OK. Can you integrate Eqns. 1 and 2 of my post #29 to get the net horizontal and vertical force components on the disk? Do you know how to work with Eqns. 1 and 2 to get the torque on the disk?fayan77 said:
No way. This does not take into account the fact that the direction of the force per unit area (as described by the unit vector) changes with angular position on the disk. That's why I thought it would be better if you switched to Cartesian coordinates. L'd like to see what you get for the x and y components of the force Fx and Fy when you integrate over the area of the disk.fayan77 said:
No. Your formulation of the polar one is incorrect because it does not take into account the change in direction of the differential force over the area of the disk. Force is a vector, and if, in your system, its direction varies with position, it is imperative that you account in summing its contribution to the net force. If you want to use polar coordinates, you can express the unit vector in the theta direction as follows in terms of the (constant) unit vectors in the x and y directions (and then integrate using polar coordinates):fayan77 said:
After you show that the net force is zero, let’s talk about how to get the torque (which is not zero).fayan77 said:
You'll see. At each location on the disk, the differential force dF on each differential area is oriented perpendicular to the moment arm drawn from the origin r. Therefore, its differential contribution to the torque is $$dT=\left[\mu \left(\frac{\Omega r}{a}+\frac{\Omega r}{b}\right)rdrd\theta\right] r=\mu \left(\frac{1 }{a}+\frac{1}{b}\right)\Omega r^3drd\theta$$What does this give you when you integrate with respect to r and theta?fayan77 said:
Suppose you have two parallel forces acting on a rigid body. The forces are equal in magnitude, but opposite in direction. What is the net force acting on the body?fayan77 said:
Yes. Please answer my questions.fayan77 said:
Correct. Now, in our problem, for each differential element of area on the disk, there is a corresponding differential area element located 180 degrees away in which the force per unit area is parallel and opposite in direction. Do you see that?fayan77 said:
Well, you already showed by integrating that the net force is equal to zero. Of course, we expected that from the symmetry of the system.fayan77 said:
Yes. You just integrate the equation in post 49, pulling out the 2pi.fayan77 said:
