Last Time : Start Rotational Motion (now thru mid-Nov)

1. Last Time: Start Rotational Motion (now thru mid-Nov) Basics: Angular Speed, Angular Acceleration • Today: Review, Centripetal Acceleration, Newtonian Gravitation • HW #6 due Tuesday, Oct 19, 11:59 p.m. • Exam #2 Thursday, Oct 21 On material through (including) today’s Lecture Tuesday’s (Oct 19) material will not be on Exam #2

2. Rotational Motion Under Constant Angular Acceleration Just like for linear motion under constant acceleration, we have kinematic equations for rotational motion. They are analogous … Linear Motion with Constant a Rotational Motion with Constant α

3. Example: 7.14 + Extra An electric motor rotating a wheel at a rate of 100 rev/min is turned off. Assume the wheel has a constant angular acceleration of –2.00 rad/s2. (a) How long does it take for the wheel to come to a stop ? (b) Through how many radians has the wheel turned during the time interval found in (a) ? Extra: Suppose the wheel has a radius of 0.20 m. At t = 2.0 s, what is the magnitude of the tangential speed and acceleration at a point on the edge of the wheel ?

4. Centripetal Acceleration Consider a car driving in a circle at a constant speed of 25 mph. The speed is constant, but what about the velocity ? • Recall: Velocity is a VECTOR. • The car’s speed is constant, but the velocity vector is constantly changing direction as it moves around the circle ! • If the velocity is changing, there must be an acceleration.

5. Centripetal Acceleration Initial Final Δs Δθ r r Note the direction of Δv Δθ (inwards…) Note: For constant speed, and differ only in their direction. Their magnitudes are the same, vi = vf = v. Recall:

6. Centripetal vs. Tangential Acceleration Magnitude: At all points along the circle, ac points radially inwards ac For circular motion, centripetal acceleration is always present, even if the speed is not changing. It is the acceleration associated with the constant change in direction. Recall: If the speed is changing, there is a tangential acceleration. Magnitude: vt at vt at

7. Circular Motion: Total Acceleration + at ac at Centripetal Tangential Always exists for circular motion For circular motion, only exists if speed is changing Magnitude of Total Acceleration for Circular Motion Direction of Total Acceleration Vector for Circular Motion Must do vector addition of ac and at

8. Angular Quantities are VECTORS Technically, the angular speed of rotation is a vector: ω . The direction is given by the “Right Hand Rule”. Point your thumb so that your fingers wrap around in the direction of motion. Your thumb then points in the direction of ω .

9. What Causes Centripetal Acceleration ? Well, we have Newton’s Second Law of Motion: If an object is moving in a circle, it is undergoing centripetal acceleration, and so this must have been caused by a force ! This force is called the “centripetal force”. [ NOT “centrifugal force”. Take PHY 404 to learn about that … ] Magnitude [SI: Newtons] Direction: The centripetal force is always directed radially inwards, just like the centripetal acceleration ! It is perpendicular to the velocity. v Fc v What happens if the centripetal force “disappears” ?

10. Example: Car in a Circular Turn A car travels at a constant speed of 13.4 m/s (30 mph) on a level circular turn of radius 50.0 m. What minimum coefficient of static friction, μs, between the tires and roadway will allow the car to make the turn without sliding ?

11. Example: Roller Coaster ! A roller-coaster moves in a circular loop of radius R. (a) What speed must the coaster have so that it will just make it over the top of the track without any assistance from the track? (b) What speed will the coaster have at the bottom ? (c) What will be the normal force on a passenger at the bottom of the loop if R = 10.0 m ? (This is your “perceived weight”.)

12. Newtonian Gravitation Q: Is the same force responsible ? A: Yes ! Newton’s Law of Universal Gravitation If two particles with masses m1 and m2 are separated by a distance r, a gravitational force F acts along a line joining them, with magnitude G = 6.673  10−11 kg−1 m3 s−2 [ “gravitational constant” ] Force is always attractive

13. Gravitational Force m2 m1 F21 F12 attractive gravitational force The Third Law. Equal but opposite directions. Action/Reaction pair. 1 F21 = –F12 Magnitudes are equal, calculated with Newton’s Law of Universal Gravitation. 2 Gauss’s Law: The gravitational force exerted by a uniform sphere on a mass located outside the sphere is the same as if the entire mass of the sphere were concentrated at its center. 3

14. Gravitational Force on the Earth’s Surface What is the gravitational force on a mass m on the surface of the Earth ? m RE RE = 6.38  106 m ME = 5.98  1024 kg ME By Gauss’s Law: What is the acceleration ? m RE ME Plug in the numbers … a = 9.80 m/s2 = g !!

15. What if Above the Surface of the Earth ? m 1000 km Example: A mass m is 1000 km above the surface of the Earth. What is its acceleration ? RE = 6.38  106 m RE ME = 5.98  1024 kg ME By Gauss’s Law: m RE + 1000 km a = 7.33 m/s2 < g !! ME F = mg only applicable “near” the surface of the Earth !!

16. Next Class • 7.5 – 7.6 : Gravitational Potential Energy (re-visited), Kepler’s Laws of Planetary Motion