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Centripetal force and centripetal acceleration of bodies moving in uniform circular motion

If we tie an eight ounce ball to the end of a cord and whirl it in a circle we know from experience that the ball will exert a force on the cord in the form of a tension and the faster we whirl the object, the larger will be the force. In 1543 Copernicus published his theory that the earth and other planets revolved about the sun in circular orbits. Such a theory would naturally call to mind the idea of an object being whirled about on a cord and give rise to the question of what force constrained a planet to a circular orbit. Newton later answered that question with his postulate of the existence of a general attractive force existing between any two material objects in the universe i.e. his Law of Universal Gravitation.

We will now consider the laws ruling the motion of a body moving in a circle at a constant speed. Suppose we have a cord tied to an object and we whirl the object about in such a way that it slides in a circular path at a constant speed on a frictionless horizontal plane (we consider a horizontal path to eliminate the effects of gravity). Given the mass m of the object, the radius r of the circular path, and the speed v of the object, we wish to know what force F is exerted on the object in the form of tension on the string.

Newton’s First Law of Motion states: A body at rest will remain at rest and a body in motion will remain in motion at constant speed in a straight line, as long as no unbalanced force acts on it. From Newton’s First Law we understand that any object moving in a circle must have a force acting on it or else it would move in a straight line. Furthermore this force must act towards the center of the circle or else there would be a component of velocity in the direction of its path and its speed would change. The only effect of this force directed towards the center of the circle is to change the direction of the motion of the object. The law that gives the force on the object is Newton’s second law, F = ma, where F is the applied force, m is the mass of the object and a is the acceleration towards the center of the circle caused by the force. In fact, we can show that the acceleration towards the center of the circle experienced by the object is a function of only the two variables v and r and is given by the formula

Given values for v and r we can compute the acceleration a and then compute the force F on the object from the formula F = ma.

Derivation of the formula a = v^{2}/r. Consider Fig. 1
which shows the velocity of the body at points P and Q. At P
it has velocity V = V_{1} and at point Q it has velocity V = V_{2} .
Here V is a vector quantity which has a constant magnitude v.
The change in velocity ΔV is given by the vector difference V_{2}
- V_{1}. If Δt is the time required to go from P to Q, then the
average acceleration A_{ave} of the body is given by

Considering now Fig. 2, we shall show that triangle MNQ is similar to triangle OPQ.. To do this we will employ the following theorem from Euclidean geometry:

Theorem 1. Given: BAC. See Fig. 3. From point E draw perpendiculars EF and EG to AC and AB respectively. Then E = A.

This theorem is easily proven since AQG = EQF and thus ΔAQG is similar to ΔEFQ.

We now note in Fig. 2 that MN is perpendicular to OP and MQ is perpendicular to OQ so from Theorem 1 it follows that θ of Δ OPQ is equal to M of ΔMNQ. Since both triangles OPQ and MNQ are isosceles triangles, it follows that they are similar.

Denote the magnitude of V_{1} by v and the magnitude of ΔV
=V_{2} - V_{1} by Δv. Then

Dividing both sides by Δt and multiplying both sides by v, we obtain

Denoting the magnitude of the average acceleration by , we obtain

To find the instantaneous acceleration a, we take the limit of as Δt approaches zero: The points P and Q approach each other and the chord PQ approaches the length of the arc PQ. The distance PQ can be replaced by vΔt. Substituting these values into 2), we obtain

When the angle θ is very small, the vector ΔV is perpendicular to the velocity V_{1} and is directed
toward the center of the circle.

A second simpler method for deriving the formula a = v^{2}/r. There is a much
simpler method for deriving the formula using differential calculus. In Fig. 4, s = θr, where s is
the distance on the circle from the x-axis
to point P, θ is the angle in radians from
the x-axis to point P and r is the radius of
the circle. The x and y coordinates of
point P are given by

1) x = r cos θ

2) y = r sin θ

Now s = vt and θ = s/r, so θ = vt/r where v is the speed i.e. magnitude of velocity V. The above equations then become

which constitute parametric equations giving the x and y coordinates of the body P as a function of time t. Differentiating 3) and 4) with respect to time gives

Differentiating 5) and 6) with respect to time, we get the acceleration

The acceleration a is thus a vector of magnitude v^{2}/r directed in toward the center of the circle.

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