00:02
All right, here we have a fun circular slide.
00:06
And we have a student going down the slide, starting from rest at theta 1 at 20 degrees.
00:14
And then we're going to be interested at position 2, which is at 30 degrees, and position 3, which is at 90 degrees, are basically at the bottom of the slide.
00:27
And our goal is to find the tangential acceleration, the speed, and the normal force.
00:32
At position two and position three.
00:35
So let's start at position two.
00:40
Okay, so we are told that the slide is frictionless.
00:44
I'm actually going to change colors, make it a little bit easier to see.
00:48
Okay, position two.
00:50
So therefore, we don't have to worry about non -conservative forces and we can do conservation of energy.
00:56
So we'll have kinetic energy at state one plus potential energy at state one, needs to equal kinetic energy at state two plus potential energy at state two.
01:10
At first, we have zero kinetic energy, so that is easy.
01:15
That's a zero.
01:16
Our initial potential energy should be mg times h1.
01:21
Well, we have to do a little geometry to find h1.
01:24
Basically, we have similar, well, we have a little triangle here.
01:30
And notice that each of these are radius r and that theta during alternate interior angles, that theta 1 is also this theta 1.
01:41
So basically height 1 is equal to the full radius minus this distance, which is r sine theta 1.
01:49
So mg times h1, but h1 is r minus r sine theta 1, which is we can plug in values, that's 20 degrees.
02:01
And then ke after, that's what we're going to find the, i'll call it v2 squared, and our potential energy afterwards, we've gone down to position two.
02:13
So now we need height two, which is the full radius minus this distance.
02:18
So r minus r sine theta two.
02:23
Okay, so our goal at first, let's go ahead and solve for v2.
02:29
We see kind of a mess, but i'm gonna go ahead and rearrange.
02:34
Basically, i'm going to subtract, the right term from both sides, and then divide through by one half m and square root.
02:42
So all those three things, i'm going to do in one go.
02:45
So v2 then is equal to square root of.
02:49
So remember, i'm first attracting the term on the right to both sides, and that's why i get a minus.
03:01
Oops, i forgot mg over here.
03:04
Oh, you know what? let me actually go.
03:07
Back and let's let's clean it up at just a tad let's go ahead and cancel what the common masses so that we can just keep this little cleaner and then i'm going to actually um okay that works okay so now i'm going to i'm just going to clean it up a tad before i do the square root and make it hopefully more clear.
03:38
So i'm going to multiply everything by two and at the same time factor out an r.
03:42
So i'll get 2gr 1 minus sine theta 1 equals v2 squared plus 2gr 1 minus sine theta 2.
03:58
So notice what i did was i will find every term by two and i factored out the r's out of the parentheses term.
04:08
Now hopefully it makes sense.
04:09
B2 then a square root of just simply the left term minus the right term.
04:21
And all right.
04:24
So we can go ahead and we could clean it up more but since i'm going to use the calculator anyway to plug in.
04:33
Let's just go ahead and identify what goes where and then we'll plug it into our calculator.
04:39
So we'll get twice 9 .8 meters per second squared times 2 .5 meters for our radius, 1 minus sign of 20 degrees, and then minus 2 .8 meter per second squared times 2 .5 meters.
05:02
Five meters running out of room here.
05:04
So in that location, it's one minus sign of 30 degrees.
05:11
So we have to make sure we're in degree mode.
05:13
But when we plug that into our calculator, then we get v2 as 2 .782 meters per second.
05:23
Okay, so that is good to know that as our velocity or speed at position two.
05:30
And let's go ahead and then and make sure that we answer each of the question.
05:35
Remember we need to find the tangential acceleration, the speed.
05:38
Well, we did find the speed.
05:39
So this is our speed at position two.
05:44
So we did find one of them.
05:45
We just had to find the rest.
05:48
And we also, so we did speed.
05:50
We need to find now the tangential acceleration and the normal force.
05:57
Okay, so let's clear a little room.
05:59
We're going to need to look at the forces.
06:01
So let's look at position two.
06:03
And notice we're on a kind of a circle there.
06:08
We have mg down, which can be split into components, one tangential to the circle, and one perpendicular.
06:18
And we also have the normal force.
06:21
Normal force is always perpendicular to the surface.
06:25
So we, in fact, normal force, well, it didn't draw it purposely the same, but normal force to actually be a bit bigger here.
06:38
Because we need a force towards the center to keep us on the circle, centripetal force.
06:43
Okay, so now we just need to do a little bit of careful geometry to make sure we have the angle right.
06:52
So if this is theta 2 and this is the right angle here, then this is the complement and this.
06:59
Is back to being theta 2.
07:01
Therefore, this is m .g.
07:04
Cosine theta 2.
07:06
And this other is m .g.
07:08
Sine theta 2.
07:11
All right.
07:11
So notice that we can go ahead and do tangential acceleration.
07:20
We'll do newton's second law.
07:21
Some of the force tangential should make the mass accelerate tangentially.
07:26
Our only force, since we have no friction, that's tangential, is mg cosine theta 2.
07:35
And therefore, we'll be able to solve for the tangential acceleration.
07:41
It is equal to g cosine of theta 2.
07:45
So we'll plug in values...