Given the mass and the velocity of an object in circular motion with radius #r#, how do we calculate the magnitude of the centripetal force, #F#?

Question

Charlotte Aaron · Accepted Answer

Constant acceleration (change of direction) requires a constant force, which is described by #F=(mv^2)/r# if #v# is measured in #ms^-1# or #F=momega^2r# if #omega# is measured in #rads^-1#.

Acceleration is defined as the rate of change of velocity, and velocity has a direction. Since the direction of something in circular motion is always changing, it requires a constant acceleration, and Newton's Second Law indicates that constant acceleration requires a constant force. #F=ma# If we look at the instantaneous linear velocity of the object at any moment as #v# #ms^-1#, the expression for the centripetal acceleration is: #a=v^2/r# So the force acting is simply #F=(mv^2)/r# Let's look at where that expression for the acceleration comes from. The speed of the object is constant: it is not accelerating because its speed increases (speeding up) or decreases (slowing down). It is accelerating because its direction is changing constantly. If we imagine a direction vector, which is a tangent to the circle in which the object is moving, at each moment it points in a slightly different direction. Acceleration is just change in velocity divided by change in time: #a=(Delta v)/(Delta t)# This Khan Academy video offers a more-detailed explanation of how we get from there to #a=v^2/r#: https://tutor.hix.ai If, instead, we measure the radial velocity of the object in radians per second #(rads^-1)#, the expression is: #F=m omega^2 r#

Olivia Askew · Answer

undefined The magnitude of the centripetal force, $ F $, acting on an object in circular motion can be calculated using the following formula:

$$ F = \frac{{m \cdot v^2}}{r} $$

Where:
- $ F $ is the magnitude of the centripetal force,
- $ m $ is the mass of the object,
- $ v $ is the velocity of the object in circular motion, and
- $ r $ is the radius of the circular path.