Problem: If an object is to move at velocity v, in m/s, on a circle of radius r, it must be accelerated toward the center at with acceleration a=v ^ 2 /r This acceleration is necessary to hold the object in the circle. This acceleration is always perpendicular to the direction of motion, so it has no effect on the object's speed. This acceleration simply keeps turning the object so it will remain on the circle. This acceleration is called the centripetal acceleration of the object.
What is the centripetal acceleration of an object on a circle of radius 3 meters, if the object's speed is 4 meters per second?
If the object has mass 91 kg, how much force is required to provide the acceleration?
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Solution: The acceleration is v ^ 2 /r = ( 4 m/s) ^ 2 / ( 3 m) = 5.333 meters per second per second toward the center of the circle. To accelerate 91 kg at this rate requires ( 91 kg) ( 5.333 m/s/s) = 485.3 Newtons of force.
Generalized Response: The centripetal acceleration of an object moving along a circle of radius r at velocity v is a = v^2 / r; this acceleration is directed toward the center of the circle. If the object has mass m, the force required to keep the object in its circular path is F = ma = m v^2 / r; this force is also directed toward the center of the circle.
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Figure Description: The figure below shows an object moving with constant speed v on a circle of radius r. The acceleration required to keep the object moving in the circle, as opposed to the straight line along which it would travel if it had no acceleration, is a = v^2 / r, and is in the direction of the center of the circle.
If the object has mass m, the force required to keep it on its circular path will therefore be F = ma = m v^2 / r. This force will be directed in the same direction as the acceleration, toward the center of the circle.
