Sliding Off a Smooth Sphere

Transcript

0:00Introduction

Hello everyone! Let's start with a puzzle. Picture a smooth dome, a hemisphere, fixed on the ground. You place a small block at the very top, and it slides down from rest. No friction at all. Does it hug the surface all the way to the ground? It feels like it should, right? But no! Somewhere on the way, the block lifts off and flies away. Where exactly, and how fast? Questions on this come up often in JEE, so let's find out.

0:31Speed from energy

So, look at the figure. Here is a smooth hemisphere of radius R, fixed on the ground; its centre is O, the black dot. The dark blue dot is our particle, and it starts at the top with speed u. Theta, this Greek letter, is the angle of the radius from the upward vertical, the dashed line. And the notes on the right of the figure? Don't read them yet; that's where we are heading, and we will get there ourselves. Got it? Now, the surface is smooth, so there is no friction. It only pushes, with the normal force N, the arrow pointing straight out from the centre. Does N do any work? No work at all. Correct! N is perpendicular to the velocity v, the orange arrow; this upside down T sign means perpendicular. So only gravity does work, and we can use energy. With me? Now, the drop. The top is R above the centre; now it is R cos theta above, the dotted line with the label R cos theta. So the drop is R minus R cos theta, that is, R times one minus cos theta. Right? So energy gives: v squared equals u squared, plus two g R times one minus cos theta. Here g is the acceleration due to gravity. Plus, because the particle is going down and speeding up. Clear?

1:48The normal force: the radial equation

Now the normal force. The particle moves on a circle, so it needs a net force towards the centre O, okay? The weight m g, the downward arrow, has a part m g cos theta towards O, and N points the other way. So, m g cos theta minus N equals m v squared over R, where m is the mass. With me? Rearrange: N equals m g cos theta, minus m v squared over R, okay? This is the key equation of today, so write it down! Now think. As the particle slides down, theta grows. Does N grow, or shrink? It shrinks. Yes, exactly! Cos theta falls, so the first term gets smaller, and v grows, so we subtract more. Right? So at some point, N becomes zero. Look, a surface can only push; it can never pull. So the moment N becomes zero, the particle leaves the surface, just like a string going slack. Clear?

2:47From rest: it leaves at two-thirds

Now the classic case: the particle starts from rest, so u is zero. Then v squared is just two g R times one minus cos theta. Got it? Put this into the formula for N, okay? The term m v squared over R becomes two m g times one minus cos theta. Right? Now collect the cos theta terms: N equals m g times three cos theta minus two. With me? So where does it leave? Many students say: at the ground, where theta is ninety degrees; the small circle means degrees. That's the answer, right? No, no, careful! At ninety degrees, cos theta is zero, so N would be minus two m g. Negative! But a surface can only push, so that is impossible. Clear? So it leaves much earlier, where three cos theta minus two is zero: cos theta equals two thirds. The height above O is R cos theta, so two thirds of R, okay? It drops only one third of R and flies off. So that answers where. Next, how fast. Its speed there: v squared equals two g R times one third, that is, two thirds of g R, right? So v is root two thirds g R; the tick shaped sign is the square root. Remember both results!

4:03Given a push u at the top

Now, suppose we give the particle a push: a speed u at the top. Then v squared has an extra u squared, so N gets an extra minus m u squared over R, okay? N equals m g times three cos theta minus two, minus m u squared over R, and we set this equal to zero. Okay? Solve: cos theta equals one third, times two plus u squared over g R, right? With u equal to zero, we get back two thirds. And a bigger push means an earlier exit, higher up. With me? Now think. What speed at the top makes the particle leave at once, without sliding at all? Here is the trick: look at the very top, where theta is zero. There, N equals m g minus m u squared over R, right? If this is zero or less, the surface cannot hold the particle even for a moment. So u greater than or equal to root g R; that sign is greater than or equal to. Then it leaves at once. If you said root g R, very good! Clear?

5:04Example 1: where a block leaves a smooth hemisphere

Alright, now real numbers. Read it with me: a small block slides from rest off the top of a smooth hemisphere of radius four point five metres, fixed on the ground. At what height does it leave the surface, and with what speed? Take g as ten. Pause the video and try it. Done? Okay, let's check. It starts from rest, so straight away, cos theta equals two thirds. Right? The height above O is two thirds of R: two thirds times four point five, that is three metres. The hemisphere is fixed on the ground, so O is on the ground. So the block leaves three metres above the ground. With me? Now the speed. v squared is two thirds g R: two thirds times ten times four point five, that is thirty. Okay? So v is root thirty, about five point five metres per second; that is the m s to the minus one on screen. The wavy equals sign means approximately. That's it!

6:03Example 2: normal force after sliding through 30 degrees

Now a harder one. Read it with me: a particle slides from rest from the top of a smooth sphere. Find the normal force when it has moved through thirty degrees. Pause the video and try it. Done? Okay, let's check. Many students write: N equals m g cos thirty degrees, and finish. That's it, right? No, no, wait! That is true only for a particle at rest. This one moves on a circle. From rest, v squared is two g R times one minus cos theta, so m v squared over R is two m g times one minus cos theta. Okay? So N equals m g cos theta, minus two m g times one minus cos theta: m g times three cos theta minus two. Same as before. Clear? Now put theta equal to thirty degrees. Cos thirty is about zero point eight six six, and three times that is two point five nine eight. So N is m g times two point five nine eight minus two. With me? That is about zero point six zero m g. So it is much less than m g cos thirty; part of the weight is busy turning the particle round the circle. And N is still positive, so the particle is still on the surface. Perfect!

7:19Quick revision: cos theta = 2/3 and the push case

So, quick revision. Speed from energy; then N equals m g cos theta minus m v squared over R, okay? From rest, it leaves where cos theta is two thirds: two thirds of R above the centre, with speed root two thirds g R, right? With a push u, cos theta is one third, times two plus u squared over g R; and if u is root g R or more, it leaves at once. Now go and try the practice problems on this page. You can do them. All the best!