Momentum, Work and Energy in Mechanical Systems

1. Momentum, Work and Energy in Mechanical Systems Prof Bill Easson http://www.see.ed.ac.uk/~bille

2. Introduction • So far, we have explicitly used the force and Newton’s 2nd Law to analyse dynamic systems. However, we do not always know the force, and there are other methods available. Here we deal with two of these • Momentum • Energy

3. Momentum NII : resultant force = rate of change of momentum where m is constant If a constant force, F, acts on a mass m for a time t, then where v1 and v2 are initial and final velocities

4. Momentum (2) What happens if F varies with time? We get a series of products where the time intervals t are small enough for the forces to be  constant over each interval. Maths gives us an easy way to write this sum:

5. Momentum (3) If F is constant then we should get the same result as before F t1 t2 t

6. Momentum (4) If F is a known function of t, then we can solve Suppose a force acting in the direction, x, gets larger with time, t. F=kt F t t1 t2 In this case the velocity varies as the square of time, t

7. before FBDs during A B A B A after B A B Momentum (5) Even if the equation of the force is not known, the momentum can sometimes be used to solve dynamical problems where there are no external forces Consider two balls colliding: uA uB vA vB Although we do not know the value of the forces during the collision, we know that they must be equal and opposite

8. Momentum (6) The total momentum after the collision is equal to the total momentum before the collosion. This turns out to be true however many bodies there are and is called the Law of Conservation of Momentum

9. Momentum conservation Consider a truck carrying a large gun which fires shells of mass 20kg. The truch + gun together weigh 5 Tonnes. If the shell is fired at 300ms-1 at an angle of 60o to the horizontal, what is the recoil velocity of the truck? y u2 x u1 u1=u2=0 Before After

10. Momentum conservation As the truck can only move horizontally, we only consider x values Before: m1=5000 kg m2=20 kg u1=0 ms-1 u2=0 ms-1 After: m1=5000 kg m2=20 kg v1=? v2=300cos60o=150 ms-1 Momentum before is zero, so momentum after must also be zero m1v1+m2v2=0 v1=-m2v2/m1 =-20.150/5000 v1=-0.6 ms-1

11. Energy (and Work) Here we look at the effect of applying a force, F, over a distance, x. • The work done is the energy transferred, by the force, to the body – units are joules(J). • For a constant force, as in the figure • W=F x d F x1 x2 x

12. Work and Power If F varies with x, as in the figure, then we add all the small elements to get the area under the curve, F Power is the rate at which work is done and is measured in Watts (W) (since integration is the opposite of differentiation) x1 x2 x

13. Potential Energy In dynamics, potential energy is work done against gravity, often called gravitational potential energy. Consider a mass being raised at a constant speed on a cable Before After T T h mg U=mgh mg

14. Spring Energy Empirically (from experiment), we find that F=kx F(x) x F(x) This is referred to as spring potential energy

15. Kinetic Energy Consider Newton’s second law in the integral for work: This is the work done accelerating a body It is normally called kinetic energy

16. Conservation of Energy In many of the problems you encounter mechanical energy will be conserved, and this can be a useful tool. T+U=const Example: A ball of mass 1kg is dropped from a balcony of height 10m. Ignoring drag, calculate its velocity just before it hits the ground. energy before =energy after v=14ms-1

17. Elastic Collisions If the collision is elastic, then kinetic energy is conserved, and momentum is always conserved, so we have two equations CoM: (1) CoE: (2) For an in-line (1D) collision, divide equation 2 by equation 1: and

18. Elastic collisions - examples Substituting the equations for v1 and v2 back into the momentum equation gives: Snooker (m1= m2; u2=0): (Second ball moves off with speed of first. First ball stops. Ignores spin.) v2= u1=0 and v1= u2 Ball hitting wall ( m2 >> m1 ; u2=0) v2= u2=0 and v1= -u1 (Wall remains stationary, ball bounces back with same velocity) Golf club hitting ball ( m1 >> m2 ; u2=0) (Ball goes off at twice clubhead speed and clubhead velocity remains the same) v2= 2u1 and v1= u1

19. Snookered Consider two smooth, hard balls colliding. During the collision, which is very short, friction has a negligible effect. So the force can only be transferred by the impulse acting normal to the surface at the collision point. FA uA FB uB=0 In snooker, all the balls have the same diameter and mass, which makes the problem easy to solve. The velocity of ball B after impact must be in the direction of the line of action.

20. Snookered Case 1: In line y uA uB=0 x As the line of action is in the x direction, there can be no motion in the y direction. The energy equation gives

21. Snookered Energy (1) Momentum (2) Substituting for vA from equ 1 into equ 2, So, vB=uA and vA=0 The second ball moves off at the speed of the first, and the first ball becomes stationary

22. Snookered Balls not in line of action: y vA  uA  x uB=0 vB We know that the momentum equation must be satisifed in x and in y, since it is a vector equation. The collision is elastic, so energy is conserved. We also know that B can only move off on the line of action, . x-momentum is conserved no y-momentum before, therefore no momentum after

23. Snookered Energy: For any given angle of action, , and initial velocity, uA, we have three unknowns, vA, vB and the angle of the first ball, . However, we also have three equations so, by substitution and some algebra, we find that the white ball must move off at exactly 90o to the red ball. 90o This enables a simple program to be written to simulate a snooker/billiards table.