00:01
In this question, we have given a thin spherical shell of radius 5 semi and mass 2 kgis placed at the top of incline and the surface of the incline makes an angle of 25 degree with the horizontal and it is told that the shell is released from the rest and rolls without slipping.
00:18
Then, in the first part, we have to find the acceleration of center of mass during the rolling motion.
00:24
So we have given data for this question, that is, mass of spherical shell.
00:30
That is m is 2 kilograms and we have given the radius that is r is equals to 5 semi we have given the angle of inclination theta is 25 degree so first of all i am drawing the diagram for this question so this is the inclined surface and that this is the height of the inclined surface that is h and we have given that from at this point there is a spherical shell which is rolling down and it rolls and reaches at the bottom.
01:16
So we can say that the mass of this spherical shell is m.
01:20
So we can say that the weight is acting vertically downward that is mg and this angle is theta.
01:26
So we can say that the component of weight in this direction is mg sine theta.
01:31
And as it rolls down there is a frictional force which opposes this motion and let this is f.
01:37
So we can say that first i am calculating the center of the center of the the moment of inertia of spherical shell about center of mass is given by that is i -c -o -m this is equal to 2 by 3 m r square so we can say this will come out to be 2 by 3 into mass is 2 kilogram into radius is 5 into 10 to the power minus 2 in meters this is semi so we can say this is 5 into 10 to the power minus 2 meter so finally we can say this will be written as this will come out to be 1 by 3 into 10 to the power minus 2 kilogrammetre square...