Understanding Motion and Energy: Hooke's Law & Conservation Principles

1. Lecture 14 • Chapter 10 • Understand the relationship between motion and energy • Define Potential Energy in a Hooke’s Law spring • Exploit conservation of energy principle in problem solving • Understand spring potential energies • Use energy diagrams Goals: Assignment: • HW6 due Tuesday Oct. 26th • For Monday: Read Ch. 11

2. Energy -mgDy= ½ m (vy2 - vy02) -mg(yf – yi) = ½ m ( vyf2 -vyi2 ) A relationship between y- displacement and change in the y-speed squared Rearranging to give initial on the left and final on the right ½ m vyi2+ mgyi = ½ m vyf2 + mgyf We now definemgy = Uas the “gravitational potential energy”

3. Energy • Notice that if we only consider gravity as the external force then the x and z velocities remain constant • To ½ m vyi2+ mgyi = ½ m vyf2 + mgyf • Add ½ m vxi2 + ½ m vzi2and ½ m vxf2 + ½ m vzf2 ½ m vi2+ mgyi = ½ m vf2 + mgyf • where vi2 ≡ vxi2 +vyi2 + vzi2 ½ m v2 = K terms are defined to be kinetic energies (A scalar quantity of motion)

4. Reading Quiz • What is the SI unit of energy A erg B dyne C N s D calorie E Joule

5. Inelastic collision in 1-D: Example 1 • A block of massMis initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. • What is the initial energy of the system ? • What is the final energy of the system ? • Is energy conserved? x v V before after

6. Inelastic collision in 1-D: Example 1 What is the momentum of the bullet with speed v ? • What is the initial energy of the system ? • What is the final energy of the system ? • Is momentum conserved (yes)? • Is energy conserved? Examine Ebefore-Eafter v No! V x before after

7. Elastic vs. Inelastic Collisions • A collision is said to be inelastic when “mechanical” energy ( K +U ) is not conserved before and after the collision. • How, if no net Force then momentum will be conserved. Kbefore Kafter • E.g. car crashes on ice: Collisions where objects stick together • A collision is said to be elastic when both energy & momentum are conserved before and after the collision. Kbefore = Kafter • Carts colliding with a perfect spring, billiard balls, etc.

8. Kinetic & Potential energies • Kinetic energy, K = ½ mv2, is defined to be the large scale collective motion of one or a set of masses • Potential energy, U, is defined to be the “hidden” energy in an object which, in principle, can be converted back to kinetic energy • Mechanical energy, EMech, is defined to be the sum of U and K.

9. Energy • If only “conservative” forces are present, the total mechanical energy (sum of potential, U, and kinetic energies, K) of a systemis conserved For an object in a gravitational “field” ½ m vyi2+ mgyi = ½ m vyf2 + mgyf U ≡ mgy K ≡ ½ mv2 Emech = K + U Emech = K + U= constant • K and U may change, but Emech = K + Uremains a fixed value. Emech is called “mechanical energy”

10. m h1 h2 v Example of a conservative system: The simple pendulum. • Suppose we release a mass m from rest a distance h1 above its lowest possible point. • What is the maximum speed of the mass and where does this happen ? • To what height h2 does it rise on the other side ?

11. Example: The simple pendulum. • What is the maximum speed of the mass and where does this happen ? E = K + U = constant and so K is maximum when U is a minimum. y y=h1 y=0

12. Example: The simple pendulum. • What is the maximum speed of the mass and where does this happen ? E = K + U = constant and so K is maximum when U is a minimum E = mgh1at top E = mgh1 = ½ mv2 at bottom of the swing y y=h1 h1 y=0 v

13. Example: The simple pendulum. To what height h2 does it rise on the other side? E = K + U = constant and so when U is maximum again (when K = 0) it will be at its highest point. E = mgh1 = mgh2 or h1 = h2 y y=h1=h2 y=0

14. 1 3 2 h Potential Energy, Energy Transfer and Path • A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless • The ball is dropped • The ball slides down a straight incline • The ball slides down a curved incline After traveling a vertical distance h, how do the three speeds compare? (A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell

15. 1 3 2 h Potential Energy, Energy Transfer and Path • A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless • The ball is dropped • The ball slides down a straight incline • The ball slides down a curved incline After traveling a vertical distance h, how do the three speeds compare? (A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell

16. Ub=mgh Recall that “g” is the source of the centripetal acceleration and N just goes to zero is the limiting case. Also recall the minimum speed at the top is Car has mass m U=mg2R h ? R ExampleThe Loop-the-Loop … again • To complete the loop the loop, how high do we have to let the release the car? • Condition for completing the loop the loop: Circular motion at the top of the loop (ac = v2 / R) • Exploit the fact thatE = U + K = constant ! (frictionless) (A) 2R (B) 3R (C) 5/2 R (D) 23/2 R y=0 U=0

17. ExampleThe Loop-the-Loop … again • Use E = K + U = constant • mgh + 0 = mg 2R + ½ mv2 mgh = mg 2R + ½ mgR = 5/2 mgR h = 5/2 R h ? R

18. Variable force devices: Hooke’s Law Springs • Springs are everywhere, (probe microscopes, DNA, an effective interaction between atoms) • In this spring, the magnitude of the force increases as the spring is further compressed (a displacement). • Hooke’s Law, Fs = - kDu Ds is the amount the spring is stretched or compressed from it resting position. Rest or equilibrium position F Du

19. 8 m 9 m Whatis the spring constant “k” ? 50 kg Exercise Hooke’s Law (A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m

20. Exercise 2Hooke’s Law 8 m 9 m Whatis the spring constant “k” ? SF = 0 = Fs – mg = k Du - mg Use k = mg/Du = 500 N / 1.0 m Fspring 50 kg (A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m mg

21. m Force vs. Energy for a Hooke’s Law spring • F = - k (x – xequilibrium) • F = ma = m dv/dt = m (dv/dx dx/dt) = m dv/dx v = mv dv/dx • So - k (x – xequilibrium) dx = mv dv • Let u = x – xeq. & du = dx 

22. m Energy for a Hooke’s Law spring • Associate ½ ku2 with the “potential energy” of the spring • Ideal Hooke’s Law springs are conservative so the mechanical energy is constant

23. Emech K Energy U y Energy diagrams • In general: Ball falling Spring/Mass system Emech K Energy U 0 0 u = x - xeq

24. U U Equilibrium • Example • Spring: Fx = 0 => dU / dx = 0 for x=xeq The spring is in equilibrium position • In general: dU / dx = 0 for ANY function establishes equilibrium stable equilibrium unstable equilibrium

25. Recap Assignment: • HW6 due Tuesday Oct. 26th • For Monday: Read Ch. 11