1. CHAPTER 7 Rotational Motion and the Law of Gravity Angular Speed and Angular Acceleration s = arc length Θ = arc angle (radians) Θ = s ; where r = radius r

2. TermSymbolUnits Angular Displacement ∆Θ radians Angular Speed ωradians/sec (Average) Angular Acceleration  radians/sec 2 Average Angular Speed ω radians/sec

3. ω = Θ 2 – Θ 1 t 2 – t 1 ω = ∆Θ ∆t Average Angular Speed Average rate at which the arc angle is changing ω = lim ∆Θ ∆t ∆t  0 (Instantaneous) Angular Speed The rate at which the arc angle is changing at a particular instant in time  = ω 2 – ω 1 t 2 – t 1  = ∆ ω ∆t Average Angular Acceleration The average rate at which the angular speed is changing NOTE: You are not required to work with changing rates of acceleration. Therefore,  = 

4. Linear/ Angular Comparison TermLinearAngular Displacement ∆x m ∆Θ rad Velocity v m/s ω rad/sec Acceleration a m/s 2  rad/sec 2 Formulas: ∆x = ½(v o +v)t ∆x = v o t + ½at 2 v = v o + at v 2 = v o 2 + 2a∆x N/A ∆Θ = ½(ω o +ω)t ∆Θ = ω o t + ½  t 2 ω = ω o +  t ω 2 = ω o 2 + 2  ∆Θ s = rΘ NOTE: All points on a rotating disc have the same values for ω and .

5. Example Problem Example Problem (Rotating Disc) A disc of radius 2.50m accelerates to an angular velocity of 1.35 radians/sec in 12.5 seconds. Calculate (a) the disc’s angular acceleration; (b) the number of revolutions completed during the acceleration; (c) the distance traveled by a point on the edge of the disc. Identify the variables ω o =  = ω = t = ∆Θ = Choose Appropriate Formulas ω = ω o +  t  =.108 rad/sec 2 [ ] ∆Θ = t ω o + ω 2 ∆Θ = ω o t + ½  t 2 ∆Θ = 8.44 rad 1 revolution = 2  rad = 6.2 rad 8.44 rad 1 rev 6.28 rad # rev = 1.34rev ∆s = r∆Θ ∆s = 21.1m 0.00 rad/sec? 1.35 rad/sec12.5 sec ?

6. v t = r ω Tangential Speed Tangential Speed (v t ) v t = ∆s ∆t ∆s = r ∆Θ v t = r ∆Θ ∆t Tangential Acceleration (  t ) a t = r  NOTE: Not every point on a rotating disc has the same tangential speed or tangential acceleration. arc length of the movement time to complete the movement

7. Centripetal Acceleration Centripetal Acceleration (center-seeking) Question: The tangential velocity v t illustrated above is clearly changing. How can you tell it is changing? Answer: Its direction is changing. Its direction is changing. a = ∆v = v f – v i = v f + -v i ∆t ∆t ∆t The illustration on the right clearly shows that ∆v points towards the center of rotation.

8. Centripetal Acceleration Centripetal Acceleration (a c ) ∆v = ∆sfrom similar triangles v r ∆v = v r ∆sand ∆v = a c ∆ta c ∆t = ∆s vrvr a c = v ∆s and ∆s = v r ∆t ∆t a c = = Centripetal Acceleration v 2 r

9. Centripetal Centripetal Force: The net force that causes an object to accelerate towards its center of rotation (i.e. go in a circle) F c = ma c F c is not a new force F c will be either one or a combination of more than one of the following forces: FgFg FNFN FfFf F a (T)

10. Example Problem Example Problem (Vertical loop) A rollercoaster contains a loop 20m in diameter. a) What speed will cause a passenger to feel “weightless” at the top of the loop? b) If this same person has a mass of 90kg what weight will they experience at the bottom of the loop? Sketch the Force Diagrams Analysis: Analysis: The person is “weightless” if no forces other than gravity are acting upon them (i.e. F N =0). Hence, F g is F c Analysis: Analysis: The only forces acting on the person are the normal force and force of gravity. F c = F net(center) = F N – F g F N = F c + F g = mv 2 + mg r F N = 1764 Newtons mg = mv 2 r v = √rg v = 9.9m/s

11. Example Problem Example Problem (Banked Curve) a) For a car traveling with a speed v around a curve of radius r, determine the formula for the angle at which the road should be bank so that no friction is required. b) what is this angle for an expressway off-ramp of radius 50m at a design speed of 50km/hr? Sketch the Force Diagram Part a:Analysis Since there is no friction, there is no component of friction directed to the center (right). Only the component F N sinΘ of the normal force is directed to the center. F N sinΘ is the net force towards the center = F c

12. F N sinΘ = F c F N sinΘ = mv 2 r mg = F N cosΘ mg · sin Θ = mv 2 cosΘ r mg tanΘ = mv 2 r tanΘ = v 2 gr Θ = tan -1 ( ) v 2 gr Part b: r = 50m v = 50km 1000m 1hr hr km 3600s = 14m/s g = 9.8 Θ = 22°

13. Newton’s Universal Law of Gravitation Every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. F g = G m 1 m 2 r 2 G = universal gravitational constant G = 6.67 x 10 -11 N ·m 2 /kg 2

14. Force of Earth’s Gravity Force of Earth’s Gravity (on Earth’s Surface) F g = G m earth m object r earth 2 F g = · mass object [ ] G m earth r earth 2 F g = · mass object (6.67 x10 -11 N·m 2 /kg 2 )(5.98x10 24 kg) (6.37x10 6 m) 2 F g = 9.8 x mass object F g = m obect g

15. Velocity of an Earth Satellite Analysis: What is the only Force acting on the satellite?Answer: FgFg If satellite is in circular orbit, then… F g = F c mg = mv s 2 v s = velocity of satellite r v s = √grg = GM earth r 2 v s = GM earth · r r 2 √ v s = GM earth r √

16. Gravitational Potential Energy PE = -GM E m r r = distance of object from Earth’s center from Calculus PE = zero at infinite distance from Earth PE = some negative value closer to the Earth Escape Speed Speed at surface of Earth where the object’s KE 1 + PE 1 is sufficient to reach infinite distance from Earth (PE 2 =0J) with a final speed of 0m/s (KE 2 =0J) From Conservation of Energy: KE 1 + PE 1 = KE 2 + PE 1 mv 2 + -GM earth m 2 R earth = 0 + 0 mv 2 + -GM earth m 2 R earth v escape = [ ] 1/2 2GM earth R earth