- #1
Taniaz
- 364
- 1
Motion in a vertical circle*
1. Homework Statement
Case 1: When a particle P of mass m moves in a vertical circle centered on O, its motion is governed by two forces, its weight mg and a force R directed radially inwards constraining it to move in a circle radius r. Resolving parallel to an perpendicular to OP (radially inwards and tangentially respectively) where OP is at an angle theta to the downward vertical, and applying Newton's Second Law, we get:
R-mg cos(theta) = mv^2/r and mg sin(theta) = -m dv/dt
Case 2:
If the particle is constrained to circular motion due to R radially outwards, the forces are directed differently as follows (theta is measured with respect to the upwards vertical)
mg cos (theta) - R = mv^2/r and mg sin (theta) = m dv/dt
2.Relevant equations
As above3.The attempt at a solution
After drawing the relevant diagrams, I was able to derive the 1 equation of both cases but I was a little unsure on how they got the second equation in both cases. I know they've use Newton's second law in this case but I don't know how they got the negative sign in the first case and a positive sign in the second case. Also, why did the angle change when the radial force changed direction? How can a force be radially outward?
1. Homework Statement
Case 1: When a particle P of mass m moves in a vertical circle centered on O, its motion is governed by two forces, its weight mg and a force R directed radially inwards constraining it to move in a circle radius r. Resolving parallel to an perpendicular to OP (radially inwards and tangentially respectively) where OP is at an angle theta to the downward vertical, and applying Newton's Second Law, we get:
R-mg cos(theta) = mv^2/r and mg sin(theta) = -m dv/dt
Case 2:
If the particle is constrained to circular motion due to R radially outwards, the forces are directed differently as follows (theta is measured with respect to the upwards vertical)
mg cos (theta) - R = mv^2/r and mg sin (theta) = m dv/dt
2.Relevant equations
As above3.The attempt at a solution
After drawing the relevant diagrams, I was able to derive the 1 equation of both cases but I was a little unsure on how they got the second equation in both cases. I know they've use Newton's second law in this case but I don't know how they got the negative sign in the first case and a positive sign in the second case. Also, why did the angle change when the radial force changed direction? How can a force be radially outward?