hint: work in the train frame. Done? Okay. Many students say: dropped straight down, so it lands directly below. Is that right? No, no, careful! Just before release, the ball moves along with the train, so in the train frame it starts from rest. After release, it has acceleration g down, and a backwards, because of the pseudo force. Okay? Now the vertical part. It falls one point two five metres from rest, with g taken as ten metres per second squared. So the time t is the square root of two times one point two five, over ten. That is the root of zero point two five: zero point five seconds, the small s on screen. And in this same half second, it also accelerates backwards at five. How far back does it drift? Half times five times zero point five squared. That is half times five times zero point two five: zero point six two five metres. So it lands zero point six two five metres behind the point directly below its release. Not directly below! Remember effective gravity? It slants backwards in the train, and so does the ball's path. Very good!

7:48Quick revision: pseudo force, effective gravity, one frame

So, quick revision. In a frame with acceleration a nought, add a pseudo force, minus m a nought, to every body. Weight and pseudo force together give effective gravity, g minus a nought, and plumb lines and dropped bodies follow it. And in the ground frame there is no pseudo force: never mix the two frames in one equation. Now go and try the practice problems on this page. All the best!