Physics 111: Lecture 11, Pg 1 Physics 111: Lecture 11 Today’s Agenda l Review l Work done by variable force in 3-D ç Newton’s gravitational force l Conservative forces & potential energy l Conservation of “total mechanical energy” ç Example: pendulum l Non-conservative forces ç friction l General work/energy theorem l Example problem

Physics 111: Lecture 11, Pg 2 Work by variable force in 3-D: F l Work dW F of a force F acting through an infinitesimal  r displacement  r is: F  r dW = F.  r l The work of a big displacement through a variable force will be the integral of a set of infinitesimal displacements: F  r W TOT = F.  rF rrrr 

Physics 111: Lecture 11, Pg 3 Work by variable force in 3-D: Newton’s Gravitational Force dr l Work dW g done on an object by gravity in a displacement dr is given by: Fdr rr rrr dW g = F g. dr = (-GMm / R 2 r). (dR r + Rd  ) dW g = (-GMm / R 2 ) dR (since r.  = 0, r. r = 1) ^ ^ ^ r  ^ ^ dr Rd  dR R FFgFFg m M dd ^ ^ ^^

Physics 111: Lecture 11, Pg 4 Work by variable force in 3-D: Newton’s Gravitational Force l Integrate dW g to find the total work done by gravity in a “big” displacement: W g = dW g = (-GMm / R 2 ) dR = GMm (1/R 2 - 1/R 1 ) F F g (R 1 ) R1R1 R2R2 F F g (R 2 ) R1R1 R2R2 R1R1 R2R2 m M

Physics 111: Lecture 11, Pg 5 Work by variable force in 3-D: Newton’s Gravitational Force not on the path taken l Work done depends only on R 1 and R 2, not on the path taken. R1R1 R2R2 m M

Physics 111: Lecture 11, Pg 6 Lecture 11, Act 1 Work & Energy l A rock is dropped from a distance R E above the surface of the earth, and is observed to have kinetic energy K 1 when it hits the ground. An identical rock is dropped from twice the height (2R E ) above the earth’s surface and has kinetic energy K 2 when it hits. R E is the radius of the earth. çWhat is K 2 / K 1 ?(a)(b)(c) 2 RERE RERE 2R E

Physics 111: Lecture 11, Pg 7 Lecture 11, Act 1 Solution Since energy is conserved,  K = W G. RERE RERE 2R E Where c = GMm is the same for both rocks

Physics 111: Lecture 11, Pg 8 Lecture 11, Act 1 Solution l For the first rock: RERE RERE 2R E l For the second rock: So:

Physics 111: Lecture 11, Pg 9 Newton’s Gravitational Force Near the Earth’s Surface: Suppose R 1 = R E and R 2 = R E +  y but we have learned that So: W g = -mg  y R E +  y M m RERE

Physics 111: Lecture 11, Pg 10 Conservative Forces: l We have seen that the work done by gravity does not depend on the path taken. R1R1 R2R2 M m h m W g = -mgh

Physics 111: Lecture 11, Pg 11 Conservative Forces: conservative l In general, if the work done does not depend on the path taken, the force involved is said to be conservative. l Gravity is a conservative force: l Gravity near the Earth’s surface: l A spring produces a conservative force:

Physics 111: Lecture 11, Pg 12 Conservative Forces: l We have seen that the work done by a conservative force does not depend on the path taken. W1W1 l The work done can be “reclaimed.” W2W2 W1W1 W2W2 W 1 = W 2 W NET = W 1 - W 2 = 0 l Therefore the work done in a closed path is 0.

Physics 111: Lecture 11, Pg 13 Lecture 11, Act 2 Conservative Forces l The pictures below show force vectors at different points in space for two forces. Which one is conservative ? (a) (b) (c) (a) 1 (b) 2 (c) both (1)(2) y x y x

Physics 111: Lecture 11, Pg 14 Lecture 11, Act 2 Solution l Consider the work done by force when moving along different paths in each case: (1)(2) W A = W B W A > W B

Physics 111: Lecture 11, Pg 15 Lecture 11, Act 2 l In fact, you could make money on type (2) if it ever existed: çWork done by this force in a “round trip” is > 0! çFree kinetic energy!! W = 15 J W = 0 W = -5 J W = 0 W NET = 10 J =  K

Physics 111: Lecture 11, Pg 16 Potential Energy l For any conservative force F we can define a potential energy function U in the following way: ç The work done by a conservative force is equal and opposite to the change in the potential energy function. l This can be written as: Fr W = F. dr = -  U  Fr  U = U 2 - U 1 = -W = - F. dr  r1r1 r2r2 r1r1 r2r2 U2U2 U1U1

