1. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 5: Uniform Circular Motion Chapter Goal: To learn how to solve problems about uniform circular motion. Uniform circular motion: the motion of an object traveling at a constant speed on a circular path.

2. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Uniform Circular Motion The velocity vector shows the direction of motion at any point on the circle. For uniform circular motion the velocity vectors at each point are tangent to the circle and are the same length Period (T) – the time it takes to make one revolution around the circle. T is in seconds, v is in m/s.

3. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example Problem Your roommate is working on his bicycle. The bicycle wheel has a radius of 0.3 meters. As he spins the bicycle wheel, you notice that a stone is stuck in the tread. It goes by 3 times every second. a. What is the period of the stone? b.What is the speed of the stone? Hint: the stone makes 3 revolutions per second (rps). The period is how many seconds it takes to make one revolution.

4. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example Problem - Answer Your roommate is working on his bicycle. The bicycle wheel has a radius of 0.3 meters. As he spins the bicycle wheel, you notice that a stone is stuck in the tread. It goes by 3 times every second. a. What is the period of the stone? T = 1/3 rps =.333s b.What is the speed of the stone? v = 2πr/T = 2π (.3m)/(.333s) = 5.7m/s

5. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Determining the direction of acceleration for uniform circular motion using vector subtraction Draw a velocity vector on top of each dot, tangent to the circle. Starting with 0, take 2 adjacent velocity vectors and do a graphical vector subtraction to find the direction of ∆v and therefore.

6. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Acceleration of Uniform Circular motion For uniform circular motion, the acceleration always points towards the center of the circle. This is called centripetal (“center- seeking”) acceleration. Is centripetal acceleration equal to ∆v/ ∆t? Yes, but if the magnitude of v remains constant, how do we evaluate ∆v ? Using trig, it can be shown that a c = ∆v/ ∆t = v 2 /r.

7. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Rank in order, from largest to smallest, the centripetal accelerations (a r ) a to (a r ) e of particles a to e.

8. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. (a r ) b > (a r ) e > (a r ) a = (a r ) c > (a r ) d Rank in order, from largest to smallest, the centripetal accelerations (a r ) a to (a r ) e of particles a to e.

9. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Determining the direction of acceleration for uniform circular motion using Newton’s 2 nd Law ΣF = ma tells us the net force vector and the acceleration vector are in the same direction (since mass is a scalar). ball on a string ball around a track

10. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. FBD for Uniform Circular Motion Problems – The radial/vertical (rz) coordinate system r axis – similar to a true horizontal x axis, except positive is always towards the center of the circle, regardless of whether it is left or right. z-axis – similar to a true vertical y axis. velocity is tangent to the circle (perpendicular to both r and z axes. Note: if circular motion is vertical, the r axis and z axis are the same!

11. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. How to analyze motion problems for uniform circular motion, using Newton’s 2 nd Law ΣF = ma If the object is in uniform circular motion then acceleration is a centripetal acceleration, a c : –|a c | = v 2 /r –direction is towards the center of the circle (positive r axis). –|ΣF| = ma c = m v 2 /r –ΣF must be in the positive r direction.

12. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Car turning a “circular” corner at constant speed Is there a net force on the car as it negotiates the turn? Where did it come from?

13. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Car turning a “circular” corner at constant speed When the driver turns the wheel the tires turn. To continue along a straight line, the car must overcome static friction and slide. If the static friction force is less than the maximum, the tire cannot slide and so has no choice but to roll in the direction of the turn. f s-max = μ s |n|

14. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Car turning a “circular” corner at constant speed – example problem What is the maximum speed at which a 1500 kg car can make a turn around a curve of radius 50 m on a level road without sliding out of the turn (skidding)? Recall toward the center of the circle is positive r, even if it is to the left in this example

15. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Car turning a “circular” corner at constant speed – example problem What is the maximum speed at which a 1500 kg car can make a turn around a curve of radius 50 m on a level road without sliding out of the turn (skidding)? 1. draw a free body diagram with appropriate axes and lists knowns and “finds”

16. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Car turning a “circular” corner at constant speed – example problem 2.Newton’s 2 nd Law in component form. Start with z axis since there is no acceleration: ΣF z = ma z = 0 n – F G = 0, therefore n = mg

17. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Car turning a “circular” corner at constant speed – example problem 2.Newton’s 2 nd Law in component form. Now do forces in the r direction: 3.ΣF r = ma c = mv 2 /r. The only force in the r direction is the frictional force, therefore: f s = mv 2 /r v max occurs at f s-max. Therefore: μ s |n| = mv 2 /r or μ s mg = mv 2 /r

18. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Car turning a “circular” corner at constant speed – example problem Note the mass term drops out, as all forces depend on the mass. Solve for v max : v = (u s rg) -1/2 = 22 m/s, about 45 mph Slow down!