Mixed Problems: Set 2

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Take a ring and a disc of the same mass, rolling side by side at the same speed. Same mass, same speed, so the same kinetic energy, right? Hmm, not so fast! One of them carries more, and soon you will know exactly how much. Today: three quick ideas from Set two of these mixed problems, then two solved problems, including a ladder about to slip. Problems like these come up often in JEE Main. So, let's begin.

0:34Idea 1: the energy of a rolling body

Idea one is exactly our puzzle; it is a problem from this set. Here, a ring and a disc of the same mass roll with the same speed. Find the ratio of their kinetic energies. Okay? Rolling is two motions at once: the centre moves with speed v, and the body spins with angular speed omega, this Greek letter. So K, the kinetic energy, is half capital M v squared, plus half ICM omega squared. Capital M is the mass; ICM, with small c m below, is the moment of inertia about the axis through the centre of mass. Right? For rolling without slipping, v equals omega R, with capital R the radius. Then K is half capital M v squared, times one plus k squared over R squared, where k is the radius of gyration. That bracket depends only on the shape. Clear? Now think. In a ring, all the mass sits at the rim. So, which has the bigger bracket? Ring or disc? The ring, yes! For a ring, k squared over R squared is one; for a disc, only half. That is why the ring carries more. Okay? Same mass, same speed, so half capital M v squared cancels, and only the brackets remain: one plus one is two, one plus half is three halves. K ring to K disc is two to three halves, that is, four to three. Puzzle solved! The ring carries one third more, all of it spin. Got it?

1:59Idea 2: a hinged rod falls: use energy

Idea two. Here, a uniform rod, sixty centimetres long and hinged at one end, is released from the horizontal. Find the speed of its centre after it turns through thirty degrees; the little circle means degrees. It turns about a fixed hinge, so we use energy. But how far does the weight fall? Careful! The weight acts at the centre, L over two from the hinge. So the centre falls L over two, times sine thirty degrees: L over four. Right? The rod turns about its end, so I is one third m L squared, not one twelfth. Small m is its mass, capital L its length. With me? So, m g times L over four equals half, times one third m L squared, times omega squared; the dot means times. The double arrow means so: the mass cancels, and omega squared is three g over two L, which, with g as ten and L as point six metres, is twenty five. Okay? So omega is five radians per second. And v c m, the speed of the centre, is omega times L over two: five times point three, one point five metres per second. That's it!

3:11Idea 3: a torque acting for a time

Idea three. Here, a solid sphere of mass one kilogram and radius point one metre spins at fifty radians per second about a diameter. Friction puts a constant torque of point zero one newton metre on it. Find the time to stop, and the number of revolutions. Okay? For a solid sphere about a diameter, I is two fifths capital M R squared. Two fifths, times one, times point one squared: point zero zero four kilogram metre squared. Right? Careful: capital L here is angular momentum, not a length. L equals I omega: point zero zero four times fifty, point two kilogram metre squared per second. Clear? The torque acts for a time t, about this fixed axis. So what does torque times time give you? The change in angular momentum, exactly! Angular impulse. Tau, this Greek letter, is the torque: tau t equals delta L, where capital delta, the triangle, means change in. The sphere stops, so all of L is lost: t equals point two over point zero one, twenty seconds. With me? And the turns? The torque is constant, so omega falls steadily to zero. So theta, the angle turned, is half of fifty, times twenty: five hundred radians. Divide by two pi; the wavy equals sign means approximately: about seventy nine point six, nearly eighty revolutions. Got it?

4:40Example 1: solid sphere rolling down a 30 degree incline

Now, Example one. Here, a solid sphere rolls from rest down an incline seven metres long, at thirty degrees. Find its speed at the bottom. Pause the video, and try it first. Done? Okay, let's check. It is pure rolling, that is, rolling without slipping, so friction does no work: the contact point does not move. So energy is conserved. Now, the height. Seven metres, right? No, no, careful! Seven metres is the length along the slope. So h, the height, is seven times sine thirty degrees: three point five metres. Okay? For a solid sphere, k squared over R squared is two fifths. So v squared equals two g h, over one plus two fifths: ten sevenths g h. Idea one's bracket again. Right? Ten sevenths, times ten, times three point five: fifty. So v is root fifty, that is, five root two, about seven point zero seven metres per second. Clear?

5:43Example 2: how far up can a person climb the ladder

Now a harder one. Here, a uniform ladder, ten metres long and weighing four hundred newtons, rests at sixty degrees to the floor, against a smooth wall. Mu, the coefficient of friction at the floor, is point three. How far up can a person of eight hundred newtons climb before it slips? Pause and try. Hint: torques about the foot. Done? Look at the figure. The smooth wall only pushes straight out: N w, small w for wall. At the floor, N f, small f for floor, pushes up, and friction, small f, points towards the wall. Capital W is the ladder's weight, at its middle, and theta is the angle with the floor. On the right, it solves the ladder alone; we add the person, a distance x along the ladder from the foot. Okay? Vertically, only the floor pushes up, so N f carries both weights: four hundred plus eight hundred, twelve hundred newtons. Right? It is about to slip, so friction is at its maximum, f max: point three times twelve hundred, three hundred and sixty newtons. Clear? Horizontally, N w balances friction, so it is three hundred and sixty newtons too. Now, torques. About which point? The foot, yes! N f and friction both act there, so they drop out. That is the trick: take torques where the most unknown forces act. Now, the ladder's weight acts five metres along it. Is its lever arm five sine sixty, or five cos sixty? Five cos sixty! The weight is vertical, so its lever arm is the horizontal distance from the foot; for the person, x cos sixty. But N w is horizontal, so its lever arm is the height of the top, ten sine sixty. So, N w times ten sine sixty equals four hundred times five cos sixty, plus eight hundred x cos sixty. With me? The left side is about three thousand one hundred and eighteen; the right, one thousand plus four hundred x. So x is about five point three metres. Got it?

7:48Quick revision: rolling KE, hinged rod, impulse, ladder

So, quick revision. Rolling energy: half capital M v squared, times one plus k squared over R squared; only the shape matters. For a hinged rod, use energy, with the drop of the centre and I about the hinge. Torque for a time gives tau t equals delta L, and for a ladder, take torques about the foot. Now go and try the practice problems on this page. All the best!