Physics 111: Lecture 11, Pg 17 Gravitational Potential Energy 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 W g = -mg  y The change in potential energy of this object is therefore:  U = -W g = mg  y yy m W g = -mg  y j

Physics 111: Lecture 11, Pg 18 Gravitational Potential Energy So we see that the change in U near the Earth’s surface is:  U = -W g = mg  y = mg(y 2 -y 1 ). arbitrary constant l So U = mg y + U 0 where U 0 is an arbitrary constant. l Having an arbitrary constant U 0 is equivalent to saying that we can choose the y location where U = 0 to be anywhere we want to. y1y1 m W g = -mg  y j y2y2

Physics 111: Lecture 11, Pg 19 Potential Energy Recap: l For any conservative force we can define a potential energy function U such that: l The potential energy function U is always defined only up to an additive constant. ç You can choose the location where U = 0 to be anywhere convenient. Fr  U = U 2 - U 1 = -W = - F. dr S1S1 S2S2

Physics 111: Lecture 11, Pg 20 Conservative Forces & Potential Energies (stuff you should know): ForceF Work W(1-2) Change in P.E  U = U 2 - U 1 P.E. function U Fj F g = -mg j Fr F g = r F s = -kx ^ ^ -mg(y 2 -y 1 ) mg(y 2 -y 1 ) mgy + C

Physics 111: Lecture 11, Pg 21 Lecture 11, Act 3 Potential Energy l All springs and masses are identical. (Gravity acts down). çWhich of the systems below has the most potential energy stored in its spring(s), relative to the relaxed position? (a) (a) 1 (b) (b) 2 (c) (c) same (1) (2)

Physics 111: Lecture 11, Pg 22 Lecture 11, Act 3 Solution l The displacement of (1) from equilibrium will be half of that of (2) (each spring exerts half of the force needed to balance mg) (1) (2) d 2d 0

Physics 111: Lecture 11, Pg 23 Lecture 11, Act 3 Solution (1) (2) d 2d 0 The potential energy stored in (1) isThe potential energy stored in (2) is The spring P.E. is twice as big in (2) !

Physics 111: Lecture 11, Pg 24 Conservation of Energy l If only conservative forces are present, the total kinetic plus potential energy of a system is conserved. constant!!! l If only conservative forces are present, the total kinetic plus potential energy of a system is conserved. E = K + U is constant!!! l Both K and U can change, but E = K + U remains constant. E = K + U  E =  K +  U = W +  U = W + (-W) = 0 using  K = W using  U = -W

Physics 111: Lecture 11, Pg 25 Example: The simple pendulum l Suppose we release a mass m from rest a distance h 1 above its lowest possible point. ç What is the maximum speed of the mass and where does this happen? ç To what height h 2 does it rise on the other side? v h1h1 h2h2 m

Physics 111: Lecture 11, Pg 26 Example: The simple pendulum l Kinetic+potential energy is conserved since gravity is a conservative force (E = K + U is constant) l Choose y = 0 at the bottom of the swing, and U = 0 at y = 0 (arbitrary choice) E = 1 / 2 mv 2 + mgy v h1h1 h2h2 y y = 0

Physics 111: Lecture 11, Pg 27 Example: The simple pendulum l E = 1 / 2 mv 2 + mgy. ç Initially, y = h 1 and v = 0, so E = mgh 1. ç Since E = mgh 1 initially, E = mgh 1 always since energy is conserved. y y = 0

Physics 111: Lecture 11, Pg 28 Example: The simple pendulum l 1 / 2 mv 2 will be maximum at the bottom of the swing. l So at y = 0 1 / 2 mv 2 = mgh 1 v 2 = 2gh 1 v h1h1 y y = h 1 y = 0

Physics 111: Lecture 11, Pg 29 Example: The simple pendulum l Since E = mgh 1 = 1 / 2 mv 2 + mgy it is clear that the maximum height on the other side will be at y = h 1 = h 2 and v = 0. l The ball returns to its original height. y y = h 1 = h 2 y = 0

Physics 111: Lecture 11, Pg 30 Example: The simple pendulum l 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 / 2 mv 2 + mgy = K + U = constant. y Bowling

Physics 111: Lecture 11, Pg 31 Example: The simple pendulum l 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 / 2 mv 2 + mgy. v h1h1 h2h2 y y = 0

Physics 111: Lecture 11, Pg 32 Example: The simple pendulum l E = 1 / 2 mv 2 + mgy. ç Initially, y = 0 and v = 0, so E = 0. ç Since E = 0 initially, E = 0 always since energy is conserved. y y = 0

Physics 111: Lecture 11, Pg 33 Example: The simple pendulum l 1 / 2 mv 2 will be maximum at the bottom of the swing. l So at y = -h 1 1 / 2 mv 2 = mgh 1 v 2 = 2gh 1 v h1h1 y y = 0 y = -h 1 Same as before!

