Mixed Problems: Set 3 (Relative and Circular Motion)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Look at a wall clock. The minute hand is six centimetres long, and the second hand is eight centimetres long. The second hand goes round sixty times faster than the minute hand, doesn't it? So its tip also moves sixty times faster. Right? Hmm, keep your answer ready. This is Set three of our mixed problems: relative motion and circular motion. And look, in JEE Main these ideas come again and again, all mixed up, so you must spot the right tool yourself. So, let's begin.

0:36Tool 1: relative velocity, subtract the vectors

So, tool one, with the first problem of this set. Car A moves north at twenty metres per second, and car B moves east at fifteen; on screen, m s to the power minus one means metres per second. Find the velocity of B relative to A, okay? Imagine you are sitting in car A, and looking out at the other car. Which way does car B seem to go? Hmm, keep that picture in your head. Now, e cap and n cap, the letters with little caps on top, are unit vectors pointing east and north. Many students just put the two velocities together: fifteen e cap plus twenty n cap. Is that correct? No, no, careful! Relative means subtract. v B A vector equals v B vector, minus v A vector; the little arrow on top means a vector, and the small B slash A below means B relative to A, okay? So it is fifteen e cap, minus twenty n cap. And your picture from car A? See, you are rushing north, so car B seems to go east and also slide south. That south slide is the minus twenty n cap. Now its size. The two parts are at right angles, so use Pythagoras. Root of fifteen squared plus twenty squared is root six hundred and twenty five: twenty five metres per second. The tick-shaped sign is the square root, and the small two on top means squared. Clear? And the direction: fifteen towards the east, and twenty towards the south. Theta, the Greek letter on screen, is the angle from east towards south, and tan theta equals twenty by fifteen, that is, four by three. So theta is tan inverse four by three, south of east; the small minus one means inverse. That's it! Twenty five metres per second, at tan inverse four by three, south of east.

2:25Tool 2: two accelerations on a circle

Now, tool two: circular motion. Here, a particle moves on a circle of radius two metres, and s, the arc length it covers, equals t cubed, in metres. Find its acceleration at t equals one second. First, the speed. How do you get the speed from s? Differentiate, exactly! v equals d s by d t, the rate at which s changes with time. That is three t squared, so at t equals one, three metres per second. Differentiate once more: d v by d t is six t. This is a sub t, the tangential acceleration; the small t below stands for tangential. It acts along the velocity, and changes the speed. At t equals one, it is six metres per second squared; that is the m s to the power minus two on screen, okay? Many students stop right here. The acceleration is six metres per second squared, done, right? No, no, careful! The particle is moving on a circle, isn't it? Its velocity keeps turning, so there is also a centripetal acceleration, towards the centre. So, a sub c, with the small c for centripetal, equals v squared by r: three squared by two, that is, nine by two, four point five metres per second squared. Now, a sub t points along the circle, and a sub c points to the centre. They are at right angles, so it is Pythagoras again. The total acceleration is root of a sub t squared plus a sub c squared: root of thirty six plus twenty point two five, that is, seven point five metres per second squared. Never just add the two parts.

4:03Example 1: speed ratio of minute and second hand tips

Got it? Alright, remember our clock from the start? Let's finish it. Example one. The minute and second hands of a clock are six and eight centimetres long; c m means centimetres. Find the ratio of the speeds of their tips. Pause the video, and try it first. Done? Okay, let's check. The second hand goes round sixty times faster, so the tips are in the ratio one is to sixty; the colon on screen means is to. Right? Only half right! One turn takes the minute hand sixty minutes, but the second hand only one minute; m i n on screen means minutes. So their angular speeds, omega, are in the ratio one is to sixty. The small m and s below omega mean the minute hand and the second hand, okay? But omega tells us how fast a hand turns, not how fast its tip moves. For the tip, use v equals omega r, where r is the length of the hand. So v m is to v s equals one into six, is to sixty into eight. The cross means into. With me? That is six is to four hundred and eighty, which is one is to eighty. Eighty times, not sixty! See, puzzle solved: the second hand turns sixty times faster, and it is also longer, so its tip moves eighty times faster. Exactly!

5:25Example 2: bolt falling in a lift rising at steady speed

Now a harder one, JEE level. Here, a lift rises at a steady two metres per second. Then a bolt comes loose from its ceiling, two point five metres above the floor. How long does it take to hit the floor? Pause and try. Hint: sit inside the lift. Done? Okay, let's check. Many students think: the floor is rising, and the bolt first moves up with the lift. So the two metres per second must change the answer, right? No, no, careful! Sit inside the lift, and remember: relative means subtract. Just before it comes loose, the bolt moves up at two metres per second, along with the lift. So u rel, its initial velocity relative to the lift, is two minus two, zero. The small r e l below u means relative, okay? And the relative acceleration, a rel? The bolt falls with g, the acceleration due to gravity. The lift moves steadily, so its acceleration is zero. So a rel equals g minus zero, just g. Inside the lift, the bolt simply drops from rest, through two point five metres. Nice and simple. So, distance equals half g t squared, with g equal to ten. Two point five equals half into ten into t squared, that is, five t squared. So t squared is zero point five, and t is root zero point five: about zero point seven one seconds. The wavy equals sign means approximately. And look, the two metres per second simply cancelled out: two minus two, zero! Sitting in the right frame turned a tricky problem into a simple drop. That is the power of relative motion.

7:09Quick revision: relative velocity and total acceleration

So, quick revision. Relative motion: v B A vector equals v B vector minus v A vector, and when it helps, sit on one body. On a circle: v equals omega r, and a sub c equals v squared by r, towards the centre. And when the speed changes, the total acceleration is root of a sub t squared plus a sub c squared, never their plain sum. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!