00:01
So according to the question, let's suppose it is given that an inclined surface which is inclined at an angle of 37 degree.
00:10
Now there is a block which is attached by a cord and it is go over to that of the pulley which is connected right over here having the radius of r.
00:22
Ok so this cord is go over through this pulley.
00:26
Now the mass of this block is 12 kg.
00:29
So according to question it is given it is moving with an acceleration of 2 .0 m per second square in this direction.
00:37
So in the first part we need to calculate the tension in the rope.
00:41
So let us start with this.
00:42
So first of all we need to resolve the forces whatever will be exerting.
00:47
So let's say if mass is m for the block then in downward direction there will be a force that is mg.
00:53
If this angle is 37 degree so with corresponding to perpendicular this angle will be theta which should be equals to the 37 of degree.
01:03
So this mg would have the two components one will be along the inclined plane that is horizontal component mg sine of theta and the second one will be the vertical component mg cosine of theta.
01:16
While the tension in the rope will be in upward direction that is t and along this it will be t.
01:22
Ok so let's say this point to be as p here.
01:27
So now applying the newton's law so we can see that the net force is equals to the mass into acceleration.
01:37
Now net force along this will be nothing but mg sine of theta.
01:44
Now theta will be 37 of degree minus of tension t which should be equals to the mass into acceleration that is ma.
01:53
So from here this tension t will be comes out equals to mg sine of 37 of degree minus of m times of a.
02:02
So let's substitute all the required values that is mass is 12 kg multiplied by value of t will be 9 .8 multiplied by sine 37 is known that is 3 by 5 minus of mass is 12 multiplied by acceleration is given 2.
02:17
So after simplifying and solving tension will comes out 46 .56 of newton's.
02:23
So this will be the required answer for a part.
02:26
Now in b part we have to calculate moment of inertia of the wheel.
02:30
So let's suppose the angular acceleration of the pulley is nothing but what alpha.
02:35
So by using this we can clearly say that for no slipping the acceleration a should be equals to alpha times of the radial distance that is r.
02:47
Let's say in the bracket because of no slipping.
02:51
So from here we can say that the tangential acceleration of ap should be equals to alpha times of a.
02:58
That means whatever we have written this is the tangential acceleration...