00:10
So m1 rests here is being pulled by t1.
00:15
There is a pulley.
00:17
The pulley is being pulled by t1 over here and t2 over here.
00:24
And then the upward t2, who is the m2 block.
00:35
Now there is m2g force here, m1g force here, normal force here, and there is no friction.
00:46
So this move down with the acceleration a, this moves to the right with an acceleration a, this moves with an angular acceleration of alpha equals a over r where r is the radius of the pulley and then the mass of the pulley is also known.
01:10
So for the m1 drop, c1 equals m one a for drop 2, k2, k2, ab to g minus k 2, equals m to a and for the pulley it is higher so p2 minus g1 times r will be the torque which equals i alpha uh i of the alpha is a over r so authentically we can write it as this equals a over r square so now if we add all of the up we will get m to g equals m1 plus m2 plus i is the moment of inertia of the desk so that's half m r squared and then r square goes here thanks one over r square thanks a so a will be m2 g over m2 over 2 so, 12 and 515, 12 and 5, 17 plus 21, 12 and 5 is 17, 12 and 5 is 17, and this is 1, 18, and m2 is 5...