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LecturesWeek6 - Today Finish examples with Potential Energy...

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Today .... • Finish examples with Potential Energy • Power ....... didn’t get to it last lecture • that finishes Chapter 7. 1
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Ex. Circular “Loop-the-Loop” From what height H should the car be released at rest in order to just barely make it around the loop without falling from the track? (Neglect friction, drag, and the size of the car.) R H Worked out on transparency Answer: H= 5 R/2. For R=15cm, H=37.5cm In this example we needed both Energy ideas and F=ma.
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Ex. If a sliding friction force of constant magnitude f k is present, what is the block’s speed just as it leaves the spring after being compressed a distance D? D k m Q: Suppose 2f k D > kD 2 . How should this be interpreted physically? v = kD 2 - 2f k D m f k = μ k mg A: The block stops before the spring returns to its relaxed length. (on transparency)
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Bungee jumping problem You want to bungee-jump from a 500m bridge. You know that a typical bungee cord has a spring constant of about 100 N/m. Neglecting air resistance and friction, estimate the length of the bungee cord you should use. 500m DEMO: Bungee
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L H m k Ex. Bungee Jumping — Body of mass m hanging from an ideal (F = kx) Bungee cord of relaxed length L and spring constant k . K U g U s y = 0 0 1 2 mv 2 2 v 2 mgH 0 mg(H - L) 0 0 0 1 2 k(H-L) 2 (Neglect the size of the body.) Jump! 1 2 3 2 parameters to be determined for each body m : L & k total drop = H Stretch! Stop! DEMO: Bungee
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Bungee jumping problem You want to bungee-jump from a 500m bridge. You know that a typical bungee cord has a spring constant of about 100 N/m. Neglecting air resistance and friction, estimate the length of the bungee cord you should use. 500m Equating total energies at (1) and (3) gives: mgH = 1/2 k (H-L) 2 for L given: H = 500m, k=100N/m, m=60 kg, Answer = 423 m.
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Stability and the shape of the potential One can learn a lot just from the shape of the potential energy U as a function of x. For example: Could represent the force between two molecules, for instance. U(x) x
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shape of potential ......... To gain intuition: consider ball rolling on a curved track, where the curve is given by a function y(x). Since U(x) = mg y(x), The shape of the potential is identical to the shape of the track! Demo: Stein’s bomb. Can easily answer questions such as: How much energy E does it take to get out of the valley: U(x) x E
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Roller Coasters — Gravity Machines Lift Hill Δ K + Δ U g = W by friction & drag U g = mgy Track is a graph of gravitational PE vs. position. After Lift Hill: 0 over short distances
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Equilibrium Equilibrium is reached when no force is acting on the body = 0 Stable Equilibrium : any minimum in a potential energy curve Unstable Equilibrium : any maximum in a potential energy curve F x ( x ) = dU ( x ) dx
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Equilibrium • Stable Unstable Neutral U(x) F (x) x x x 0 > 0 < 0 U(x) F (x) x x x 0 > 0 < 0 U(x) F (x) x x x 0 F = 0 x
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Equilibrium U(x) x Spring Gravitation U(x) y F(x) x F(x) y 1/2 k x 2 -k x mgy -mg No Equilibrium
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h H ?!?
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