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Circular Motion & Rotation Chapter 7: Circular Motion & Rotation “I shall now recall to mind that the motion of the heavenly bodies is circular, since the motion appropriate to a sphere is rotation in a circle.” — Nicolaus Copernicus Objectives 1. Explain the acceleration of an object moving in a circle at con- stant speed. 2. Define centripetal force and recognize that it is not a special kind of force, but that it is provided by forces such as tension, gravity, and friction. 3. Solve problems involving calculations of centripetal force. 4. Calculate the period, frequency, speed and distance traveled for objects moving in circles at constant speed. 5. Differentiate between translational and rotational motion of an object. 6. Describe the rotational motion of an object in terms of rotational position, velocity, and acceleration. 7. Use rotational kinematic equations to solve problems for objects rotating at constant acceleration. 8. Utilize the definitions of torque and Newton’s 2nd Law for Ro- tational Motion to solve static equilibrium problems. 9. Apply principles of angular momentum, torque, and rotational dynamics to analyze a variety of situations. 10. Explain what is meant by conservation of angular momentum. 11. Calculate the rotational kinetic energy and total kinetic energy of a rotating object moving through space. Chapter 7: Circular Motion & Rotation 163 Now that you’ve talked about linear and projectile kinemat- ics, as well as fundamentals of dynamics and Newton’s Laws, you have the skills and background to analyze circular motion. Of course, this has obvious applica- tions such as cars moving around a circular track, roller coasters moving in loops, and toy trains chug- ging around a circular track under the Christmas tree. Less obvious, however, is the application to turning objects. Any object that is turning can be thought of as moving through a por- tion of a circular path, even if it’s just a small portion of that path. With this realization, analysis of circular mo- tion will allow you to explore a car speeding around a corner on an icy road, a running back cutting upfield to avoid a blitzing linebacker, and the orbits of planetary bodies. The key to understand- ing all of these phenomena starts with the study of uniform circular motion. Uniform Circular Motion The motion of an object in a circular path at constant speed is known as uniform circular motion (UCM). An object in UCM is constantly chang- ing direction, and since velocity is a vector and has direction, you could say that an object undergoing UCM has a constantly changing velocity, even if its speed remains constant. And if the velocity of an object is changing, it must be accelerating. Therefore, an object undergoing UCM is constantly ac- celerating. This type of acceleration is known as centripetal acceleration. 7.01 Q: If a car is accelerating, is its speed increasing? 7.01 A: It depends. Its speed could be increasing, or it could be acceler- ating in a direction opposite its velocity (slowing down). Or, its speed could remain constant yet still be accelerating if it is trav- eling in uniform circular motion. Just as importantly, you’ll need to figure out the direction v of the object’s acceleration, since acceleration is a vec- tor. To do this, draw an object moving counter-clock- v0 wise in a circular path, and show its velocity vector at two different points in time. Since acceleration is the rate of change of an object’s velocity with respect to time, you can determine the direction of the object’s acceleration by finding the direction of its change in velocity, Δv. 164 Chapter 7: Circular Motion & Rotation To find its change in velocity, Δv, recall that Δv = v − v0 . v v0 Therefore, you can find the difference of the vectors v and v0 graphically, which can be re-written as Δv = v + (−v0 ) . v -v0 Recall that to add vectors graphically, you line them up tip-to-tail, then draw the resultant vector from the starting point (tail) of the first vector to the ending point (tip) of the last vector. v -v0 a So, the acceleration vector must point in the direction shown above. If this vector is shown back on the original circle, lined up directly between the initial and final velocity vector, it’s easy to see that the acceleration vector points toward the center of the circle. v v0 ac You can repeat this procedure from any point on the circle... no matter where you go, the acceleration vector always points toward the center of the circle. In fact, the word centripetal in centripetal acceleration means “center-seeking!” So now that you know the direction of an object’s acceleration (toward the center of the circle), what about its magnitude? The formula for the magni- tude of an object’s centripetal acceleration is given by: v2 a = c r Chapter 7: Circular Motion & Rotation 165 7.02 Q: In the diagram below, a cart travels clockwise at constant speed in a horizontal circle. At the position shown in the diagram, which arrow indicates the direction of the centripetal acceleration of the cart? (A) A (B) B (C) C (D) D 7.02 A: (A) The acceleration of any object moving in a circular path is toward the center of the circle. 