1. PHY 231 1 PHYSICS 231 Lecture 18: equilibrium & revision Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom

2. PHY 231 2 gravitation Only if an object is near the surface of earth one can use: F gravity =mg with g=9.81 m/s 2 In all other cases: F gravity = GM object M planet /r 2 with G=6.67E-11 Nm 2 /kg 2 This will lead to F=mg but g not equal to 9.8 m/s 2 (see Previous lecture!) If an object is orbiting the planet: F gravity =ma c =mv 2 /r=m  2 r with v: linear velocity  =angular vel. So: GM object M planet /r 2 = mv 2 /r=m  2 r Kepler’s 3 rd law: T 2 =K s r 3 K s =2.97E-19 s 2 /m 3 r: radius of planet T: period(time to make one rotation) of planet Our solar system!

3. PHY 231 3 Previously Translational equilibrium:  F=ma=0 The center of gravity does not move! Rotational equilibrium:   =0 The object does not rotate Mechanical equilibrium:  F=ma=0 &   =0 No movement! Torque:  =Fd Center of Gravity:

4. PHY 231 4 examples: A lot more in the book! Where is the center of gravity?

5. PHY 231 5 Weight of board: w What is the tension in each of the wires (in terms of w)? w T1T1 T2T2 0 Translational equilibrium  F=ma=0 T 1 +T 2 -w=0 so T 1 =w-T 2 Rotational equilibrium   =0 T 1 0-0.5*w+0.75*T 2 =0 T 2 =0.5/0.75*w=2/3w T 1 =1/3w T 2 =2/3w

6. PHY 231 6  s =0.5 coef of friction between the wall and the 4.0 meter bar (weight w). What is the minimum x where you can hang a weight w for which the bar does not slide? w snsn n w T TyTy TxTx (x=0,y=0) Translational equilibrium (Hor.)  F x =ma=0 n-T x =n-Tcos37 o =0 so n=Tcos37 o Translational equilibruim (vert.)  F y =ma=0  s n-w-w+T y =0  s n-2w+Tsin37 o =0  s Tcos37 0 -2w+Tsin37 0 =0 1.00T=2w Rotational equilibrium:   =0 xw+2w-4Tsin37 0 =0 so w(x+2-4.8)=0 x=2.8 m

7. PHY 231 7 Tips for study Look through the lecture sheets and pick out the summaries to get a good overview Make an overview for yourself (about 1 Letter size paper) Read the chapters in the book to make sure that your overview contains all the main issues. Study the examples given in the lectures Study the problems in LON-CAPA Study the worked-out examples in the book Practice the previous midterm exams. Practice problems from the book

8. PHY 231 8 Revision: chapter 5  Work: W=Fcos(  )  x Energy transfer  Power: P=W/  tRate of energy transfer  Potential energy (PE) Energy associated with position.  Gravitational PE: mgh Energy associated with position in grav. field.  PE stored in a spring: 1/2kx 2 x is the compression of the spring k is the spring constant  Kinetic energy KE: 1/2mv 2 Energy associated with motion  Conservative force: Work done does not depend on path  Non-conservative force: Work done does depend on path  Mechanical energy ME: ME=KE+PE  Conserved if only conservative forces are present KE i +PE i =KE f +PE f  Not conserved in the presence of non-conservative forces (KE i +PE i )-(KE f +PE f )=W nc

9. PHY 231 9 example k=100 N/m m=1 kg 1 cm A pendulum is pushed with initial velocity 0.1 m/s from a height of 1 cm. How far does it compress the spring? (assume m does not rise significantly after hitting the spring) Conservation of ME: (mgh+1/2mv 2 +1/2kx 2 ) initial = (mgh+1/2mv 2 +1/2kx 2 ) final 1*9.8*0.01+0.5*1*0.1 2 +0.=0.+0.+0.5*100*x 2 so x=0.045 m

10. PHY 231 10 Saving electricity A ‘smart’ student decides to save energy by connecting his exercise treadmill to his laptop battery. If it takes 70 J to move the belt on the treadmill by 1 meter and 50% of the generated energy is stored in the battery, how ‘far’ must the student run to use his 100 W laptop for free for 2 hours? Work done by student: W=70*d J Energy given to the battery 0.5W=35*d J 100 W laptop for 2 hours: 100 J/s*3600*2 s=7.2E+5 J 720 KJ 7.2E+5=35*d so d=(7.2E+5)/35=2.1E+4 m =21km!!!

11. PHY 231 11 Non-conservative vs conservative case 45 0 A block of 1 kg is pushed up a 45 o slope with an initial velocity of 10 m/s. How high does the block go if: a) there is no friction b) if the coefficient of kinetic friction is 0.5. A) Conservation of ME: (mgh+1/2mv 2 ) initial = (mgh+1/2mv 2 ) final 0.+0.5*1*10 2 =1*9.8*h+0. So h=5.1 m B) Energy is lost to friction: (mgh+1/2mv 2 ) initial = (mgh+1/2mv 2 ) final +W friction [W=F  x=  n  x=  mgcos(45 o )h/sin(45 o )=0.5*1*9.8*h=4.9h] 0.+0.5*1*10 2 =1*9.8*h+0.+4.9h so h=3.4 m h

12. PHY 231 12 Chapter 6 Momentum p=mv F=  p/  t Impulse (the change in momentum)  p= F  t 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 (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 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

