00:01
Hello and welcome to this video's solution of numerate.
00:03
So here we are given a pulley having mass capital m which is 6 kgs and its radius r is 0 .28 and it is mounted on a horizontal frictionless axis that is used to lift a block as shown in this diagram.
00:18
The block is having mass small m which is 1 kg and is hung from a massless string that is wrapped around the rim of a pulley and the constant force f which is 30 newton is applied at the free end of the string right.
00:31
So this is the system so this is the force f here you have the mass hanging the tangent is t sort of like this t and the same here will be happening on the pulley on down the downward side right.
00:51
So weight is mg now based on this you have to calculate the upward acceleration of the block right.
00:59
So the net force on the block is ma right.
01:03
So here you write ma that is equal to t minus mg right or t will be equal to ma plus mg is the tension right.
01:14
Now you have got this force f right and based on this there will be angular acceleration alpha right on the pulley.
01:25
So what you can do is you can write down the torque equation so the net torque tau will be equal to i alpha right.
01:33
The net torque tau will be given by the force f minus the tension t times the radius of the pulley right.
01:39
So this is the net torque that is acting on the pulley.
01:41
The moment of inertia of the pulley is half mr square right...