Set 2 (Power, Vertical Circles and Potential-Energy Curves)

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Picture a wind turbine giving twenty kilowatts in a wind of eight metres per second. Next day the wind blows at four metres per second, half the speed. So the turbine gives half the power, ten kilowatts, right? Not even close! Hold that thought. Today, Set two of the mixed problems: power, vertical circles and potential energy curves, which come up often in JEE Main. Four quick tools, then two full problems. Let's go.

0:35Tool 1: power is force times speed

So, tool one: power. Power is the rate of doing work. When a force F pushes along the motion, the power P equals F times v, force times speed. Okay? Now, a car moves at a steady speed. So tell me, how big is the engine's driving force? Equal to the total resisting force, yes, exactly! Steady speed means zero acceleration, so the net force is zero.

1:02Tool 2: wind power goes as v³

Right? Now, back to our turbine. Air of density rho, the Greek letter like a small p, blows at speed v through the area capital A swept by the blades. Each second a mass rho A v passes, and each kilogram carries half v squared. So the kinetic energy arrives at the rate half rho A v cubed. With me? The turbine extracts a fraction, so its power is proportional to v cubed; the sign like an open eight on its side means proportional to. So, half the wind speed, half the power, right? No, no, careful! The arrow shows the speed falling from eight to four metres per second, written m s to the power minus one. Four over eight, cubed, is one eighth, and twenty times one eighth is two point five kilowatts, k W on screen. Not ten! Puzzle solved. Got it?

1:55Tool 3: will the bob go round?

Tool three: a vertical circle. In the figure, a bob goes round the centre O on a string of length small l. Point A is the lowest point, with speed u; B is level with O, and C is the top. Theta is the angle from the lowest point, T the tension, m g the weight, and the orange arrow the velocity v. At the bottom, T minus m g gives the centripetal force, so T equals m u squared over l, plus m g. Okay? Now think. Where is the string most likely to go slack? At the top, yes! The string can pull, but never push. So it stays taut only if v top, the speed at the top, is at least root g l; this sign means greater than or equal to. Energy then needs u at least root five g l at the bottom, for a bob going round freely under gravity; the figure lists it too. The figure also gives v squared and T at any angle theta, and T A minus T C, bottom minus top, equal to six m g. We will check that last one in Example two. Got it? Now a problem from this set: zero point four kilograms on a zero point eight metre string, six metres per second at the bottom. Tension: zero point four times thirty six, over zero point eight, is eighteen, plus m g, four. Twenty two newtons, N for newtons, a nice tight string. So it surely goes round, right? No, no, wait! Never judge by the bottom tension. Use the test: u squared is thirty six, but five g l is five times ten times zero point eight, forty. Thirty six is less than forty, so the string slackens before the top. Clear?

3:42Tool 4: reading a potential-energy curve

Last tool: potential energy curves. The figure plots U of x, the potential energy at position x. Only a conservative force acts, so the total energy, capital E, the orange dashed line, stays fixed. E equals K plus U, with capital K the kinetic energy. So K equals E minus U, the gap between line and curve; the double arrow means so. With me? So, where does the particle move fastest? At the bottom of the valley, exactly! Where U is least, K is greatest, like point A, the stable point. B is unstable, C neutral, and x one and x two are turning points, where K is zero. And F equals minus d U over d x: the force points downhill. Right? Next, from this set: a one kilogram particle, U equals ten x squared minus forty x, in joules, released at rest at x equals zero. There U and K are both zero, so E is zero. Okay? Where is U least? Set d U over d x, the slope of the curve, to zero. Twenty x minus forty is zero, so x is two. There U is forty minus eighty: minus forty joules, capital J for joules. So K max, written with a small max, is zero minus minus forty: forty joules. Half times one times v squared is forty, so v is root eighty, about eight point nine metres per second; the wavy equals sign means approximately. Clear?

5:22Example 1: power of a car climbing a slope

Now, Example one, the easier one. Read it with me: a twelve hundred kilogram car climbs a slope with sine theta equal to one over thirty, at a steady fifteen metres per second, against a resistance of eight hundred newtons. Take g as ten, and find the engine's power. Pause the video and try it. Done? Okay, let's check. Power is force times speed, and the resistance is eight hundred newtons. So eight hundred times fifteen, twelve kilowatts, right? Oh no, careful! Look at the figure. Theta is the angle of the slope, capital R the resistance, and the orange arrow F the engine's push. But gravity also pulls the car back: m g sine theta, the part of the weight along the slope. Twelve hundred times ten times one over thirty: four hundred newtons. Okay? Steady speed, so F balances both: four hundred plus eight hundred, twelve hundred newtons, as the lines beside the figure say. With me? Now P equals F times v: twelve hundred times fifteen, eighteen thousand watts, capital W for watts. That is eighteen kilowatts. Forget the slope, and you lose a third of the answer!

6:40Example 2: tension at top and bottom of a vertical circle

Now a harder one, JEE level. Read it: a two kilogram stone on a one metre string has eight metres per second at the bottom of a vertical circle. Find the tensions at the bottom and top, with g as ten. Pause the video and try it. Done? Okay, let's check. First, our test: does it go round? u squared is sixty four, and five g l is fifty. Sixty four is bigger, so yes, it goes round. Right? At the bottom: m u squared over l, plus m g. Two times sixty four, over one, is one hundred and twenty eight, plus twenty: one hundred and forty eight newtons. Clear? Now the top. It has risen two l, two metres, so energy gives v squared equals u squared minus four g l. Sixty four minus forty: twenty four. Now, at the top, which way do the tension and the weight point? Both downwards, towards the centre, yes! Together they give the centripetal force, so T equals m v squared over l, minus m g. Two times twenty four is forty eight, minus twenty: twenty eight newtons. Got it? Now a topper's check. One hundred and forty eight minus twenty eight is one hundred and twenty newtons. And six m g is six times two times ten, also one hundred and twenty! Going round freely, the bottom tension is always six m g more, just as the figure in tool three said. Perfect!

8:18Quick revision: P = Fv, wind power, root 5gl, U(x) curve

So, quick revision. Power is force times speed; at a steady speed, the driving force equals the total resistance. Wind power goes as v cubed, and a bob on a string goes round only if u is at least root five g l. On a potential energy curve, K equals E minus U, so the speed is greatest where U is least. Now try the rest of this set on this page. All the best!