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Phys 207 Announcements z z Hwk 6 is posted online; submission deadline = April 4 Exam 2 on Friday, April 8th Today’s Agenda z z z z z Freshman Interim Grades Review Work done by variable force in 3-D Í Newton’s gravitational force Conservative forces & potential energy Conservation of “total mechanical energy” Í Example: pendulum 1 Work by variable force in 3-D: z Work dWF of a force F acting through an infinitesimal displacement ∆r is: F ∆r . dW = F ∆r z The work of a big displacement through a variable force will be the integral of a set of infinitesimal displacements: ∫. WTOT = F ∆r 2 Page 1 Work by variable force in 3-D: Newton’s Gravitational Force z Work dWg done on an object by gravity in a displacement dr is given by: . . ^ ^ (dR r^ + Rdθ θ) dWg = Fg dr = (-GMm / R2 r) dWg = (-GMm / R2) dR . . ^ (since r^ θ = 0, r^ r^ = 1) dR Rdθ Fg ^ θ r^ dr m dθ R M 3 Work by variable force in 3-D: Newton’s Gravitational Force z Integrate dWg to find the total work done by gravity in a “big” displacement: ∫ R2 ∫ R2 Wg = dWg = (-GMm / R2) dR = GMm (1/R2 - 1/R1) R1 R1 Fg(R2) m R2 Fg(R1) R1 M 4 Page 2 Work by variable force in 3-D: Newton’s Gravitational Force z Work done depends only on R1 and R2, not on the path taken. ⎛ 1 1 ⎞⎟ − Wg = GMm⎜⎜ ⎟ ⎝ R2 R1 ⎠ m R2 R1 M 5 Lecture 16, Act 1 Work & Energy z A rock is dropped from a distance RE above the surface of the earth, and is observed to have kinetic energy K1 when it hits the ground. An identical rock is dropped from twice the height (2RE) above the earth’s surface and has kinetic energy K2 when it hits. RE is the radius of the earth. ÍWhat is K2 / K1? (a) 2 (b) 3 2 2RE (c) 4 3 RE RE 6 Page 3 Lecture 16, Act 1 Solution z Since energy is conserved, ∆K = WG. ⎛ 1 1 ⎞ WG = GMm ⎜⎜ − ⎟⎟ R R ⎝ 2 1⎠ ⎛ 1 1 ⎞ ⎟⎟ ∆K = c ⎜⎜ − R R ⎝ 2 1⎠ Where c = GMm is the same for both rocks 2RE RE RE 7 ⎛ 1 1 ⎞ ⎟⎟ ∆K = c ⎜⎜ − R R ⎝ 2 1⎠ Lecture 16, Act 1 Solution ⎛ 1 1 ⎞ 1 1 ⎟⎟ = c ⋅ ⋅ − R 2R 2 RE ⎝ E E ⎠ z For the first rock: K1 = c ⎜⎜ z ⎟⎟ = c ⋅ ⋅ − For the second rock: K 2 = c ⎜⎜ 3 RE ⎝ R E 3R E ⎠ ⎛ 1 So: 1 ⎞ 2 1 2 K2 3 4 = = K1 1 3 2 2RE RE RE 8 Page 4 Newton’s Gravitational Force Near the Earth’s Surface: z Suppose R1 = RE and R2 = RE + ∆y ⎛ R − R1 ⎞ ⎛ (R + ∆y ) − (RE ) ⎞ ⎛ ⎞ ⎟ = −GMm ⎜ E ⎟ ≅ −m⎜ GM ⎟ ∆y Wg = −GMm ⎜⎜ 2 2 ⎟ ⎜ (R + ∆y )(R ) ⎟ ⎜ R ⎟ ⎝ R1R2 ⎠ ⎝ ⎝ E ⎠ ⎠ E E ⎛ GM ⎞ but we have learned that ⎜ 2 ⎟ = g ⎝ RE ⎠ So: Wg = -mg∆y m RE+ ∆y RE M 9 Conservative Forces: z We have seen that the work done by gravity does not depend on the path taken. m ⎛ 1 1 ⎞⎟ − Wg = GMm⎜⎜ ⎟ ⎝ R2 R1 ⎠ R2 R1 M m h Wg = -mgh 10 Page 5 Conservative Forces: z In general, if the work done does not depend on the path taken (only depends the initial and final distances between objects), the force involved is said to be conservative. z ⎛ 1 1 ⎞⎟ − Gravity is a conservative force: Wg = GMm⎜⎜ ⎟ ⎝ R2 R1 ⎠ z Gravity near the Earth’s surface: W g = − mg ∆ y z A spring produces a conservative force: Ws = − 1 k (x22 − x12 ) 2 11 Conservative Forces: z We have seen that the work done by a conservative force does not depend on the path taken. W2 W1 = W2 z Therefore the work done in a closed path is 0. WNET = W1 - W2 = W1 - W1 = 0 W1 W2 W1 12 Page 6 Lecture 16, Act 2 Conservative Forces z The pictures below show force vectors at different points in space for two forces. Which one is conservative ? (a) 1 (b) 2 y (c) both y x x (1) (2) 13 Lecture 16, Act 2 Solution z Consider the work done by force when moving along different paths in each case: W A = WB W A > WB (1) (2) Page 7 14 Lecture 16, Act 2 z In fact, you could make money on type (2) if it ever existed: ÍWork done by this force in a “round trip” is > 0! ÍFree kinetic energy!! WNET = 10 J = ∆K W=0 W = 15 J W = -5 J Note: NO REAL FORCES OF THIS TYPE EXIST, SO FAR AS WE KNOW W=0 15 Potential Energy z For any conservative force F we can define a potential energy function U in the following way: W = ∫ F.dr = -∆U Í The work done by a conservative force is equal and opposite to the change in the potential energy function. r2 z U2 This can be written as: r2 ∆U = U2 - U1 = -W = - ∫F.dr r1 r1 U1 16 Page 8 Gravitational Potential Energy z We have seen that the work done by gravity near the Earth’s surface when an object of mass m is lifted a