00:01
Hello friends as soon in the figure two rotating rod in a vertical plane are connected by the slider block p of negligible mass the rod attached at a having the mass of 0 .8 kg the mass of a here we write 0 .8 kg and its length is 160 millimeter that is 0 .16 meter mass of bp rod is given 1 kg and having the length 200 mm that is 0 .2 meter the friction between the block p and a .e is negligible the motion of the system is controlled by applying the couple on var ae so m applied on ae it is given that angular velocity of rod bp is 20 radian per second in clockwise direction and angular acceleration is 80 radian per second square in clockwise direction we have to calculate first part moment applied at a and component of force exerted on ae by p exerted by p on ae.
02:31
Let us start solving it.
02:34
First write the position vector of different points.
02:40
Position vector of pa is 0 .2.
02:59
Sign of 30 downward that is 0 .1 meter in downward direction.
03:10
Position vector of p with respect to b is 0 .2 meter down to the horizontal line by 30 degree position vector of e with respect to a is downward by 0 .16 meter kinematics angular velocity of bp is 20 radian per second in clock by direction and angular acceleration of bp is 80 radian per second square in clock by direction.
04:15
Velocity analysis rod bp.
04:32
Valacity of p is defined as cross product of angular velocity and position vector.
04:39
Angular velocity of bp cross position vector of p with respect to b substituting the value we will get 20 into 0 .2 meter above the horizontal plane y angle 60 degree so we can write velocity of p point to be 4 meter per second in that direction up above the horizontal line y 6 degree.
05:27
Rod a .e.
05:32
Use a frame of reference rotating with angular velocity omega a .e to b.
06:07
Omega a .e.
06:09
In clockwise, anti -clockwise direction.
06:14
The caller slides on the rod with relative velocity of p with respect to a .e.
06:46
Is u in upper direction so velocity of p can be written as velocity of p -dess plus velocity of p with respect to ae velocity of p -dess is defined as cross product of angular velocity and position vector so a e cross r p a vector plus u it can be written as velocity of p b you will get point one angular velocity of a e towards right plus u in upward direction equate the expression for bp using triangle construction for vector addition this is u vector this is bp vector and this is minus point 1 omega a that is 4 meter per second so we can write minus 0 .1 omega a .e is equal to 4 cost of 60 so angular velocity of a .e we will get 20 radiant per second in clock by direction and u is to 4 sign of 60 that is 3 .41 sorry 461 meter per second so you can say velocity of p with respect to a .e having the value 3 .4641 meter per second in upper direction now we will do acceleration analysis vp acceleration of p can be defined as cross product of angular velocity and position vector and omega square bp r p b b b b b b b b b b b b b b bactur substituting the value 80 radian per second square into 0 .2 meter in the direction so in figure plus 20 square 0 .2 meter in the direction up along the horizontal line by 30 degree.
11:40
So acceleration of p point would be 16 meter per second square in the direction above to the horizontal line by 60 degree plus 80 meter per second square in the direction above the horizontal line by 30 degree.
12:19
Rod a .e.
12:23
Angular acceleration of ae is alpha ae in counterclockwise direction and acceleration of p with respect to ae is derivative of u in upper direction.
12:48
So acceleration of p can be written as ap does plus acceleration of p with respect to ae plus acceleration of p with respect to ae plus twice.
13:13
Of omega a .e having the cross vector with velocity of p with respect to a .e.
13:25
Here, a p -des having the value, alpha cross a .e vector with p .a.
13:45
Minus omega square a .e.
13:51
Into position vector of p.
13:54
With respect to a.
13:57
Substituting the value we will write 0 .1 into alpha a .e in the direction of right plus 20 square into 0 .1 in vertically upper direction.
14:20
So acceleration of p -des, you can write 0 .1 alpha a .e towards left plus 40 upward meter per second square value calculating the value of omega cross a .e into velocity of p with respect to ae here you have to take the cross product 2 into 20 into 3 .4641 towards right.
15:05
So it will be 138 .564 towards right meter per second square.
15:16
Equating the two expression of acceleration at p and resolving into its component, equate the two expression for acceleration at p and equating or and resolving into its component.
15:45
So if we take horizontal component right direction to be positive, it will be minus 16 cost of 60 plus 80 cost of 30 is equal to 0 .1 alpha ae plus 138 .564.
16:24
So from here angular acceleration of ae we will get 772 .82 radian per second square in clockwise direction.
16:44
Now mass weight and moment of inertia mass weight and moment of inertia.
16:59
Mass of ae is given 0 .8 kg...