00:01
In this question, a stone is dropped at t equals zero, and then a second stone with twice the mass is dropped from the same point at t equals 100 milliseconds.
00:14
We are to calculate how far below the release point is the center of mass of the two stones at t equals 300 milliseconds.
00:22
And in part b, how fast the center of mass of the two stone system is moving at that same time as well.
00:31
So to start this off, we know that both of the stones are, if they're being dropped, you know, straight.
00:40
If they're being dropped, they're going to move straight down.
00:43
So we really only need to worry about the y direction here.
00:47
So to start off, i'm going to calculate for each stone at 300 milliseconds how far they have moved in down in the y direction.
00:57
So we know from kinematics that we can use the false.
01:01
Following formula, the distance traveled is the initial speed times time plus one half a t squared.
01:11
And for both of our stones, the initial speed is zero.
01:14
So we can just go ahead and cancel that out.
01:16
And we can just use a simplified formula.
01:19
Distance is equal to one half a t squared.
01:23
So for stone number one, we know that the acceleration is just 9 .81, acceleration due to gravity.
01:37
And at three 300 milliseconds, it has been moving that whole time.
01:42
So the time that it's been moving is 0 .3 seconds.
01:48
And we're going to square that based on the formula.
01:52
Once we put that in our calculator, we're going to get 0 .4 -4145 meters.
02:00
I'm just going to keep a lot of decimals at this point.
02:03
And then we're going to do the same thing for stone number two.
02:09
So d2 is 1ā2 .9 .81.
02:13
And for the time, since it was only dropped at 100 milliseconds, at 300 milliseconds, it's only been moving for 200 milliseconds.
02:21
So we need to put 0 .2 seconds as the time here.
02:26
And so that's going to give us a distance traveled of 0 .1962 meters.
02:33
Okay, so to find the position of the center of mass, i'm going to just use a typical center of mass formula.
02:51
Mass 1 times d1 plus mass 2 times d2 and then divide by the overall mass.
02:59
Now we weren't given the masses in this, so we do have to be a little bit careful here.
03:03
We're going to use the information that the second mass is twice the first mass.
03:08
So m2 is equal to 2 times m1.
03:11
So i'm going to go ahead and sub in for the big m.
03:15
That's the total mass.
03:16
That's going to be m1 plus m2.
03:18
But i also know that m2 is 2 times m1 so that's just going to be m1 plus 2m1 and at the same time let's also sub in for the distance traveled by the first mass and then for m2 we will sub in 2 times m1 and then d2 0 .1962 so what we can do is multiply this 2 here and then add together the m1s since those are like terms...