distance ∆y is Wg = -mg ∆y z The change in potential energy of this object is therefore: ∆U = -Wg = mg ∆y j m ∆y Wg = -mg ∆y 17 Gravitational Potential Energy z So we see that the change in U near the Earth’s surface is: ∆U = -Wg = mg ∆y = mg(y2 -y1). z So U = mg y + U0 where U0 is an arbitrary constant. z Having an arbitrary constant U0 is equivalent to saying that we can choose the y location where U = 0 to be anywhere we want to. j m y2 y1 Wg = -mg ∆y 18 Page 9 Potential Energy Recap: For any conservative force we can define a potential energy function U such that: z S2 ∆U = U2 - U1 = -W = - ∫F.dr S1 The potential energy function U is always defined only up to an additive constant. z can choose the location where U = 0 to be anywhere convenient. Í You 19 Conservative Forces & Potential Energies (stuff you should know): ^ Fg = -mg j Fg = − GMm ^ r R2 Fs = -kx Change in P.E ∆U = U2 - U1 Work W(1-2) Force F mg(y2-y1) -mg(y2-y1) ⎛ 1 1 ⎞⎟ GMm⎜⎜ − ⎟ ⎝ R2 R1 ⎠ − 1 k (x22 − x12 ) 2 P.E. function U mgy + C ⎛ 1 GMm 1 ⎞⎟ − GMm⎜⎜ − − +C ⎟ R ⎝ R2 R1 ⎠ 1 k (x22 − x12 ) 2 1 2 kx + C 2 (R is the center-to-center distance, x is the spring stretch) 20 Page 10 Lecture 16, Act 3 Potential Energy z All springs and masses are identical. (Gravity acts down). ÍWhich of the systems below has the most potential energy stored in its spring(s), relative to the relaxed position? (a) 1 (b) 2 (c) same (1) (2) 21 Lecture 16, Act 3 Solution z The displacement of (1) from equilibrium will be half of that of (2) (each spring exerts half of the force needed to balance mg) 0 d 2d (1) (2) 22 Page 11 Lecture 16, Act 3 Solution 1 2 2 The potential energy stored in (1) is 2 ⋅ kd = kd 2 The potential energy stored in (2) is 1 2 k (2d) = 2kd2 2 The spring P.E. is twice as big in (2) ! 0 d 2d (1) (2) 23 Conservation of Energy z If only conservative forces are present, the total kinetic plus potential energy of a system is conserved, i.e. the total “mechanical energy” is conserved. Í(note: E=Emechanical throughout this discussion) E=K+U ∆E = ∆K + ∆U ⇒ using ∆K = W = W + ∆U = W + (-W) = 0 ⇒ using ∆U = -W E = K + U is constant!!! z z Both K and U can change, but E = K + U remains constant. But we’ll see that if non-conservative forces act then energy can be dissipated into other modes (thermal,sound) 24 Page 12 Example: The simple pendulum z Suppose we release a mass m from rest a distance h1 above its lowest possible point. Í What is the maximum speed of the mass and where does this happen? Í To what height h2 does it rise on the other side? m h1 h2 v 25 Example: The simple pendulum z z Kinetic+potential energy is conserved since gravity is a conservative force (E = K + U is constant) Choose y = 0 at the bottom of the swing, and U = 0 at y = 0 (arbitrary choice) E = 1/2mv2 + mgy y y=0 h1 h2 v Page 13 26 Example: The simple pendulum z E = 1/2mv2 + mgy. Í Initially, y = h1 and v = 0, so E = mgh1. Í Since E = mgh1 initially, E = mgh1 always since energy is conserved. y y=0 27 Example: The simple pendulum z 1/2mv2 z will be maximum at the bottom of the swing. 1/ mv2 = mgh So at y = 0 v2 = 2gh1 2 1 v = 2 gh1 y y = h1 y=0 h1 v Page 14 28 Example: The simple pendulum z z Since E = mgh1 = 1/2mv2 + mgy it is clear that the maximum height on the other side will be at y = h1 = h2 and v = 0. The ball returns to its original height. y y = h1 = h2 y=0 29 Example: The simple pendulum z The ball will oscillate back and forth. The limits on its height and speed are a consequence of the sharing of energy between K and U. E = 1/2mv2 + mgy = K + U = constant. y 30 Page 15 Example: The simple pendulum z We can also solve this by choosing y = 0 to be at the original position of the mass, and U = 0 at y = 0. E = 1/2mv2 + mgy. y y=0 h1 h2 v 31 Example: The simple pendulum z E = 1/2mv2 + mgy. Í Initially, y = 0 and v = 0, so E = 0. Í Since E = 0 initially, E = 0 always since energy is conserved. y y=0 32 Page 16 Example: The simple pendulum z 1/2mv2 z will be maximum at the bottom of the swing. 1/ mv2 = mgh So at y = -h1 v2 = 2gh1 2 1 v = 2 gh1 y Same as before! y=0 y = -h1 h1 v 33 Example: The simple pendulum z z Since 1/2mv2 - mgh = 0 it is clear that the maximum height on the other side will be at y = 0 and v = 0. The ball returns to its original height. y y=0 Same as before! 34 Page 17