00:01
Hello, so today we will be looking at problem number 14 from chapter 8, and these set of questions will refer to a similar problem that we've done before from problem 4, and this is the mass on the rod problem.
00:16
So let me first draw that out for you.
00:18
So in this problem, basically we're just going to be using the conservation of energy principle that we've been using for the last couple of problems to solve this.
00:29
So the first question is it wants to know what initial speed the ball has to be given in order for it to go from its original position, which i've marked with the dark black ball, to its top position here, which is the white ball.
00:46
And let's keep in mind this is the initial position.
00:49
Now, let's remember that this is a problem that we've already done, so it has measurements given to us.
00:57
So we know that the mass of the ball, m, is equal to 0 .341 kilograms.
01:05
And we know that the length of this rod l is equal to 0 .452 meters.
01:13
Okay.
01:14
So now let's go on to solve this problem now.
01:16
So what is the beginning speed the ball needs to go from here to here? well, very simple.
01:22
This is conservation of energy.
01:24
So let's write out what we have.
01:25
We have our initial energy and our final energy because the conservation of energy states that the sum of your initial energies is equal to the sum of your final energies.
01:35
So what do we have for our initial energy? well, we know that it must start out with some kind of kinetic energy in order for it to move.
01:45
Now, does it have potential energy? no, because it's at ground level.
01:50
We're assuming this is ground level, right? and now up here, since it's reaching the top of the arc, at that moment, velocity is equal to zero.
01:59
But we do have gravitational potential energy just because it is changing heights.
02:05
So now let's simply write out what we have.
02:08
So one half, m v initial squared is equal to m g, h.
02:13
Notice something here, right? since there's a direct proportional relationship, m cancels out.
02:19
And this is very useful.
02:22
Because we can basically solve for our initial velocity without having to input mass at all.
02:28
So now let's just solve for the initial.
02:32
Well, that's equal to 2gh, and v initial is equal to the square root of 2gh.
02:38
And all you have to do is substitute your h value, which in this case is obviously l, right? because if it travels from the initial state to the top of the spin, then you know that it is at height of l because that's the mass that the rod is.
02:58
Very simple.
02:59
And now b is asking a similar problem.
03:03
It's asking for what the speed is at the lowest point...