00:01
Hello students, in this question we have given a schematic diagram of cable drum which can be treated as cylinder.
00:10
Now from the figure we have the equation of motion based on the free body diagram acted vertically i .e.
00:19
Summation of fy is equal to m a gy.
00:28
Here this can be written as 300 minus t that will be equal to 300 divided by 32 .2 a gy.
00:37
Here the reaction force is a gy which is equal to a.
00:42
Now the moment along the mass center of the drum g is given as, now taking the moment we have m a is equal to i g z a.
00:54
Here the moment of inertia i g z and here the acceleration is a.
01:01
Now we obtain moment as 3 t is equal to i g alpha.
01:10
Now to find the moment of inertia of the rectangular plate about the support a that is given as half m r square into alpha.
01:25
Now we have mass that is equal to 300 upon 32 .2 kg.
01:31
This will be given as mass is 18 .63 kg.
01:37
Now we have also given diameter that is equal to 18.
01:42
So the radius will be equal to 1 .5 divided by 2 that will be equal to 0 .75 feet.
01:51
Now substituting all the values here we get 3 into t that is here we have 3 t.
01:59
3 into 300 is equal to now substituting the values here we have mass 18 .63 into r square 0 .75 square into alpha.
02:14
Now after solving this we get alpha is equal to 171 .76.
02:21
Now to determine the time required to rotate the cable drum three revolutions we proceed this will be omega minus omega, omega square minus omega square is equal to 2 alpha theta minus theta naught...