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Calculate the Tension:

Calculate the Tension:. Fy= ma =..?. 0. T = mg/ cos  = 620*10/cos40. T cos  - mg = 0. Calculate the Drag:. Fx= ma =..?. Draw FBD. T sin  - D = 0. T. D = T sin  = mg sin / cos  = mg tan. D. Can  ever reach 90 º ? . Only if drag became infinitely large!. W.

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Calculate the Tension:

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  1. Calculate the Tension: Fy= ma =..? 0 T = mg/ cos  = 620*10/cos40 T cos  - mg = 0 Calculate the Drag: Fx= ma =..? Draw FBD T sin  - D = 0 T D = T sin  = mgsin /cos  = mg tan D Can  ever reach 90º ? Only if drag became infinitely large! W

  2. What speed will allow a given mass to revolve at a given angle? More speed should increase the angle. Derive a formula relating speed to angle in the situation below.

  3. T should depend on …… ……M and . Now derive an equation Reality check: as m increases, T increases. As  increases…. Fy= ma =..? 0 T cos  - mg = 0 T = mg/ cos  …cos decreases and T increases What speed will allow a given mass to revolve at a given angle? More speed should increase the angle. Derive a formula. Fx= ma =..? …mv2/r T = mv2/r sin  = mv2/L sin2 Tsin = mv2/r R = Lsin  V = (TL sin2 /m)

  4. How big a  is needed to round the curve without skidding? Depends… ..on r, v, m “radius of curvature” Derive a formula for  = f(m,v,r) Fy= ma Fx= ma =..? …mv2/r f = mv2/r = N But I don’t want N as a variable… N-mg = 0 f = mv2/r = mg N = mg  = v2/rg What questions can you do to reality check this equation? High v and tight turns (small r) * What v and r require the highest coef of friction? * Is the equation dimensionally consistent? Yes! No unit = (m/s)2/ m(m/s2)

  5. What’s the advantage to banking a roadway at the turns? The normal force actually provides a centripetal component Yes, but it wouldn’t be car racing…..it would be the luge! Is this possible without friction? But the luge is progressively banked. If this were a car on ice at constant we will see that only one possible speed will maintain equilibrium. Go too fast and you slide up and over….too slow and you slide down hill! Let’s first do the unreal frictionless analysis…then the more real case of friction adding a second centripetal force to keep the car in the circle.

  6. Find the velocity necessary to stay on track in terms of m,g, r and  : Fc= mv2/r Nsin = mv2/r Fy=0 Now get N out of equation and introduce m Ncos - mg = 0 Nsin = mv2/r N = mg/cos How would introducing friction change the resulting v needed to stay on track? mg sin /cos = mv2/r mg tan = mv2/r rmg tan/m= v2 A new  term would lower the v needed to stay on track. rg tan = v

  7. N f Fy=0 mg Ncos + f sin - mg = 0 Fc = mv2/r And f = N Nsin - f cos = mv2/r Nsin - N cos = mv2/r Ncos + N sin - mg = 0 Factor out N, then solve for N, then substitute N away N( cos +  sin) - mg = 0 N (sin -  cos) = mv2/r v = r N (sin -  cos)/m N = mg/ ( cos +  sin) v = {r mg { (sin - cos)/m( cos +  sin)} Wow !!!

  8. v = {r g{ (sin - cos)/( cos +  sin)} Reality check: This equation should, when frictionless, reduce to rg tan = v It does! So we trust are new equations because they are MATHEMATICALLY CONSISTANT with the other equations we have developed so far. Now use this awesome new equation to calculate the minimum speed that a car would have to go to avoid slipping on a curve of 10 meter radius if the μ between the tire and rod were 0.5 and the bank were 30⁰. v = {10x10{ (sin30 – 0.5 cos30)/( cos30 + 0.5 sin30)} v = {10 { (.5 – 0.433)/( 0.866+ 0.25)} = 10 { (.067)/(1.116)} = 2.4 m/s

  9. Toy plane circling on a string at constant speed r Lift is normal to the wing’s lower surface. Create a formula relating the tension in the string to speed, m, Flift, & R

  10. Toy plane circling on a string at constant speed Lift is normal to the wing’s lower surface. r Create formulas relating speed, m, Flift,T,  & R: Fy= ma =..? 0 Fx= ma =..? …mv2/r Flift cos 2 – mg – T sin 1= 0 T cos 1 + Flift sin 2 = mv2/r

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