Uniform Circular Motion

A particle is in uniform circular motionif 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. However, actually 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 relation 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

(centripetal acceleration) (4-32)

where is the radius of the circle and 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 ) in time

(period). (4-33)

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. 4-32

To find the magnitude and direction of the acceleration for uniform circular motion, we consider Fig. 4-19. In Fig. 4-l9a, particle p moves at constant speed around a circle of radius . At the instant shown, p has coordinates and .

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 drawn to the particle's position. Then the angle that makes with a vertical at p equals the angle that radius makes with the axis.

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

(4-34)

Now, using the right triangle in Fig. 4-19a, we can replace with and with to write

(4-35)

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

(4-36)

Now note that the rate at which changes is equal to the velocity component . Similarly, , and, again from Fig. 4-19b, we see that and . Making these substitutions in Eq. 4-36, we find

(4-37)

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

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

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

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 or , the pilot feels heavy. At about , 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 -LOC for " -induced loss of consciousness." What is the centripetal acceleration, in units, of a pilot flying an F-22 at speed = 2500 km/h (694 m/s) through a circular arc with radius of curvature = 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. 4-32:

(Answer)

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

Velocity and Coordinate by Integration

When varies with time, we can use the relation to find the velocity as a function of time if the position is a given function of time. Similarly, we can use to find the acceleration as a function of time if the velocity is a given function of time.

We can also reverse this process. Suppose is known as a function of time; how can we find as a function of time? To answer this question, we first

Fig.3 The area under a velocity-time graph equals the displacement

consider a graphical approach. Figure 3 shows a velocity-versus-time curve for a situation where the acceleration (the slope of the curve) is not constant but increases with time. Considering the motion during the interval between times and , we divide this total interval into many smaller intervals, calling a typical one . Let the velocity during that interval be . Of course, the velocity changes during , but if the interval is very small, the change will also be very small. This displacement during that interval, neglecting the variation of , is given by

.

This corresponds graphically to the area of the shaded strip with height and width , that is, the area under the curve corresponding to the interval . Since the total displacement in any interval (say, to )is the sum of the displacements in the small subintervals, the total displacement is given graphically by the total area under the curve between the vertical lines and . In the limit, when all the become very small and their number very large, this is simply the integral of (which is in general a function of )from and . Thus is the position at time and the position at time :

(2-14)

A similar analysis with the acceleration-versus-time curve, where is in general a function of , shows that if is the velocity at time and the velocity at time , the change in velocity during a small time interval is approximately equal to , and the total change in velocity ( )during the interval is given by

Or, finally, .

Exercises

1. Velocity of a body, moving in viscous medium, is given by the equation , where - initial velocity, - constant. What are the distance and acceleration as function of time?

2. A particle moves along a straight line with velocity , where is constant. If at time the distance, traveled the particle was , determine: (a) dependence of speed and acceleration on time ( and )

3. (a) If particle’s acceleration is given by , (where is in meter/second² and in seconds), what its velocity at ? (b) What is its coordinate at s?