I've just read that the centripetal force on an electron is mv^2/r; and that this is also equal to the electrostatic force pulling the electron toward the nucleus.
Does mv^2/r hold good in all Newtonian siuations as a formula for centripetal force acting on a body?
Thanks
No. Suppose the electron is moving in some arbitrary way. Then the compinent of the acceleration in the radial direction (i.e. the direction away from the nucleus) is given by:
a = d^2r/dt^2 - (V_t)^2/r
Here V_t is the velocity of theelectron in the "tangential direction", this is the component of the velocity orthogonal to the distance vector of the electron to the nucleus.
This means that the second derivative of the radius has to be a constant for the formula for the acceleration in the radial direction to reduce to:
V_t^2/r
For V_t to be equal to V ,m requires that there is no component of the velocity in the radial direction, so r must be constant.
So, you obtain the result that if the radius is constant, then the acceleration in the radial direction is given by the V^2/r. Since F = m a, it then follows that the force in the radial dorection equals m V^2/r.
If I had made it conditional that the electron was orbiting at a constant radius, would it then be correct; and would it also be accurate in 'ordinary' situations, eg spinning a stone around on a string?
Yes, if the radius is constant, then it is correct.
Thanks, CI
Yes, the formula for centripetal force, mv^2/r, holds true in all Newtonian situations as a formula for the centripetal force acting on a body.
To understand why this formula is applicable, let's break it down step by step:
1. Centripetal Force: Centripetal force is the force that acts towards the center of a circular path, and it keeps an object in circular motion. In other words, it is the force required to maintain an object moving in a circular path instead of moving in a straight line.
2. Mass and Velocity: In the formula mv^2/r, "m" represents the mass of the object, and "v" represents its velocity. The mass of an object determines its inertia, while its velocity represents the speed at which it is moving in the circular path.
3. Radius: The "r" in mv^2/r represents the radius of the circular path. The radius is the distance from the center of the circle to the object.
4. Electrostatic Force: In the case of the electron, the electrostatic force between the negatively charged electron and the positively charged nucleus provides the centripetal force that keeps the electron in orbit. The electrostatic force is given by Coulomb's law, which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of their distance.
So, the equation mv^2/r is appropriate for the electron's circular motion because the electrostatic force between the electron and the nucleus provides the centripetal force needed to maintain the circular orbit. In essence, the electrostatic force is equal in magnitude and opposite in direction to the centripetal force, allowing the electron to stay in a stable orbit around the nucleus.
However, it's important to note that while mv^2/r holds true for circular motion, it may not be applicable for other types of motion, such as non-uniform circular motion or motion along a curved path. In such cases, additional factors like tangential acceleration or other forces may need to be considered.