## SAT Physics Circular Motion - Uniform Circular Motion

UNIFORM CIRCULAR MOTION
Uniform circular motion involves objects moving at a constant speed but with changing velocity. When an object moves at constant speed, the magnitude of velocity is also constant. How then is velocity changing in circular motion? Velocity is a vector quantity, meaning it consists of both magnitude and direction. Objects moving in circular motion are continually changing direction. As a result, the velocity of the object is changing even though it moves at a constant speed.
Like velocity, the acceleration of an object in uniform circular motion has constant magnitude but has changing direction. Acceleration is the rate of change in velocity (the change in velocity during a time interval). If velocity has a constant magnitude, the acceleration will also have a constant value. However, the direction of acceleration constantly changes as an
object moves in circular motion. In uniform circular motion, the direction of the acceleration vector is toward the center of the circle. This acceleration is said to be uniform: having a constant magnitude and applied in the same manner (toward the center) at all times.

Period and Frequency
All linear motion quantities are based on the linear meter. However, the linear meter is not very useful when describing motion that does not follow a straight path. All circles have one thing in common. Objects moving in circles return to the same location every time they complete one cycle (one circle, one revolution, one rotation, and so on). A cycle consists of one circumference, and this is the basis for all circular motion quantities.
The time to complete one cycle is known as the period, T. The period of a circling object can be calculated by dividing the time of the motion, t, by the number of cycles completed during time t. The number of cycles does not have any units, and the units of period are seconds.
The frequency of an object is the number of cycles an object completes during one second. Think of the frequency as how frequently the object is cycling. Mathematically, frequency is the inverse of the period. The units of frequency are inverse seconds (1/s or s-1). These units are also known as Hertz (Hz). Any of these units may be used on exams, and the formula for frequency is simply the inverse of the formula for the period. The relationship between the period and frequency is expressed in the following equation: Period and Frequency
An object completes 20 revolutions in 10 seconds. Determine the period and frequency of this motion.

WHAT'S THE THICK?

A revolution is another way to indicate a cycle, and a cycle is simply an event.The number of cycles is just the count of an event, and therefore it has no units.Period is the time for one complete cycle. Time has the units of seconds, and therefore time must be in the numerator. Tangential Velocity and Centripetal Acceleration
Velocity is not constant in uniform circular motion. However, when any moving object is paused (frozen for an instant of time), it will have an instantaneous velocity vector with a specific magnitude and direction. When objects follow a curved path, the instantaneous velocity is tangent to the motion of the object. Thus, the instantaneous velocity is referred to as the tangential velocity. Several instantaneous velocity vectors are shown for the object circling in Figure 6.1. Figure 6.1. Instantaneous velocity vectors

In Figure 6.1, it is apparent that although the direction of the tangential velocity is con­tinually changing, its magnitude remains constant. The magnitude of the tangential velocity is also equal to the speed of the circling object. They are both determined using a modified version of the constant speed formula: As previously stated, circular motion is based on one complete cycle. In one complete cycle, an object travels one circumference, d = 2nr, in one period, t=T.
One important aspect of tangential velocity involves objects leaving the circular path. Forces are responsible for creating circular motion. If the forces causing circular motion stop acting, the object will leave the circular path. When this happens, the tangential velocity becomes the initial velocity for the object’s subsequent motion. If no forces act on the object, it will move in a straight line matching the tangential velocity at the time of release, as shown in Figure 6.2(a). However, if another force acts on the object, the object will become subject to the new force. In Figure 6.2(b), the object leaves the circle and is acted upon by gravity, causing projectile motion. Figure 6.2. Tangential velocity. In (a) the object is initially circling
horizontally on a frictionless surface. In (b) the object is initially
circling vertically at a distance h above the surface.

Since circling objects have a changing velocity vector, they are continuously accelerating. This type of acceleration is known as centripetal acceleration, ac Centripetal means “center seeking.” Centripetal acceleration is directed toward the center of the circle, as shown in Figure 6.3. Figure 6.3 centripital acceleration

Even though circling objects are accelerated toward the center of the circle, their tangential velocities prevent them from ever reaching the center. Since the acceleration vectors lie along the radii of the circle, the centripetal acceleration may also be referred to as the radial acceleration. The centripetal acceleration can be determined with the following formula: Tangential Velocity and Centripetal Acceleration
Determine the acceleration of an object experiencing uniform circular motion.It is moving in a circle with a radius of 10 meters and a frequency of 0.25 Hertz.

