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6.2Section

6.2 Circular MotionCircular Motion

● Explain why an object moving in a circle at a constant speed is accelerated.

● Describe how centripetal acceleration depends upon the object’s speed and the radius of the circle.

● Identify the force that causes centripetal acceleration.

In this section you will:

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6.2Section

6.2 Circular MotionCircular Motion

Suppose that you were driving a car with the steering wheel turned in such a manner that your car followed the path of a perfect circle with a constant radius. And suppose that as you drove, your speedometer maintained a constant reading of 10 mi/hr. In such a situation as this, the motion of your car could be described as experiencing uniform circular motion.

Uniform circular motion is the motion of an object in a circle with a constant or uniform speed. 

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6.2Section

6.2 Circular MotionCircular Motion

The distance of one complete cycle around the perimeter of a circle is known as the circumference.

With a uniform speed of 5 m/s, a car could make a complete cycle around a circle that had a circumference of 5 meters. At this uniform speed of 5 m/s, each cycle around the 5-m circumference circle would require 1 second. At 5 m/s, a circle with a circumference of 20 meters could be made in 4 seconds; and at this uniform speed, every cycle around the 20-m circumference of the circle would take the same time period of 4 seconds

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6.2Section

6.2 Circular MotionCircular Motion

Circumference = 2*pi*Radius

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6.2Section

6.2 Circular MotionCircular Motion

The Direction of the Velocity VectorObjects moving in uniform circular motion will have a constant speed. But does this mean that they will have a constant velocity?

The direction of the velocity vector is directed in the same direction that the object moves. Since an object is moving in a circle, its direction is continuously changing.

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6.2Section

6.2 Circular MotionCircular Motion

AccelerationAs mentioned, an object moving in uniform circular motion is moving in a circle with a uniform or constant speed. The velocity vector is constant in magnitude but changing in direction. Therefore the object is accelerating in a circular motion path.

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6.2Section

6.2 Circular MotionCircular Motion

The Centripetal Force RequirementAs mentioned earlier, an object moving in a circle is experiencing an acceleration. Even if moving around the perimeter of the circle with a constant speed, there is still a change in velocity and subsequently an acceleration.

This acceleration is directed towards the center of the circle. And in accord with Newton's second law of motion, an object which experiences an acceleration must also be experiencing a net force. The direction of the net force is in the same direction as the acceleration. So for an object moving in a circle, there must be an inward force acting upon it in order to cause its inward acceleration. This is sometimes referred to as the centripetal force requirement.

The word centripetal (not to be confused with the F-word centrifugal) means center seeking. For object's moving in circular motion, there is a net force acting towards the center which causes the object to seek the center.

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6.2Section

6.2 Circular MotionCircular Motion

The acceleration of an object moving in a circle is always in the direction of the net force acting on it, there must be a net force toward the center of the circle. This force can be provided by any number of agents.

When an Olympic hammer thrower swings the hammer, the force is the tension in the chain attached to the massive ball.

Centripetal Acceleration

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6.2Section

6.2 Circular MotionCircular Motion

When an object moves in a circle, the net force toward the center of the circle is called the centripetal force.

Centripetal Acceleration

To analyze centripetal acceleration situations accurately, you must identify the agent of the force that causes the acceleration. Then you can apply Newton’s second law for the component in the direction of the acceleration in the following way.

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6.2Section

6.2 Circular MotionCircular Motion

Centripetal Acceleration

The net centripetal force on an object moving in a circle is equal to the object’s mass times the centripetal acceleration.

Newton’s Second Law for Circular Motion

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6.2Section

6.2 Circular MotionCircular Motion

When solving problems, it is useful to choose a coordinate system with one axis in the direction of the acceleration.

For circular motion, the direction of the acceleration is always toward the center of the circle.

Centripetal Acceleration

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6.2Section

6.2 Circular MotionCircular Motion

Rather than labeling this axis x or y, call it c, for centripetal acceleration. The other axis is in the direction of the velocity, tangent to the circle. It is labeled tang for tangential.

Centripetal force is just another name for the net force in the centripetal direction. It is the sum of all the real forces, those for which you can identify agents that act along the centripetal axis.

Centripetal Acceleration

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6.2Section

6.2 Circular MotionCircular Motion

According to Newton’s first law, you will continue moving with the same velocity unless there is a net force acting on you.

A Nonexistent Force

The passenger in the car would continue to move straight ahead if it were not for the force of the car acting in the direction of the acceleration.

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6.2Section

6.2 Circular MotionCircular Motion

The so-called centrifugal, or outward force, is a fictitious, nonexistent force.

A Nonexistent Force

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6.2 Section CheckSection Check

Explain why an object moving in a circle at a constant speed is accelerating.

Question 1

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6.2Section

6.2 Section CheckSection Check

Acceleration is the rate of change of velocity, the object is accelerating due to its constant change in the direction of its motion.

Answer 1

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6.2 Section CheckSection Check

What is the relationship between the magnitude of centripetal acceleration (ac) and an object’s speed (v)?

Question 2

A.

B.

C.

D.

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6.2 Section CheckSection Check

Reason: From the equation for centripetal acceleration:

Centripetal acceleration always points to the center of the circle. Its magnitude is equal to the square of the speed divided by the radius of the motion.

Answer 2

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6.2 Section CheckSection Check

What is the direction of the velocity vector of an accelerating object?

Question 3

A. toward the center of the circle

B. away from the center of the circle

C. along the circular path

D. tangent to the circular path

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6.2 Section CheckSection Check

Answer 3

Reason: While constantly changing, the velocity vector for an object in uniform circular motion is always tangent to the circle. Vectors are never curved and therefore cannot be along a circular path.

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6.2Section

6.2 Section CheckSection Check

Known Information:

m = 900 kg

v = 10.0 m/s

R = 25.0 m

Requested Information:

a = ????

Fnet = ????

Sample Problem #1A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.

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6.2Section

6.2 Section CheckSection Check

To determine the acceleration of the car, use the equation a = v2 / R.

The solution is as follows:a = v2 / R a = (10.0 m/s)2 / (25.0 m)a = (100 m2/s2) / (25.0 m)a = 4 m/s2

To determine the net force acting upon the car, use the equation Fnet = m•a.

The solution is as follows.Fnet = m • a

Fnet = (900 kg) • (4 m/s2)

Fnet = 3600 N

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6.2Section

6.2 Section CheckSection Check

Sample Problem #2A 95-kg halfback makes a turn on the football field. The halfback sweeps

out a path that is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds.

Determine the speed, acceleration and net force acting upon the halfback.

Known Information:

m = 95.0 kg

R = 12.0 m

Traveled 1/4-th of the circumference in 2.1 s

Requested Information:

v = ????

a = ????

Fnet = ????

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6.2Section

6.2 Section CheckSection CheckTo determine the speed of the halfback, use the equation v = d / t where the d is one-fourth of the circumference and the time is 2.1 s. The solution is as follows:

v = d / t v = (0.25 • 2 • pi • R) / tv = (0.25 • 2 • 3.14 • 12.0 m) / (2.1 s)v = 8.97 m/s

To determine the acceleration of the halfback, use the equation a = v2 / R. The solution is as follows:

a = v2 / R a = (8.97 m/s)2 / (12.0 m)a = (80.5 m2/s2) / (12.0 m)a = 6.71 m/s2

To determine the net force acting upon the halfback, use the equation Fnet = m•a.

The solution is as follows.Fnet = m*a

Fnet = (95.0 kg)*(6.71 m/s2)

Fnet = 637 N