1. PHY 231 1 PHYSICS 231 Lecture 15: Rotations Remco Zegers Walk-in hour Tue 4-5 pm helproom

2. PHY 231 2 Types of collisions Inelastic collisionsElastic collisions Momentum is conserved Some energy is lost in the collision: KE not conserved Perfectly inelastic: the objects stick together. Momentum is conserved No energy is lost in the collision: KE conserved Conservation of momentum: m 1 v 1i +m 2 v 2i =m 1 v 1f +m 2 v 2f Conservation of KE: ½m 1 v 1i 2 +½m 2 v 2i 2 =½m 1 v 1f 2 +½m 2 v 2f 2 (v 1i -v 2i )=(v 2f -v 1f ) Conservation of momentum: m 1 v 1i +m 2 v 2i =(m 1 +m 2 )v f Chapter 6 in one slide Momentum p=mv F=  p/  t Impulse (the change in momentum)  p= F  t

3. PHY 231 3 conceptual M1M1 M2M2 Given M 1 >M 2, and M 2 initially at rest. Just after the collision: M 1 a) moves left b) moves right c) is at rest. The velocity of M 2 will be a) smaller than b) larger than c) the same as the new velocity of M 1. M 2 compresses the spring. The maximum potential energy stored in the spring will be a) equal to b) smaller than c) larger than the kinetic energy of M 2 just after the collision. The maximum potential energy stored in the spring will be a) equal to b) smaller than c) larger than the kinetic energy of M 1 before the collision.

4. PHY 231 4 elastic collision of unequal masses Given m 2 =3m 1. What is the velocity of m 1 and m 2 after the collision in terms of the initial velocity of the moving bullet if a) m 1 is originally at rest b) m 2 is originally at rest A) m 1 v 1i +m 2 v 2i =m 1 v 1f +m 2 v 2f 3m 1 v 2i =m 1 v 1f +m 2 v 2f 3v 2i =v 1f +3v 2f (v 1i -v 2i )=(v 2f -v 1f ) -v 2i = v 2f -v 1f B) m 1 v 1i +m 2 v 2i =m 1 v 1f +m 2 v 2f m 1 v 1i =m 1 v 1f +m 2 v 2f v 1i =v 1f +3v 2f (v 1i -v 2i )=(v 2f -v 1f ) v 1i = v 2f -v 1f v 2f =v 2i /2 v 1f =3v 2i /2 v 2f =v 1i /2 v 1f =-v 1i /2

5. PHY 231 5 another one 1 2 A train wagon (1) is pushed off a hill with height h and couples with wagon (2) which has the same mass (m). Can the combined system move up to the next hill of which the height is half of the height of the first hill? a) Yes b) No c) Don’t know h h/2 On the first hill: Potential energy is mgh. If the coupled train makes it up to the next hill: Pot. E: 2mgh/2=mgh. This is possible only if no energy gets lost in collision. However, collision is perfectly inelastic: no conservation of KE! So, not possible

6. PHY 231 6 quiz: extra credit The orange block collides with the blue block (same mass). Which of the following is not true? a)If the collision is elastic, the orange block will stop moving after the collision. b)If the collision is (perfectly) inelastic, the spring will be compressed less than if the collision is elastic c)If the collision is (perfectly) inelastic, the maximal potential energy stored in the spring equals the kinetic energy of the orange block before the collision

7. PHY 231 7 Radians & Radius  r Circumference=2  r s Part s=  r  in radians! 360 o =2  rad = 6.28… rad  (rad) =  /180 o  (deg)  =s/r

8. PHY 231 8 Angular speed and acceleration Average angular velocity Instantaneous Angular velocity Angular velocity : rad/s rev/s rpm (revolutions per minute)

9. PHY 231 9 example What is the angular velocity of earth around the sun? Give in rad/s, rev/s and rpm Answer: in 1 year, the earth makes one full orbit around the sun. = 2  rad/1 year 1 rev/1 year 1 rev/1 year = 2  rad/(3.2E+7 s) 1 rev/(3.2E+7 s) 1 rev/(5.3E+5 min) = 2.0E-07 rad/s 3.1E-8 rev/s 1.9E-6 rpm rad/s rev/s rpm

