00:02
Hello guys, in this problem we have a solid cylinder of mass 2 kg and the radius 0 .54 meter.
00:12
A horizontal force of magnitude 120 newton is applied on the top of the cylinder along the horizontal direction.
00:20
Now we need to find out the frictional force acting on the cylinder for pure rolling motion.
00:25
Consider the frictional force act along the horizontally backward direction that is to the left direction, has a magnitude f.
00:36
For translational motion, if we consider the acceleration of the center of mass is a -c -m, then we can write from newton's second law of motion as the net force, that is the horizontal force f minus the frictional force f equals the mass m times a.
01:01
Let's say this has equation well.
01:04
Now, consider in pure rolling motion, the angular acceleration about the center of mass is alpha.
01:12
For rotational motion, we can write the net talk about the center of mass equals the rotational inertia about the center of mass times the angular acceleration alpha.
01:30
And the net talk about the center of masses, the chalk due to the applied force f, that is f r, plus the talk due to the friction, that is, the frictional force f times r, equals the rotational inertia that is half times m are.
01:48
Square times the angular acceleration alpha.
01:52
So from this we get the horizontal force f plus the frictional force equals mr alpha over 2.
02:05
Let's say this has equation 2...