The discussion revolves around a circular motion problem involving the conservation of energy and Newton's second law. Participants are examining the forces acting on a bead threaded onto a wire in vertical circular motion, particularly focusing on the normal force and its direction.
There is an active exchange of ideas regarding the correct formulation of the forces and their implications. Some participants provide guidance on considering the direction of forces and the conditions under which the bead remains in circular motion. Multiple interpretations of the problem are being explored, particularly regarding the normal force's behavior at various angles.
Participants are working with constraints such as the requirement for the angle theta to be less than 33.7 degrees and the implications of negative values for the normal force in the context of the bead's motion.
ehild said:It is a bead on a wire. The normal force can be both outward and inward. How can the bead be in rest at the top, if two downward forces act on it?
Imagine you have this loop of wire, and you hold a bead to it at the top, just below the wire so the normal force acts downward. Release the bead, what happens?
ehild
inky said:According to the mechanics textbooks, only we consider one normal force. Not two normal force. Please see Application of Mathematics or Mechanics about vertical circular motion.
ehild said:Yes, but the roller coaster has some speed at the top. If it does not move, it falls down.
The centripetal force is the resultant of normal force and gravity.
mv^2/R = N +G. N=mv^2/R-G, which can be either positive or negative, depending on v.
Why don't you calculate the value R1 with your formula for a small angle, for example 10 degrees, instead of sticking to the statement that "For bead threaded onto ("on" in the original text) the wire , direction of normal force is toward the centre at any point"? The bead has a hole, and the wire goes through it. The wire can exert force both toward and away the centre, as it can be in contact with opposite walls of that hole in the bead.
I have to leave now. Think about what I have said.
ehild
ehild said:Velocity is a vector, it can have any directions. It is speed, the magnitude of velocity, that can not be negative.
You are right, the force is considered positive with respect to the acceleration if it has the same direction as that of the acceleration.
Now, about normal force: The (centripetal) acceleration points toward the centre. If the bead is at the top of the circle, gravity G=mg points also towards the centre so it is positive. The centripetal force is the sum of gravity G and the normal force R1. All these forces are vectors, they act along the vertical direction now. You found that R1=mv2/a-mgcos(theta). Calculate, R1 for theta =0, 30° and 40°, please, and tell me what you got.
ehild
ehild said:It is all right that R1 increases with the angle, but is it positive or negative?
You wrote that
R1= mv12/a-mgcos(θ),
and
1/2 mv12=mg(1-cos(θ)),
right?
So
R1=mg(2-3 cos(θ)).
If θ=0, R1=mg(2-3)=-mg negative.
If θ=30°, R1=mg(2-2.598)= -0.598 mg, negative.
If θ=40°, R1=mg(2-2.298)= -0.298 mg, negative.
You need the magnitude of the normal force. It can not be negative. Your formula is wrong.
The magnitude of the normal force is the negative of the values above. The correct form is
R1=|mg(2-3 cos(θ))| =mg(3 cos(θ)-2)
ehild