Uniform Circular Motion

A particle is in uniform circular motion if it travels around a circle or a circular arc at constant (uniform) speed. Although the speed does not vary, the particle is accelerating. That fact may be surprising because we often think of acceleration (a change in velocity) as an increase or decrease in speed. Remember, however, that velocity is a vector, not a scalar. Thus, even if a velocity changes only in direction, there is still an acceleration, and that is what happens in uniform circular motion.

Figure 4-18 shows the relationship between the velocity and acceleration vectors at various stages during uniform circular motion. Both vectors have constant magnitude as the motion progresses, but their directions change continuously. The velocity is always directed tangent to the circle in the direction of motion. The acceleration is always directed radially inward. Because of this, the acceleration associated with uniform circular motion is called a centripetal (meaning “center seeking”) acceleration. As we prove next, the magnitude of this acceleration  is

Fig. 4-18   Velocity and acceleration vectors for a particle in counterclockwise uniform circular motion. Both vectors have constant magnitude but vary continuously in direction.

where r is the radius of the circle and v is the speed of the particle.

In addition, during this acceleration at constant speed, the particle travels the circumference of the circle (a distance of 2πr) in time

T is called the period of revolution, or simply the period, of the motion. It is, in general, the time for a particle to go around a closed path exactly once.

Proof of Eq. 4-34

To find the magnitude and direction of the acceleration for uniform circular motion, we consider Fig. 4-19. In Fig. 4-19a, particle p moves at constant speed v around a circle of radius r. At the instant shown, p has coordinates x_p and y_p.

Recall from Section 4-3 that the velocity  of a moving particle is always tangent to the particle’s path at the particle’s position. In Fig. 4-19a, that means  is perpendicular to a radius r drawn to the particle’s position. Then the angle θ that  makes with a vertical at p equals the angle θ that radius r makes with the x axis.

The scalar components of  are shown in Fig. 4-19b. With them, we can write the velocity  as

Now, using the right triangle in Fig. 4-19a, we can replace sin θ with y_p/r and cos θ with x_p/r to write

Fig. 4-19   Particle p moves in counterclockwise uniform circular motion. (a) Its position and velocity  at a certain instant. (b) Velocity  and its components. (c) The particle’s acceleration  and its components.

To find the acceleration  of particle p, we must take the time derivative of this equation. Noting that speed v and radius r do not change with time, we obtain

Now note that the rate dy_p/dt at which y_p changes is equal to the velocity component v_y. Similarly, dx_p/dt = v_x, and, again from Fig. 4-19b, we see that v_x = −v sin θ and v_y = v cos θ. Making these substitutions in Eq. 4-38, we find

This vector and its components are shown in Fig. 4-19c. Following Eq. 3-6, we find

as we wanted to prove. To orient , we find the angle  shown in Fig. 4-19c:

Thus,  = θ, which means that  is directed along the radius r of Fig. 4-19a, toward the circle’s center, as we wanted to prove.

CHECKPOINT 6 An object moves at constant speed along a circular path in a horizontal xy plane, with the center at the origin. When the object is at x = −2 m, its velocity is −(4 m/s). Give the object’s (a) velocity and (b) acceleration when it is at y = 2 m.

Sample Problem 4-9

“Top gun” pilots have long worried about taking a turn too tightly. As a pilot’s body undergoes centripetal acceleration, with the head toward the center of curvature, the blood pressure in the brain decreases, leading to loss of brain function.

There are several warning signs to signal a pilot to ease up. When the centripetal acceleration is 2g or 3g, the pilot feels heavy. At about 4g, the pilot’s vision switches to black and white and narrows to “tunnel vision.” If that acceleration is sustained or increased, vision ceases and, soon after, the pilot is unconscious—a condition known as g-LOC for “g-induced loss of consciousness.”

What is the centripetal acceleration, in g units, of a pilot flying an F-22 at speed v = 2500 km/h (694 m/s) through a circular arc having a radius of curvature r = 5.80 km?

Solution: The Key Idea here is that although the pilot’s speed is constant, the circular path requires a (centripetal) acceleration, with magnitude given by Eq. 4-34:

If an unwary pilot caught in a dogfight puts the aircraft into such a tight turn, the pilot goes into g-LOC almost immediately, with no warning signs to signal the danger.

Fig. 4-20   Alex (frame A) and Barbara (frame B) watch car P, as both B and P move at different velocities along the common x axis of the two frames. At the instant shown, x_BA is the coordinate of B in the A frame. Also, P is at coordinate x_PB in the B frame and coordinate x_PA = x_PB + x_BA in the A frame.