00:02
So for this problem, let's first use kinematics and find our angular speeds for each of the disks.
00:10
Firstly, the motor speed will call it omega -m, the angular speed of the motor, is given as 180 rpm.
00:22
And 180 rpm is simply 6 pi radiance per second.
00:31
So we know that one minute has 60 seconds and one revolution has 2 pi radiance.
00:36
So the motor moves as an angular speed of 6 pi radiance per second.
00:42
Now we will let omega a, omega b, and omega p be the angular velocities of disk a, disc b, and the platform respectively.
00:53
And since the motor speed is the angular velocity of disk b relative to the platform, we have that the angular velocity of disk b is equal to the angular velocity of the platform.
01:07
Platform omega p plus omega m the angular velocity of the motor and this is omega p plus six pi and since the disks have the two disks have the same outer radius omega b is equal to omega a so now we have the angular velocities which you want to calculate for omega a omega b and omega p the velocity of the center of disk a we'll call it va is equal to 4 over 12 times omega p and similarly the center of disk b moves with linear velocity for over 12 times omega p so we have a system of equations here which we will come back to later so now let's first calculate our moments of inertia the moments of inertia we will use later.
02:30
So firstly for disks a and b we will find the moment of the moments of inertia for their centers, i bar a and i bar b.
02:42
And this for a disc is equal to a half times its mass which is w over g its weight over gravity times r squared.
02:53
So since these are both the same this is simply a half into its weight which is 10 over 32 .2 feet per square second.
03:04
And r is 3 over 12 all squared and so the moment of inertia for each disk is 9 .705 times 10 to the minus 3 pound second squared foot so we keep as many decimal points as we can for the moment of inertia now we want to calculate the moment of the inertia for the platform so we'll call that ip.
03:45
Ip, the platform for a flat plate is 1 over 12, w over g, where w is the weight of the platform times r squared in this case is a squared plus b squared.
04:06
And so this is the substitute of values in here, 1 over 12 times the weight of the platform which is 15 over g which is 32 .2 into a squared which is 2, which is a squared which is is 16 over 22 squared plus b squared which is 6 over 12 or half old squared and hence we get the moment of inertia of the center of the platform to be 78 .718 again keeping three decimal points times 10 to the minus 3 and that's the same unit pound second squared feet so now we have the moment of inertia for both disks a and b and for the platform p.
05:03
Now we will apply the principle of impulse and momentum for the system.
05:15
And to do this, we will draw a diagram of the situation.
05:30
Initially, the system is at rest and so there's no momentum.
05:35
Then there's an impulse imparted by the motor.
05:39
We'll call it impulse r.
05:40
Which has a y and x component on the center of the platform at point oh...