13. PHY 231 13 Inelastic collision 5 m/s 5 kg V m/s 0.1 kg An excellent, but somewhat desperate sharp shooter tries to stop a cannon ball aimed directly at him by shooting a bullet from his gun against it. With what velocity does he need to shoot the bullet to stop the cannon ball assuming that the bullet gets stuck in the ball? How much energy is released? Inelastic: only conservation of momentum. m 1 v 1i +m 2 v 2i =(m 1 +m 2 )v f so 0.1V-5*5=0 v=25/0.1=250 m/s Change in kinetic energy: Before: ½m 1 v 1i 2 +½m 2 v 2i 2 =3125+62.5=3187.5 J After: 0 J Release: 3187.5 J

14. PHY 231 14 Elastic collision h=100 m Two balls (m 1 =1 kg, m 2 =2 kg) are released on a slope and collide in the valley. How far does each go back up? 12 Step 1: calculate their velocities just before the collision: ball 1: cons. of ME: mgh=0.5mv 2 9.8*100=0.5 v 2 v=44. m/s ball 2: cons. of ME: 2*9.8*100=0.5*2*v 2 v=-44. m/s Step 2: Collision, use cons. of P and KE (simplified). m 1 v 1i +m 2 v 2i =m 1 v 1f +m 2 v 2f so 44-88=v 1f +2v 2f (v 1i -v 2i )=(v 2f -v 1f ) so 88=v 2f -v 1f 3v 2f =44 v 2f =15 v 1f =-73 Step 3: Back up the ramp: cons. of ME: ball 1: 0.5mv 2 =mgh 0.5*73 2 =9.8h h=272 m ball 2: 0.5mv 2 =mgh 0.5*15 2 =9.8h h=11.5 m

15. PHY 231 15 Bouncing ball A 0.5 kg ball is dropped to the floor from a height of 2 m. If it bounces back to a height of 1.8 m, what is the magnitude of its change in momentum? Some energy is lost in the bounce. Just before it hits the ground, its velocity is: (use conservation of ME) mgh=1/2mv 2 so v=  (2gh)=  (2*9.8*2)= 6.26 m/s After the bounce it goes back up 1.8 m. Just after it bounces back it velocity is: (use conservation of ME) mgh=1/2mv 2 so v=  (2gh)=  (2*9.8*1.8)=5.93 m/s Must be negative!! So -5.93 m/s  p=m[6.26-(-5.93)]=0.5*12.2=6.1 kgm/s

16. PHY 231 16 Chapter 7 Average angular velocity (rad/s) Instantaneous Angular velocity Average angular acceleration (rad/s 2 ) Instantaneous angular acceleration 2  rad  360 0 1 0  2  /360 rad 1 rad  360/2  deg Be aware that sometimes rev/s or rev/min is asked

17. PHY 231 17 Angular vs linear/tangential linear angular r See e.g. the bike example (lecture 16)

18. PHY 231 18 Rotational motion Angular motion  (t)=  (0)+  (0)t+½  t 2  (t)=  (0)+  t a c =v 2 /r directed to the center of the circular motion Also v=  r, so a c =  2 r Centripetal acceleration  F to center =ma c =mv 2 /r This acceleration is caused by ‘known’ force (gravitation, friction, tension…) Make sure you understand how to use Kepler’s 3rd law and the general definition of gravitational PE.

19. PHY 231 19 Whirling ball v A ball of mass 2 kg is attached to a string of 1m and whirled around a smooth horizontal table. If the tension in the string exceeds 200 N, it will break. What will be the speed at the moment the string breaks?  F to center =ma c =mv 2 /r T=2*v 2 /1=200 N v=10 m/s

20. PHY 231 20 Consider... …a child playing on a swing. As she reaches the lowest point in her swing, which of the following is true? A) The tension in the robe is equal to her weight B) The tension in the robe is equal to her mass times her acceleration C) Her acceleration is downward and equal to g (9.8 m/s 2 ) D) Her acceleration is zero E) Her acceleration is equal to her velocity squared divided by the length of the swing.  F to center =T-mg=ma c =mv 2 /L Lowest point, so no linear acceleration!!!

21. PHY 231 21 Two objects... Are rotating. One starts with an initial linear velocity of 1 m/s and rotates with a radius of 2 m. The other starts from rest and undergoes a constant angular acceleration over a circle with radius 3 m. What should its angular acceleration be, so that it overtakes (for the first time) the first object after 10 revolutions?  1 =v 1 /r 1 =1/2=0.5 rad/s  1 (t)=  1 (0)+  1 (0)t+½  1 t 2 =0.5t  1 (t)=10*2  =0.5t so t=40   2 (t)=  2 (0)+  2 (0)t+½  2 t 2 =½  2 t 2 20  =½  2 (40  ) 2 so  2 =8E-03 rad/s

22. PHY 231 22 Conical motion  mg T If the mass of the swinging object is 1 kg, and  =30 o what should the velocity of the object be so that does not sink or rise? The length of the robe is 2 m. Tcos  Tsin   Vertical direction:  F=ma Tcos  -mg=0 So T=mg/cos  =1*9.8/0.866=11.3 N Horizontal direction:  F=ma c Tsin  =ma c 11.3*0.5=ma c =1v 2 /r v 2 =11.3*0.5*2 so v=3.4 m/s

23. PHY 231 23 Chapter 8 Summary: see beginning of this lecture!

24. PHY 231 24 Opening a hatch door. ?N 1 m A person is trying to open a 1 meter, 20 kg, trap door by pulling a rope attached to its non rotating end at an angle of 30 o. With what force should he at least pull? 30 o Center of gravity of the door: halfway the door’s length: 0.5m   =-m door gd CG +F pull,perpendicular l 0=-20*9.8*0.5+F pull sin(30 o )*1 F pull =196 N