00:01
Hello guys, in this problem we have a spool, it is given that the mass of this spool equals 130 kilograms, the radius of its gyration is 0 .3 meter and a force p is applied at a point on the spool tangential to its inner circular circumference and it is given that the force p equals 550 newton's.
00:46
If coefficient of static friction equals 0 .2, coefficient of kinetic friction equals 0 .15.
00:55
We have to determine the angular acceleration alpha for the spool.
01:07
For this, let us assume the inner radius as small r and the outer radius as capital r.
01:17
Let us draw the forces, the weight of the spool act through its center and the weight a equals mass times acceleration due to gravity.
01:26
A normal force is exerted by the ground, the normal force act in upward direction and let us assume the friction force is acting in backward direction f.
01:43
For pure rolling motion, if we assume the acceleration, the tangential acceleration of the spool is a, this is a c m, then we can write the center of mass acceleration equals the outer radius r times the angular acceleration alpha, say this as equation one.
02:10
Using the rotational motion equation, we can write the net torque about the center equals the moment of inertia about the center times the angular acceleration alpha.
02:26
Apart from this, we can write the net torque as the pulling force p times its perpendicular distance, small plus the torque due to the friction, the friction force f times the outer radius r and it equals the moment of inertia m times the radius of gyration square times the angular acceleration alpha, say this as equation one...