#### Transcript 1 PHYSICS 231 Lecture 18: equilibrium & revision

PHYSICS 231 Lecture 18: equilibrium & revision Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom PHY 231 1 Previously Torque: =Fd Center of Gravity: m x m i xCG i i i m y m i i y CG i i i i 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! PHY 231 2 examples: A lot more in the book! Where is the center of gravity? PHY 231 3 T1 T2 Weight of board: w What is the tension in each of the wires (in terms of w)? w 0 PHY 231 4 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? PHY 231 5 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 PHY 231 6 Revision: chapter 5 Work: W=Fcos()x Power: P=W/t Potential energy (PE) Energy transfer Rate of energy transfer Energy associated with position. Gravitational PE: mgh Energy associated with position in grav. field. PE stored in a spring: 1/2kx2 x is the compression of the spring k is the spring constant Kinetic energy KE: 1/2mv2 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 KEi+PEi=KEf+PEf Not conserved in the presence of non-conservative forces 7 (KEi+PEi)-(KEf+PEf)=Wnc PHY 231 example m=1 kg 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) 1 cm k=100 N/m PHY 231 8 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? PHY 231 9 Non-conservative vs conservative case h 450 A block of 1 kg is pushed up a 45o 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. PHY 231 10 Chapter 6 Momentum p=mv F=p/t Impulse (the change in momentum) p= Ft Inelastic collisions Elastic collisions •Momentum is conserved •Momentum is conserved •Some energy is lost in the •No energy is lost in the collision: KE not conserved collision: KE conserved •Perfectly inelastic: the objects stick together. Conservation of momentum: Conservation of momentum: m1v1i+m2v2i=(m1+m2)vf m1v1i+m2v2i=m1v1f+m2v2f Conservation of KE: ½m1v1i2+½m2v2i2=½m1v1f2+½m2v2f2 (v1i-v2i)=(v2f-v1f) PHY 231 11 Inelastic collision 0.1 kg V m/s 5 m/s 5 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? PHY 231 12 h=100 m 1 2 Elastic collision Two balls (m1=1 kg, m2=2 kg) are released on a slope and collide in the valley. How far does each go back up? PHY 231 13 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? PHY 231 14 Chapter 7 f i Average angular t f ti t velocity (rad/s) Instantaneous lim t 0 t Angular velocity f i t f ti Average angular t acceleration (rad/s2) Instantaneous angular lim t 0 t acceleration 2 rad 3600 10 2/360 rad 1 rad 360/2 deg Be aware that sometimes rev/s or rev/min is asked PHY 231 15 Angular vs linear/tangential v r linear r angular a r See e.g. the bike example (lecture 16) PHY 231 16 Rotational motion Angular motion (t)= (0)+(0)t+½t2 (t)= (0)+t Centripetal acceleration ac=v2/r directed to the center of the circular motion Also v=r, so ac=2r Fto center=mac=mv2/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. PHY 231 17 Whirling ball 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, what will be the speed at the moment the string breaks? PHY 231 v 18 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/s2) D) Her acceleration is zero E) Her acceleration is equal to her velocity squared divided by the length of the swing. PHY 231 19 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? PHY 231 20 Conical motion T If the mass of the swinging object is 1 kg, and =30o what should the velocity of the object be so that does not sink or rise? The length of the robe is 2 m. mg PHY 231 21 Chapter 8 Summary: see beginning of this lecture! PHY 231 22 Opening a hatch door. ?N 30o 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 30o. With what force should he at least pull? 1m PHY 231 23