7.03 Q: The diagram shows the top view of a 65-kilogram student at point A on an amusement park ride. The ride spins the student in a horizontal circle of radius 2.5 meters, at a constant speed of 8.6 meters per second. The floor is lowered and the student remains against the wall without falling to the floor. Which vector best represents the direction of the centripetal ac- celeration of the student at point A? 7.03 A: (1) Centripetal acceleration points toward the center of the circle. 166 Chapter 7: Circular Motion & Rotation 7.04 Q: Which graph best represents the relationship between the mag- nitude of the centripetal acceleration and the speed of an object moving in a circle of constant radius? 7.04 A: (2) Centripetal acceleration is proportional to v2/ r. 7.05 Q: A car rounds a horizontal curve of constant radius at a constant speed. Which diagram best represents the di- rections of both the car’s velocity, v, and acceleration, a? 7.05 A: (3) Velocity is tangent to the circu- lar path, and acceleration is toward the center of the circular path. 7.06 Q: A 0.50-kilogram object moves in a horizontal circular path with a radius of 0.25 meter at a constant speed of 4.0 meters per sec- ond. What is the magnitude of the object’s acceleration? (A) 8 m/s2 (B) 16 m/s2 (C) 32 m/s2 (D) 64 m/s2 7.06 A: (D) 64 m/s2. Circular Speed So how do you find the speed of an object as it travels in a circular path? The formula for speed that you learned in kinematics still applies. d v = t You have to be careful in using this equation, however, to understand that an object traveling in a circular path is traveling along the circumference of a circle. Therefore, if an object were to make one complete revolution around the circle, the distance it travels is equal to the circle’s circumference. Chapter 7: Circular Motion & Rotation 167 Cr= 2π 7.07 Q: Miranda drives her car clockwise around a circular track of radius 30m. She completes 10 laps around the track in 2 minutes. Find Miranda’s total distance traveled, average speed, and centripetal acceleration. 7.07 A: dr=×10 21ππ=×02(30)mm= 1885 d 1885m m v == = 15.7 s t 120s v 22(15.7 m ) s m a == = 8.2 2 c rm30 s Note that her displacement and average velocity are zero. 7.08 Q: The combined mass of a race car and its driver is 600 kilograms. Traveling at constant speed, the car completes one lap around a circular track of radius 160 meters in 36 seconds. Calculate the speed of the car. d 22ππr (160m) m 7.08 A: v == ==27.9 s t t 36s Centripetal Force If an object traveling in a circular path has an inward acceleration, Newton’s 2nd Law states there must be a net force directed toward the center of the circle as well. This type of force, known as a centripetal force, can be a gravitational force, a tension, an applied force, or even a frictional force. NOTE: When dealing with circular motion problems, it is important to realize that a centripetal force isn’t really a new force, a centripetal force is just a label or grouping you apply to a force to indicate its direction is toward the center of a circle. This means that you never want to label a force on a free body diagram as a centripetal force, Fc. Instead, label the center-directed force as specifically as you can. If a tension is causing the force, label the force FT. If a frictional force is causing the center-directed force, label it Ff, and so forth. Because a centripetal force is always perpendicular to the ob- ject’s motion, a centripetal force can do no work on an object. You can combine the equation for centripetal acceleration with Newton’s 2nd Law to obtain Newton’s 2nd Law for Circular Motion. Recall that Newton’s 2nd Law states: Fmnet = a 168 Chapter 7: Circular Motion & Rotation For an object traveling in a circular path, there must be a net (centripetal) force directed toward the center of the circular path to cause a (centripetal) acceleration directed toward the center of the circular path.
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