10. PHY 231 10 Angular acceleration Definition: The change in angular velocity per time unit Average angular acceleration Instantaneous angular acceleration Unit: rad/s 2

11. PHY 231 11 Equations of motion Linear motion Angular motion X(t)=x(0)+v(0)t+½at 2 V(t)=V(0)+at  (t)=  (0)+  (0)t+½  t 2  (t)=  (0)+  t

12. PHY 231 12 example  =0 A person is rotating a wheel. The handle is initially at  =90 o. For 5s the wheel gets an constant angular acceleration of 1 rad/s 2. After that The angular velocity is constant. Through what angle will the wheel have rotated after 10s.  t First 5s:  (5)=  (0)+  (0)t+½  t 2 =  /2+ 0 + ½1(5) 2 =  /2+12.5=14.1 rad (=806 o )  (5)=  (0)+  t = 1t= 5 rad/s Next 5s:  (5)= 14.1+5t = 39.1 rad (=2240 0 =6.2 rev)

13. PHY 231 13 question An object is rotating over a circle with radius of 2 m. Its speed is 1.5  rad/s. After 4 seconds, how many rotations did the object make? a)2 b)3 c)4 d)8 e)12  (t)=  (0)+  (0)t+½  t 2 = 0+1.5  t+0 = 6  rad 2  rad is 1 rotation (360 o ), so 3 rotations

14. PHY 231 14 AngularLinear velocities V V is called the tangential velocity or the linear velocity.

15. PHY 231 15 Angularlinear acceleration velocity Change in velocity Change in velocity per time unit acceleration The linear acceleration equals the angular acceleration times the radius of the orbit

16. PHY 231 16 example a b 5m 6m The angular velocity of a is 2 rad/s. a) What is its tangential velocity? If b is keeping pace with a, b) What is its angular velocity? c) What is its linear velocity? a) v=  r = 2*6=12 m/s 12 m/s b) Must be the same: 2 rad/s  c) v=  r = 2*5=10 m/s 10 m/s

17. PHY 231 17 A rolling coin d  0 =18 rad/s  =-1.80 rad/s a)For how long does the coin roll? b)What is the average angular velocity? c) How far (l)does the coin roll before coming to rest? l a)  (t)=  (0)+  t 0=18-1.8t t=10 s b)  = (  (0)+  (10))/2=18/2=9 rad/s d=0.02 m c) v=  r = 9*0.01=0.09 m/s l=vt=0.09*10=0.9 m

18. PHY 231 18 question What is the angular speed about the rotational axis of the earth for a person standing on the surface? a)7.3x10 -5 rad/s b)3.6x10 -5 rad/s c)6.28x10 -5 rad/s d)3.14x10 -5 rad/s e) ????

19. PHY 231 19 question peddles rear wheel front gear rear gear A boy is riding his bike. If the front gear is rotating with linear velocity v 1 and angular velocity  1 then the rear gear has a linear velocity ??????? v 1 and angular velocity ???????  1. The front gear is larger in radius than the rear gear. a)equal to, equal to b)equal to, larger than c)larger than, larger than d)smaller than, larger than e)smaller than, equal to

20. PHY 231 20 gears r 1 =0.3 r 2 =0.15 r 3 =0.7 m If  1 =3 rad/s, how fast Is the bike going? b) What if r 1 =0.1 m ? V 1 =  1 r 1 =3*0.3=0.9 m/s V 1 =V 2 (because of the chain)  2 =V 2 /r 2 =0.9/0.15=6 rad/s  3 =  2 (connected) V 3 =  3 r 3 =6*0.7=4.2 m/s b) V 1 =  1 r 1 =3*0.1=0.3 m/s V 1 =V 2 (because of the chain)  2 =V 2 /r 2 =0.3/0.15=2 rad/s  3 =  2 (connected) V 3 =  3 r 3 =2*0.7=1.4 m/s