5.2.  Centripetal Acceleration

In this section we determine how the magnitude a_c of the centripetal acceleration depends on the speed v of the object and the radius r of the circular path. We will see that a_c=v²/r.

In Figure 5.2a an object (symbolized by a dot ·) is in uniform circular motion. At time t₀ the velocity is tangent to the circle at point O, and at a later time t the velocity is tangent at point P. As the object moves from O to P, the radius traces out the angle q , and the velocity vector changes direction. To emphasize the change, part b of the picture shows the velocity vector removed from point P, shifted parallel to itself, and redrawn with its tail at point O. The angle b between the two vectors indicates the change in direction. Since the radii CO and CP are perpendicular to the tangents at points O and P, respectively, it follows that a+b=90° and a+q =90°. Therefore, angle b and angle q  are equal.

Figure 5.2  (a) For an object (·) in uniform circular motion, the velocity v has different directions at different places on the circle. (b) The velocity vector has been removed from point P, shifted parallel to itself, and redrawn with its tail at point O.

As always, acceleration is the change Dv in velocity divided by the elapsed time Dt, or a=Dv/Dt. Figure 5.3a shows the two velocity vectors oriented at the angle q  with respect to one another, together with the vector Dv that represents the change in velocity. The change Dv is the increment that must be added to the velocity at time t₀, so that the resultant velocity has the new direction after an elapsed time Dt=t–t₀. Figure 5.3b shows the sector of the circle COP. In the limit that Dt is very small, the arc length OP is approximately a straight line whose length is the distance vDt traveled by the object. In this limit, COP is an isosceles triangle, as is the triangle in part a of the drawing. Since both triangles have equal apex angles q , they are similar, so that

This equation can be solved for Dv/Dt, to show that the magnitude a_c of the centripetal acceleration is given by a_c=v²/r.

Figure 5.3  (a) The directions of the velocity vector at times t and t 0 differ by the angleq . (b) When the object moves along the circle from O to P, the radius r traces out the same angle q  . Here, the sector COP has been rotated clockwise by 90° relative to its orientation in Figure 5.2.

Centripetal acceleration is a vector quantity and, therefore, has a direction as well as a magnitude. The direction is toward the center of the circle, and Conceptual Example 2 helps us to set the stage for explaining this important fact.

In Figure 5.4 an object, such as a model airplane on a guideline, is in uniform circular motion. The object is symbolized by a dot (·), and at point O it is released suddenly from its circular path. For instance, the guideline for a model plane is cut suddenly. Does the object move along the straight tangent line between points O and A or along the circular arc between points O and P? Figure 5.4  If an object (·) moving on a circular path were released from its path at point O, it would move along the straight tangent line OA in the absence of a net force. Reasoning and Solution Newton’s first law of motion guides our reasoning. An object continues in a state of rest or in a state of motion at a constant speed along a straight line unless compelled to change that state by a net force. When the object is suddenly released from its circular path, there is no longer a net force being applied to the object. In the case of a model airplane, the guideline cannot apply a force, since it is cut. Gravity certainly acts on the plane, but the wings provide a lift force that balances the weight of the plane. In the absence of a net force, then, the plane or any object would continue to move at a constant speed along a straight line in the direction it had at the time of release. This speed and direction are given in Figure 5.4 by the velocity vector v. As a result, the object would move along the straight line between points O and A, not on the circular arc between points O and P. Related Homework: Problems 4, 45

As Example 2 discusses, the object in Figure 5.4 would travel on a tangent line if it were released from its circular path suddenly at point O. It would move in a straight line to point A in the time it would have taken to travel on the circle to point P. It is as if the object drops through the distance AP in the process of remaining on the circle, and AP is directed toward the center of the circle in the limit that the angle q  is small. Thus, the object accelerates toward the center of the circle at every moment. The acceleration is called centripetal acceleration, because the word “centripetal” means “center-seeking.”

CENTRIPETAL ACCELERATION

Magnitude: The centripetal acceleration of an object moving with a speed v on a circular path of radius r has a magnitude ac given by  (5.2)  Direction: The centripetal acceleration vector always points toward the center of the circle and continually changes direction as the object moves.

The following example illustrates the effect of the radius r on the centripetal acceleration.

The bobsled track at the 1994 Olympics in Lillehammer, Norway, contained turns with radii of 33 m and 24 m, as Figure 5.5 illustrates. Find the centripetal acceleration at each turn for a speed of 34 m/s, a speed that was achieved in the two-man event. Express the answers as multiples of g=9.8 m/s2. Figure 5.5  This bobsled travels at the same speed around two curves with different radii. For the turn with the larger radius, the sled has a smaller centripetal acceleration. Reasoning  In each case, the magnitude of the centripetal acceleration can be obtained from ac=v2/r. Since the radius r is in the denominator on the right side of this expression, we expect the acceleration to be smaller when r is larger. Solution From ac=v2/r it follows that The centripetal acceleration is indeed smaller when the radius is larger. In fact, with r in the denominator on the right of ac=v2/r, the acceleration approaches zero when the radius becomes very large. Uniform circular motion along the arc of an infinitely large circle entails no acceleration, because it is just like motion at a constant speed along a straight line.

In Section 4.11 we learned that an object is in equilibrium when it has zero acceleration. Conceptual Example 4 discusses whether an object undergoing uniform circular motion can ever be at equilibrium.

Interactive LearningWare 5.1 Car A negotiates a curve at a speed of 32 m/s and experiences a centripetal acceleration of 6.4 m/s2. Car B negotiates the same curve at a speed of 16 m/s. What centripetal acceleration does it experience? Related Homework: Problem 56

A car moves at a constant speed, and there are three parts to the motion. It moves along a straight line toward a circular turn, goes around the turn, and then moves away along a straight line. In each of these parts, is the car in equilibrium? Reasoning and Solution An object in equilibrium has no acceleration, according to the definition given in Section 4.11. As the car approaches the turn, both the speed and direction of the motion are constant. Thus, the velocity vector does not change, and there is no acceleration. The same is true as the car moves away from the turn. For these parts of the motion, then, the car is in equilibrium. As the car goes around the turn, however, the direction of travel changes, so the car has a centripetal acceleration that is characteristic of uniform circular motion. Because of this acceleration, the car is not in equilibrium during the turn. In general, an object that is in uniform circular motion can never be in equilibrium. Related Homework: Problem 43

We have seen that going around tight turns (smaller r) and gentle turns (larger r) at the same speed entails different centripetal accelerations. And most drivers know that such turns “feel” different. This feeling is associated with the force that is present in uniform circular motion, and we now turn to this topic.

The car in the drawing is moving clockwise around a circular section of road at a constant speed. What are the directions of its velocity and acceleration at (a) position 1 and (b) position 2? Specify your responses as north, east, south, or west. Background: Centripetal acceleration lies at the heart of this question. In particular, the direction of the object’s velocity and the direction of its centripetal acceleration are different. For similar questions (including calculational counterparts), consult Self-Assessment Test 5.1. This test is described at the end of Section 5.3.

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