00:01
Hello friends here it is given 6 pound weight homogeneous disc of radius 3 inch spinning at constant rate of 60 radian per second as shown in the figure.
00:18
This disc is belted by the soft cvd the system would be in equilibrium or at rest when couple m not is applied for 3 second and then remove.
00:33
If angular velocity of the shaft is 18 radian per second then we have to calculate couple m -0 and dynamic reaction c &d after couple has been removed.
00:47
Let us see here.
00:51
Start solving it.
00:56
Angular velocity of the shaft, we will write all the parameter in vector form.
01:04
Angular velocity of soft, cbd and arm a .v.
01:29
Omega 2 j -cap.
01:37
Angular velocity of disk a omega 2 j cap plus omega 1 k cap angular momentum of the disk about a i x omega x i cap i j omega x k cap omega x is 0 so this term becomes 0 so you will light i y omega x sorry omega 2 j cap plus i z omega 1 k cap angular velocity of reference frame c vd vx by z omega is called to so rate of change of angular momentum of that is about a point vx y z plus omega cross h a substituting the value so we can write i by omega 2 z rate of change of omega 2 j cap plus i z rate of change of omega 1 k cap cross omega 2 j cap cross i by omega 2 j cap plus i z omega 1 k cap taking the cross product and simplifying you will get iz omega 1 omega 2 i cap i y rate of change of omega 2 j cap plus i z rate of change of omega 1 k cap i 2 i 2 is moment of inertia about x x x that is mr square by 2 omega 1 omega 2 i cap i by moment of inertia about by axis, mr, square by 4, rate of change of omega 2 with time plus i, sorry, this is i, x, and this is iz, mr, square by 2, rate of change of angular velocity with time, k -kkk.
06:15
Now we will draw the free body diagram of the disk and the soft, cvd, let us see here.
06:36
Y -axis c point cx cz d point d -z d -z d -d -x d -d -x this is the x -a -s -m -a -z m -a -z this is m -a -z this is m -o -nor that is c -bd or this distance from here to here it is b this is b and this distance is c now, we can write velocity of the mass center a.
07:50
This is a point.
07:52
So velocity, you will write omega to j -cap cross c -i -cap, that is minus -c, omega -2, omega -2, k -cap.
08:08
So, acceleration, a, you can write rate of change of angular velocity, omega -2, j -cap -c -c -c -cac -cac -ccc, plus omega 2 k cross b so this can be written as acceleration minus c rate of change of omega 2 this time k cap minus c omega 2 square i cap now submission of f to be mass into acceleration so c x i cap plus c y cap plus c y cap plus c y cap plus c y cap d x i cap d z k cap is equal to m a.
09:14
Substitute the value of a here.
09:20
C x i cap c by k cap d x i cap plus d z k cap minus m c omega 2 k cap minus mc omega square omega 2 square i cap so we can write cx plus dx 2b minus mc omega 2 square and cz plus dz equal to minus mc rate of change of.
10:03
So these two values are important for us for further calculation.
10:11
Now moment about d is equal to rate of change of angular momentum about d.
10:35
It will be rate of change of momentum rate of change of angular momentum about a plus r a d position vector cross m a this will be m o j cap plus 2b j cap cross c x i cap plus c x i cap plus c z k cap must be equal to h a at a already we have calculated mr square y 2 omega 1 omega 2 in just a moment and the last we will substitute otherwise equation becomes very lengthy so i'm just writing here in the same way now substitute this value you will get c i cap plus vj cap cross m into a a is cb to c omega 2 .k minus c omega 2 square i...