00:00Hey guys, let's learn.
00:04Lying down doesn't make you smart or rich.
00:08This is almost the same problem as the previous series.
00:12This time, a particle is inside a cylindrical shell.
00:17We are asked to calculate the magnitude of the normal force on the particle.
00:25It wouldn't hurt to watch this short animation.
00:30A cylinder is in 3D Cartesian coordinates with the axis of the cylinder exactly on the
00:37Z axis.
00:41Inside the tube is a particle.
00:45Naturally, the particle will move vertically downward.
00:53In our problem, the tube rotates about its axis at a certain angular velocity.
01:00The particle rotates along with the tube.
01:03Thus, the particle does not shift at all from its initial position.
01:11We can draw this system in a two-dimensional view.
01:16The two vertical straight lines represent the tube's envelope when viewed from the front.
01:23Now we'll analyze the forces acting on the particle.
01:28A particle with mass near the Earth's surface will experience a gravitational force directed
01:32vertically downward.
01:37When the particle touches the surface of the tube casing, there is a normal force in the
01:40normal direction.
01:44Although not stated, we are sure that the surface of the tube casing is rough.
01:49Otherwise, the balance of forces in the vertical direction will not occur.
01:56Mathematically, F minus mg is equal to zero.
02:03The normal force is always toward the axis of rotation, this force acts as a centripetal force.
02:10N is equal to m omega squared R.
02:14R here is the radius of the tube.
02:19This particle moves periodically.
02:24If so, omega is equal to 2 pi over the period.
02:30This capital T is the period of the rotation.
02:34N is equal to 4 m pi squared R over capital T squared.
02:41The value of the physical quantity can be seen on the problem sheet.
02:46If the number is like this, we must use a scientific calculator.
02:50N is approximately 6.58 newtons.
02:55It turns out this problem is not as complicated as it seems.
03:00Happy learning everyone.
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