00:01
Hello everyone, we are going to understand this question.
00:04
Here, given in the question, given, radius of pulley, radius of pulley, that is r, moment of inertia of pulley, moment of inertia of pulley, that is i, and mass of blocks, mass of blocks are ma, and maa.
00:51
Be coefficient of kinetic friction coefficient of kinetic friction that is mu now let the net acceleration be let the net acceleration net acceleration be a and tension in the string tension in string be t1 and t2 let's draw the fbd for this weight of the that is working vertically downward m a into g and normal reaction that is an friction force is in right over direction and tension in the string t1 and here it will be t2 and net acceleration is in downward direction similarly for this weight of the block that is mb into g which is in downward direction tension in the string t2 is in upward direction and net acceleration is in downward direction now applying the force balance equation, now using force balance equation, force balance equation, so we can write mb into g minus t2 is equal to mb into a.
02:48
Let this is equation 1 and t1 minus f s is equal to m a into 8.
03:01
Again we can write t1 minus mu into m a into g is equal to m a into a.
03:12
Let this is equation 2.
03:16
Now applying torque equation on the pulley, torque equation on pulley, that is t2 minus t1 into r is equal to i into alpha, where alpha is the angular acceleration.
03:46
Now we can write t2 minus t1 is equal to i into alpha upon r...