WHAT'S THE THICK?

The period is the inverse of the frequency. Solving for the tangential velocity requires the period. Solving for the centripetal acceleration requires the tangential velocity. Solve each equation in turn. Answers may be expressed in terms of n in order to avoid multiplying by 3.14.

DYNAMICS IN CIRCULAR MOTION
In order for objects to experience acceleration, a net force (sum of forces) must act in the direction of the acceleration. Objects in circular motion are being accelerated toward the center of a circular path. This implies that the net force is also directed toward the center of the circle. In uniform circular motion, the net force is known as the centripetal force, Fc In linear motion problems, the net force is represented by and Newton’s second law dictates the relationship between the net force and acceleration: = ma. In circular motion prob­lems, Fc replaces and Newton’s second law applied to circular motion becomes:
Fc = mac
Frequently, circular motion force problems involve the speed and/or tangential velocity Dynamics problems in circular motion are solved in a similar manner as all other force problems.
1. Orient the problem. Draw a force diagram. If the motion is circular, any force vectors pointing toward the center
of the circle are considered positive. Those pointing away from the center are negative.
2. Determine the type of motion. If an object is moving along a circular path, recognize that the circular net force,
Fc, is used in place of .
3. Sum the force vectors. Add all forces pointing toward the center of the circle, and then subtract all forces
pointing away from the center.
Fc  = + Ftoward center  - Faway from center
4. Substitute and solve. In this last step, substitute known equations for specific forces and substitute numerical
values to find in the solution.
Keep in mind that circular motion problems can incorporate elements from problems learned in other chapters.

Horizontal Circular Motion
A car moving at 10 meters/second completes a turn with a radius of 20 meters. Determine the minimum coefficient of friction between the tires and the road that will allow the car to complete the turn without skidding.

WHAT'S THE THICK?

Orient the problem: Picture a car making a turn. Gravity is pulling down, and the normal force is acting upward. Neither of these forces is in the direction of motion. The force acting to keep the car in the turn is friction. Without friction acting toward the center of the turn, the car would slide in a straight path following the tangential velocity.
Determine the type of motion: A turning car is in circular motion.
Sum the force vectors in the relevant direction: In circular motion Fc is used instead of and friction is pointing toward the center of the circle.
Fc = f
Substitute and solve: The normal force is acting vertically in the y-direction. It is equal and opposite the force of gravity: N = Fg = mg. Vertical Circular Motion
A 2.0-kilogram mass is attached to the end of a 1.0-meter-long string. When the apparatus is swung in a vertical circle, the mass reaches a speed of 10 meters per second at the bottom of the swing.
(A) Determine the tension in the string at the bottom of the swing.

WHAT'S THE THICK?

Orient the problem: Sketch the scenario, including force vectors. Determine the type of motion: This is circular motion.
Sum the force vectors in the relevant direction: In circular motion, Fc is used instead of . Tension is pointing to the center of the circle and is positive. The force of gravity is pointing away from the center of the circle and is negative.
Fc = T - Fg
T = Fc + Fg
Substitute and solve: (B) Determine the minimum speed at the top of the loop that allows the mass to make one complete cycle.

WHAT'S THE THICK?

The minimum speed at the top of a loop indicates a special case where the forces contributing to the overall centripetal force must be at an absolute minimum. Begin by solving the problem the same as any other force problem.

Orient the problem: Sketch the scenario, including force vectors. Determine the type of motion: This is circular motion.
Sum the force vectors in the relevant direction: Tension and gravity are both pointing toward the center of the circle. Both are set as positive.
Fc = T + Fg
Now you must consider how to make the centripetal force, Fc, as small as pos­sible. The force of gravity cannot be altered. However, tension can be reduced by swinging the mass more slowly. When the mass is slowed to the point where it is just barely completing a full circle, the tension will be zero, T = 0, at the exact instant that the mass is at the very top of the circle.
Substitute and solve. 