Part A: Mixed Problems

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a uniform rod, hinged at one end, and hold it horizontal. Now let it go. It swings down through the vertical. At that moment, how hard does the hinge push up on the rod? Just its weight, m g, right? Hmm, are you sure? We will check it in Example two. This is Part A of the unit problem bank: mixed problems from all of rotational motion, the kind that comes up often in JEE, okay? Two quick ideas first, then two solved problems. Let's begin.

0:39Idea 1: energy with a spinning cylinder

So, here is idea one. Read it with me: a zero point five kilogram block hangs from a string wound on a solid cylinder, of mass two kilograms and radius zero point one metre, on a fixed horizontal axle. Find the angular speed of the cylinder after the block has fallen one metre from rest. Okay? The block falls, so it loses potential energy. Where does all that energy go? Into two places, yes! The block speeds up, and the cylinder starts spinning. So the loss, m g h, equals half m v squared for the block, plus half I omega squared for the cylinder. Small m is the block's mass, g the acceleration due to gravity, h the height and v the block's speed; capital I is the cylinder's moment of inertia, and omega, this Greek letter, its angular speed. Right? For a solid cylinder about its axis, I is half capital M R squared, with capital M its mass and capital R its radius. And the string comes off the rim without slipping, so omega equals v over R, okay? Put these in, and half I omega squared becomes half, times capital M over two, times v squared. So, m g h equals half, times m plus capital M over two, times v squared. With me? Now the numbers, with g as ten. m g h is zero point five times ten times one: five joules. And m plus capital M over two is one point five. So five equals half times one point five times v squared; the double arrow means so: v squared is twenty over three. And v is about two point five eight metres per second; the wavy equals sign means approximately, and s to the power minus one means per second. Got it? Finally, omega equals v over R: two point five eight over zero point one. About twenty five point eight radians per second. That's it!

2:42Idea 2: a rolling sphere climbs a curve

Now, idea two. Read it with me: a solid sphere rolls without slipping on a horizontal surface at two metres per second, and then rolls up a curved surface. How high does it rise? Take g as ten. Okay? Easy, right? Like a ball thrown up: h equals v squared over two g. Four over twenty, zero point two metre: twenty centimetres. Correct? No, no, careful! That counts only the forward motion. The sphere is also spinning. So its kinetic energy, capital K, is half m v squared plus half I omega squared, with omega equal to v over R for rolling without slipping. Right? For a solid sphere, I is two fifths m R squared. Put it in: half m v squared, plus one fifth m v squared, gives seven tenths m v squared. Clear? It keeps rolling without slipping as it climbs, so all of this energy turns into height. m g h equals seven tenths m v squared. Cancel m, and h equals seven v squared over ten g. With me? v squared is four, and ten g is one hundred. So seven times four, over one hundred: zero point two eight metre, that is twenty eight centimetres. Not twenty! The spin energy is worth eight extra centimetres. Very good!

4:06Example 1: centre of mass of a plate with a corner cut out

Alright, example one, an easy Part A problem. Read it with me: from a uniform square plate of side four a, a square of side two a is removed from one corner. Find the distance of the centre of mass of the remainder from the centre of the original plate. Pause the video, and try it. Done? Okay, let's check. Let capital M be the mass of the full plate. The plate is uniform, so mass goes with area. Two a, squared, over four a, squared: four over sixteen, one quarter. So the removed square has mass one quarter of capital M, right? Put capital O at the centre of the plate. The removed square sits in a corner, so its centre is at a comma a, okay? Its distance from O is root two times a, by Pythagoras; the tick sign is the square root. Clear? Now the key idea. The whole plate is the remainder plus the removed square. And the whole plate has its centre of mass right at O, its centre. With me? So one quarter of the mass is removed. Then the shift is one quarter of root two a, right? Oh no, careful! About the centre, the two parts must balance. The remainder has mass three quarters capital M, and its centre sits at a distance d on the other side. So three quarters M times d equals one quarter M times root two a. Got it? Cancel capital M, and d equals root two a over three. And which way? Away from the hole, towards the opposite corner. Always divide by the mass that is left. Perfect!

5:50Example 2: hinge force when the rod passes the vertical

Now, example two, and it is our opening puzzle. Read it with me: a uniform rod of mass m and length capital L is hinged at one end and held horizontal. Find the force on the hinge when the rod passes through the vertical position, after release. Pause the video, and try it. Done? Okay, let's check. At the vertical, the rod hangs straight down, just like a rod at rest. So the hinge only holds up its weight, m g, right? No, no, wait! The rod is not at rest; it is swinging. So first, how fast? From horizontal to vertical, its centre falls L over two. And it turns about a fixed hinge at its end, so I is one third m L squared. Okay? Energy: m g times L over two equals half, times one third m L squared, times omega squared; the dot means times. So omega squared is three g over L, right? Now, at the vertical, what torque does gravity give about the hinge? Zero, exactly! The weight acts straight down through the hinge. So tau, the torque, is zero, and alpha, the angular acceleration, is zero too. Clear? But the centre still moves in a circle of radius L over two. So the net upward force, capital N minus m g, where capital N is the hinge force, equals m omega squared times L over two. With me? Put in omega squared equal to three g over L, and the L cancels: N minus m g is three halves m g. Got it? So N is five halves m g, two and a half times the weight, upwards! Not m g at all. The hinge must hold up the weight and also pull the centre round its circle. Puzzle solved!

7:46Quick revision: rolling sphere rise, cut plate, hinge force

So, quick revision. Energy with rotation: add half I omega squared to half m v squared. For a rolling solid sphere, that makes seven tenths m v squared, so h is seven v squared over ten g. Piece removed from a plate: balance about the old centre, and divide by the mass that is left. And a swinging rod at the vertical: N minus m g equals m omega squared times L over two, not N equal to m g. Now try the practice problems on this page. You will find them easy, I promise. All the best!