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revised Equation of motion for average acceleration
deleted 1 characters in body
Mar
3
answered Equation of motion for average acceleration
Feb
12
revised Shouldn't the change in kinetic energy be more in a moving elevator from a stationary frame of reference?
The one-half shouldn't be there.
Feb
12
suggested suggested edit on Shouldn't the change in kinetic energy be more in a moving elevator from a stationary frame of reference?
Feb
12
comment What is the proof that a force applied on a rigid body will cause it to rotate around its center of mass?
You didn't give any arguments for why the axis should go through the center of mass even in the special case of a sphere.
Feb
12
comment What is the proof that a force applied on a rigid body will cause it to rotate around its center of mass?
I appreciate you writing the answer but consider removing it. This does not answer the question. You just gave an analysis of a force acting on a sphere and then you gave the expression for the kinetic and rotational energy for the sphere. That is all.
Feb
12
revised What is the proof that a force applied on a rigid body will cause it to rotate around its center of mass?
edited tags
Feb
10
comment Force exerted on a cone
The cone not moving horizontally may seem contrary to intuition but that's because your fingers end up applying a force perpendicular to the surface of the cone instead. It's hard to apply a force directly perpendicular (along $y$-axis) to the cone, but if you do manage it, it will not move. I think you may say that the actual effective force is not (of course) going to be along the $y$-axis, but then there would be no point of drawing the forces in that direction in the first place if they are going to act perpendicular to the surface instead.
Feb
10
comment Force exerted on a cone
I'm sorry but I don't find this correct. Forces perpendicular to the cone can't make it move in the horizontal direction. The only answer to this question is when the forces are acting perpendicular to the surface, to which you do get the same result that you got. (That of course, as you can see is because the work calculation is same for both the cases).