00:01
For both of these problems, you're considering rotational motion.
00:07
So things like torque, moment of inertia, angular velocity, angular acceleration, angular displacement will be the useful physical quantities to think about.
00:21
For problem 17, we have this system of two cylinders that are currently at rest.
00:30
They've got strings wrapped around them and forces are pulling down on those strings.
00:36
We're also given a bit of information so we know the radius of the smaller cylinder, which i'm going to consider cylinder one because it has the force f1 acting on it.
00:48
But the radius of that cylinder is 20 centimeters.
00:51
I'm going to go ahead and convert that to meters.
00:55
For the second cylinder, r2, is 50 centimeters or half of a meter.
01:07
We're also told the value of the masses of each of these.
01:13
So mass one is 30 kilograms and mass two is 50 kilograms.
01:22
And then lastly, we're also told the magnitude of those two forces.
01:27
So f1 is 98 newtons.
01:31
And f2 is 49 newton's.
01:36
I'll point out that some of the numbers and the problem, specifically m1, m2, and f1 have an exclamation mark after the number and before the unit.
01:57
I believe that this is just a typo.
02:02
The only other thing that it could mean is factorial.
02:08
So, for example, three with an exclamation mark like that is three factorial and is equal to three times two times one.
02:16
Or four factorial would be four times three times two times one.
02:22
But if that were the case, then you would have some very, very, very, very, very, very, very large numbers compared to some normal.
02:33
So i think that that is just a typo.
02:40
And so i am ignoring it.
02:43
If you talk to your teacher, they might be able to give you a bit more clarification on what those exclamation marks mean.
02:53
But we're going to run under the assumption that they are a typo and move straight into part where we're asked to find the net torque on the system of cylinders.
03:07
So torque is generally equal to rf sine theta.
03:15
And for something like this, where there is a force that is tangent to a circle, r will be the radius of the circle, f will be that force, and sine of theta will just be one because the force and the radius will be perpendicular to each other.
03:34
So for cylinder one, we would get a torque that is r1f1 in magnitude.
03:45
And then for cylinder two, we would get a torque that is r2f2 in magnitude.
03:54
So if we were to find the net torque, we want to add those up.
03:58
And now we have to pay attention to the direction.
04:01
So we're told to imagine we are looking from the left, and negative torques will be clockwise, and positive torques will be counterclockwise.
04:13
So what i'm thinking about t1, and i imagine pulling down on that string, i can imagine that that cylinder will want to rotate counterclockwise, so it will be a positive torque.
04:29
And for the second cylinder, if you pull down on f1, it wants to rotate clockwise if you're viewing it from the left.
04:40
And so that will be a negative torque.
04:45
And so now it's just a matter of actually plugging in the values given in the problem.
04:51
And when you do so, you should end up with a net torque of negative 4 .9 newton meters.
05:02
Part b is to say if the cylinders move, what direction would they rotate? and in part a, we got a negative value for the torque.
05:17
And so that tells us that the torque on this system is negative, causing it to want to rotate clockwise, which means that it should rotate clockwise if you're looking.
05:32
Looking from the left since this is the direction of the net torque and it was at rest before it starts moving.
05:43
I'll point out that obviously if it were already rotating in the opposite direction, the net torque wouldn't necessarily give us the direction of motion, but because it starts at rest, the net torque will tell us the direction of the motion.
06:00
So for b, the answer is clockwise.
06:07
I'm going to erase the work that i just did in a and b, although i'll go ahead and write the answers up here, just to make a little bit of extra space to finish off this problem.
06:28
So in part c, we now want to figure out what is the angular acceleration of the system.
06:38
So i'm going to use sort of the rotational equivalent of newton's second law to figure this out.
06:47
That is that the sum of the torques or the net torque is equal to i times alpha, where i is the moment of inertia.
06:55
So we need to look up the moment of inertia for a disk in order to be able to solve this.
07:03
Excuse me.
07:04
And for a disk, i is equal to one half mr squared.
07:11
So now that we have that information, we can plug it into this expression and solve for alpha.
07:17
And we'll get that alpha, the angular acceleration, is equal to the sum of the torques, which we just found in part a, divided by.
07:28
And we have to find the total moment of inertia, which will just be the sum of the moments of inertia for each of these disks.
07:36
So we'll have one half m1r1 squared plus one half m2r2 squared.
07:46
And once you plug in all of the values, i'm going to leave an intermediate step here.
07:53
So you end up with negative 4 .9 over 6.
07:59
Oops, over 6 .85 radians per second squared, which is equivalent to about negative .72 radians per second squared.
08:16
So my answer is negative .0 .72, but i wanted to stop at an intermediate step just in case we need to use this in future parts because if so we want to use this rather than this...