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Lecture 13 Goals: • Chapter 10 • Understand the relationship between motion and energy Define Potential Energy in a Hooke’s Law spring Develop and exploit conservation of energy principle in problem solving Chapter 11 Understand the relationship between force, displacement and work Assignments: HW6, due tomorrow, HW7 due Wednesday, Mar. 10 For Thursday, Read all of Chapter 11 Physics 207: Lecture 13, Pg 1 Newton’s Laws rearranged From motion in the y-dir, Fy = m ay and let accel. be constant y(t) = y0 + vy0 Dt + ½ ay Dt2 Dy = y(t)-y0= vy0 Dt + ½ ay Dt2 vy (t) = vy0 + ay Dt Dt = (vy - vy0) / ay and eliminating Dt yields ay Dy= ½ (vy2 - vy02 ) / ay -mg Dy= ½ m (vy2 - vy02 ) Physics 207: Lecture 13, Pg 2 Energy -mg Dy= 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 define mgy as the “gravitational potential energy” Physics 207: Lecture 13, Pg 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 vzi2 and ½ m vxf2 + ½ m vzf2 ½ m vi2 + mgyi = ½ m vf2 + mgyf where vi2 ≡ vxi2 +vyi2 + vzi2 ½ m v2 terms are defined to be kinetic energies (A scalar quantity of motion) Physics 207: Lecture 13, Pg 4 Inelastic collision in 1-D: Example 1 A block of mass M is 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 Physics 207: Lecture 13, Pg 5 Inelastic collision in 1-D: Example 1 mv What is the momentum of the bullet with speed v ? 1 1 2 mv v mv 2 2 1 What is the final energy of the system ? (m M )V 2 2 Is momentum conserved (yes)? mv M 0 (m M )V What is the initial energy of the system ? Examine Ebefore-Eafter 1 1 1 1 m 1 m mv 2 [( m M )V]V mv 2 (mv) v mv 2 1 2 2 2 2 mM 2 mM ( v No! before ) V after x Is energy conserved? Physics 207: Lecture 13, Pg 6 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. Physics 207: Lecture 13, Pg 7 Energy If only “conservative” forces are present, the total mechanical energy (sum of potential, U, and kinetic energies, K) of a system is conserved For an object in a gravitational “field” ½ m vyi2 + mgyi = ½ m vyf2 + mgyf K ≡ ½ mv2 U ≡ mgy Emech = K + U Emech = K + U = constant K and U may change, but Emech = K + U remains a fixed value. Emech is called “mechanical energy” Physics 207: Lecture 13, Pg 8 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 ? m h1 h2 v Physics 207: Lecture 13, Pg 9 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 Physics 207: Lecture 13, Pg 10 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 = mgh1 at top E = mgh1 = ½ mv2 at bottom of the swing y y=h1 y=0 h1 v Physics 207: Lecture 13, Pg 11 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 Physics 207: Lecture 13, Pg 12 Example The 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 that E = U + K = constant ! (frictionless) Ub=mgh U=mg2R h? y=0(A) 2R U=0 Recall that “g” is the source of Car has mass m the centripetal acceleration and N just goes to zero is the limiting case. Also recall the minimum speed at the top is R (B) 3R (C) 5/2 R (D) 23/2 R v gR Physics 207: Lecture 13, Pg 13 Example The Loop-the-Loop … again Use E = K + U = constant mgh + 0 = mg 2R + ½ mv2 mgh = mg 2R + ½ mgR = 5/2 mgR v gR h = 5/2 R h? R Physics 207: Lecture 13, Pg 14 Example Skateboard What speed will the skateboarder reach halfway down the hill if there is no friction and the skateboarder starts at rest? Assume we can treat the skateboarder as a “point” Assume zero of gravitational U is at bottom of the hill m = 25 kg .. R=10 m 30° R=10 m y=0 Physics 207: Lecture 13, Pg 15 Example Skateboard What speed will the skateboarder reach halfway down the hill if there is no friction and the skateboarder starts at rest? Assume we can treat the skateboarder as “point” Assume zero of gravitational U is at bottom of the hill m = 25 kg .. R=10 m Use E = K + U = constant Ebefore = Eafter 0 + m g R = ½ mv2 + mgR / 2 30° mgR/2 = ½ mv2 R=10 m gR = v2 v= (gR)½ y= 5 m v = (10 x 10)½ = 10 m/s Physics 207: Lecture 13, Pg 16 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 1. The ball is dropped 2. The ball slides down a straight incline 3. The ball slides down a curved incline After traveling a vertical distance h, how do the three speeds compare? 1 2 3 h (A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell Physics 207: Lecture 13, Pg 17 Example Skateboard What is the normal force on the skate boarder? N m = 25 kg 60° .. mg R=10 m 30° R=10 m Physics 207: Lecture 13, Pg 18 Example Skateboard Now what is the normal force on the skate boarder? N m = 25 kg 60° .. R=10 m 30° mg S Fr = mar = m v2 / R = N – mg cos 60° R=10 m N = m v2 /R + mg cos 60° N = 25 100 / 10 + 25 10 (0.87) N = 250 + 220 =470 Newtons Physics 207: Lecture 13, Pg 19 Elastic vs. Inelastic Collisions 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. vi Physics 207: Lecture 13, Pg 20 Elastic vs. Inelastic Collisions A collision is said to be inelastic when energy is not conserved before and after the collision, but momentum is conserved. Kbefore Kafter Car crashes, collisions where objects stick together, etc. Physics 207: Lecture 13, Pg 21 Variable force devices: Hooke’s Law Springs Springs are everywhere, (probe microscopes, DNA, an effective interaction between atoms) Rest or equilibrium position F Ds In this spring, the magnitude of the force increases as the spring is further compressed (a displacement). Hooke’s Law, Fs = - k Ds Ds is the amount the spring is stretched or compressed from it resting position. Physics 207: Lecture 13, Pg 23 Exercise 2 Hooke’s Law 8m 9m What is the spring constant “k” ? 50 kg (A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m Physics 207: Lecture 13, Pg 25 Exercise 2 Hooke’s Law 8m 9m What is the spring constant “k” ? Fspring 50 kg (A) 50 N/m SF = 0 = Fs – mg = k Ds - mg Use k = mg/Ds = 500 N / 1.0 m (B) 100 N/m (C) 400 N/m (D) 500 N/m mg Physics 207: Lecture 13, Pg 26 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 uf vf ui vi ku du mv dv 1 2 ku | 2 uf ui ku ku 1 2 ku mv ku mv 2 f 1 2 m 2 i 1 2 2 i 1 2 1 2 2 f 2 f 1 2 2 i 1 2 1 2 2 vf mv vi 2 mv f | 1 2 2 mvi Physics 207: Lecture 13, Pg 27 Lecture 13 Assignment: HW6 due Wednesday For Thursday: Read all of chapter 11 Physics 207: Lecture 13, Pg 28