The centripetal force acting on a particle is given by F = mv2/r.?
The centripetal force acting on a particle is given by F = mv2/r. If the centripetal force and mass are kept constant, decreasing the radius of the particle will mean that the particle's velocity must decrease.
- AenimaLv 69 years agoFavorite Answer
True, it must decrease.
F is inversely proportional to the radius, as the radius increases, the force will decrease. As the force increases, the radius decreases.
F is also directly proportional to the v^2. As velocity increases, F increases.
The only way for F to remain constant when r decreases is if v decreases as well. F will become larger when r decreases and at the same time will become bigger when v increases, making the final F constant.
For example, initially we have a force F, radius r, mass m and velocity of v.
F = mv^2/r
Now, we want the force to stay constant and decrease the radius as well.
Ff = mvf^2/rf. Let Ff be the force after the radius changes to a smaller amount, vf the velocity and rf the final radius.
Let rf = r/2. (half original radius.) We want Ff to be equal to F (to remain constant). so:
Ff = F = mv^2/r = mvf^2/(r/2)
mv^2/r = 2mvf^2/r
r's cancel, m's cancel:
v^2 = 2vf^2
v^2/2 = vf^2
root(v^2/2) = vf
vf = v/(root2) = 0.707*v
By changing the velocity to 0.707 times the original velocity, you must cut the radius in half to keep the original force constant.
^fixed typo XD
Also the converse argument would work too. By cutting the radius in half, you must decrease in velocity to 0.707 times the original velocity in order to keep the force constant.
- Steve4PhysicsLv 79 years ago
This is correct. But in practice this rarely (never?) happens, because the force is never constant. E.g. if the force is caused by attraction (gravitational or electrostatic) then the force changes with radius (proportional the the inverse of the radius squared typically). If the force is mechanical (e.g. tension) it would self adjust (think of whirling a mass at the end of a spring).
I can't think of any situation where the force would be kept constant - any suggestions?
- Anonymous9 years ago