Physics 111: Lecture 11, Pg 34 Example: The simple pendulum l Since 1 / 2 mv 2 - mgh = 0 it is clear that the maximum height on the other side will be at y = 0 and v = 0. l The ball returns to its original height. y y = 0 Same as before! Galileo’s Pendulum

Physics 111: Lecture 11, Pg 35 Example: Airtrack & Glider l A glider of mass M is initially at rest on a horizontal frictionless track. A mass m is attached to it with a massless string hung over a massless pulley as shown. What is the speed v of M after m has fallen a distance d ? d M m v v

Physics 111: Lecture 11, Pg 36 Example: Airtrack & Glider l Kinetic+potential energy is conserved l Kinetic+potential energy is conserved since all forces are conservative. Choose initial configuration to have U=0.  K = -  U d M m v Glider

Physics 111: Lecture 11, Pg 37 Problem: Hotwheel l A toy car slides on the frictionless track shown below. It starts at rest, drops a distance d, moves horizontally at speed v 1, rises a distance h, and ends up moving horizontally with speed v 2. ç Find v 1 and v 2. h d v1v1 v2v2

Physics 111: Lecture 11, Pg 38 Problem: Hotwheel... K+U energy is conserved, so  E = 0  K = -  U Moving down a distance d,  U = -mgd,  K = 1 / 2 mv 1 2 l Solving for the speed: h d v1v1

Physics 111: Lecture 11, Pg 39 Problem: Hotwheel... l At the end, we are a distance d - h below our starting point.  U = -mg(d - h),  K = 1 / 2 mv 2 2 l Solving for the speed: h d v2v2 d - h

Physics 111: Lecture 11, Pg 40 Non-conservative Forces: l If the work done does not depend on the path taken, the force is said to be conservative. l If the work done does depend on the path taken, the force is said to be non-conservative. l An example of a non-conservative force is friction. çWhen pushing a box across the floor, the amount of work that is done by friction depends on the path taken. » Work done is proportional to the length of the path!

Physics 111: Lecture 11, Pg 41 Non-conservative Forces: Friction Suppose you are pushing a box across a flat floor. The mass of the box is m and the coefficient of kinetic friction is  k. F D The work done in pushing it a distance D is given by: W f = F f D = -  k mgD. D F f = -  k mg

Physics 111: Lecture 11, Pg 42 Non-conservative Forces: Friction Since the force is constant in magnitude and opposite in direction to the displacement, the work done in pushing the box through an arbitrary path of length L is just W f = -  mgL. l Clearly, the work done depends on the path taken. l W path 2 > W path 1 A B path 1 path 2

Physics 111: Lecture 11, Pg 43 Generalized Work/Energy Theorem: l Suppose F NET = F C + F NC (sum of conservative and non- conservative forces). l The total work done is: W NET = W C + W NC The Work/Kinetic Energy theorem says that: W NET =  K.  W NET = W C + W NC =  K  W NC =  K - W C But W C = -  U So W NC =  K +  U =  E

Physics 111: Lecture 11, Pg 44 Generalized Work/Energy Theorem: l The change in kinetic+potential energy of a system is equal to the work done on it by non-conservative forces. E=K+U of system not conserved!  If all the forces are conservative, we know that K+U energy is conserved:  K +  U =  E = 0 which says that W NC = 0, which makes sense.  If some non-conservative force (like friction) does work, K+U energy will not be conserved and W NC =  E, which also makes sense. W NC =  K +  U =  E

Physics 111: Lecture 11, Pg 45 Problem: Block Sliding with Friction A block slides down a frictionless ramp. Suppose the horizontal (bottom) portion of the track is rough, such that the coefficient of kinetic friction between the block and the track is  k. ç How far, x, does the block go along the bottom portion of the track before stopping? x d  k

Physics 111: Lecture 11, Pg 46 Problem: Block Sliding with Friction... Using W NC =  K +  U As before,  U = -mgd W NC = work done by friction = -  k mgx.  K = 0 since the block starts out and ends up at rest. W NC =  U-  k mgx = -mgd x = d /  k x d kk

Physics 111: Lecture 11, Pg 47 Recap of today’s lecture l Work done by variable force in 3-D (Text: 6-1) ç Newton’s gravitational force (Text: 11-2) l Conservative Forces & Potential energy (Text: 6-4) l Conservation of “Total Mechanical Energy” (Text: 7-1) ç Examples: pendulum, airtrack, Hotwheel car l Non-conservative forces (Text: 6-4) ç friction l General work/energy theorem (Text: 7-2) l Example problem l Look at Textbook problems Chapter 7: # 9, 21, 27, 51, 77