0:00
Hi there.
00:01
So for this problem, we are told that a thin cylindrical ring starts from res at a height that is h1 and that is equal to 75 meters.
00:15
And the ring has a radius that is equal to 38 centimeters.
00:23
And a mass that is also given capital m and that is equal to 6.
00:30
Kilograms.
00:32
So for part b of this problem, we are asked about if the ring rolls without sleeping, that condition means that the angular speed is equal to the linear velocity divided by the radius.
00:51
That's a condition for rolling without sleeping all the way to the point two.
00:57
What is the rinse energy at point two in terms of the height 2 and the speed 2? so at the point 2, as you can see from the picture, we are going to have three types of energy.
01:16
Potential energy because we are at a height h2, kinetic energy because we are going to have a speed b2, and a rotational energy because this thin cylindrical ring is going to be rotating in that position.
01:36
So with that said, we're going to have that at that point, we have potential energy that is the mass times the acceleration due to gravity times the high two, plus the kinetic energy that is one over two times the mass, times the speed 2 to the squared.
01:59
And this plus the rotational energy that we know is 1 over 2 times the moment of inertia times the angular speed square.
02:11
So first, and we are going to define what is the moment of inertia in this case, because it is a thin cylindrical ring.
02:21
That means that the moment of inertia is equal to the mass times the radius, square.
02:27
You can search for that value and you will find that is this.
02:31
And now we substitute that in here alone with the condition of rolling without sleeping.
02:37
That is in this case, we're going to have that that is b2 divided by the radius.
02:44
So if we substitute that in here, we're going to have that this is the following.
02:49
Is 1 over 2 times the moment of inertia.
02:54
That is this.
02:57
Times the angular speed that is b2 divided by the radius and all of that to the square.
03:05
So we can cancel this with this because this is going to be to the squared.
03:11
And as you can see, we're going to have that this is, this is simply one over two.
03:17
The mass times the speed two to the square.
03:20
So it's going to be the same as this.
03:22
So we're going to have to sum one over two plus one over two of the same thing.
03:29
So we're going to have that the energy at the point two is the potential energy plus the mass times the speed two to the square.
03:41
So that's a solution for par b of this problem.
03:47
Now, for par c, we are asked about, given that h2, we are now given some values for h2 is equal to five meters and what is the velocity of the ring at the point two.
04:13
So we need to calculate the speed two.
04:20
Now, what we are going to do in here is to apply conservation of energy between the points one and two.
04:29
So we're going to set that the energy at the point one is equal to the energy at the point two.
04:35
So the energy at the point one, as you can, as you already put the solution in part a, is the mass times acceleration due to gravity times the height one.
04:48
And the energy two is this value that we obtain from this.
04:53
The mass times the acceleration due to gravity times the height two.
04:58
And this plus the mass times the speed, to the square.
05:05
So we need to solve for the sp2.
05:09
As you can see, we can simplify some themes in here.
05:12
We can cancel all the masses because they are in all the terms in all the terms in both sides of this equation.
05:21
And with that said, we're going to obtain that we can pass this to the other side.
05:27
So we're going to have the acceleration due to gravity, h1 minus h2.
05:35
And we take the square root of this to get rid of this square in here, so that is the expression for the speed too.
05:44
So we just need to simply substitute those values in here...