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

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.

T should depend on …… ……M and . Now derive an equation  Fy= ma =..? 0 T cos  - mg = 0 T = mg/ cos  Reality check: as m increases, T increases. As  increases…. …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 =..? …mv 2 /r Tsin  = mv 2 /rR = Lsin  T = mv 2 /r sin  = mv 2 /L sin 2  V =  (TL sin 2  /m)

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

What’s the advantage to banking a roadway at the turns? Is this possible without friction? Yes, but it wouldn’t be car racing…..it would be the luge! 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! The normal force actually provides a centripetal component 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.

 Fc= mv 2 /r Nsin  = mv 2 /r Find the velocity necessary to stay on track in terms of m,g, r and  : Now get N out of equation and introduce m  Fy=0 Ncos  - mg = 0 N = mg/cos  Nsin  = mv 2 /r mg sin  /cos  = mv 2 /r mg tan  = mv 2 /r rmg tan  /m = v 2  rg tan  = v How would introducing friction change the resulting v needed to stay on track? A new  term would lower the v needed to stay on track.

f N mg  Fc = mv 2 /r Nsin  - f cos  = mv 2 /r  Fy=0 Ncos  + f sin  - mg = 0 And f =  N Nsin  -  N cos  = mv 2 /rNcos  +  N sin  - mg = 0 Factor out N, then solve for N, then substitute N away N( cos  +  sin  ) - mg = 0 N = mg/ ( cos  +  sin  ) N (sin  -  cos  ) = mv 2 /r v =  r N (sin  -  cos  )/m v =  {r mg { (sin  -  cos  )/m ( cos  +  sin  )} Wow !!!

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 30 ⁰. 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. v =  {10x10{ (sin30 – 0.5 cos30)/( cos sin30)} v = {10  { (.5 – 0.433)/( )} = 10  { (.067)/(1.116)} = 2.4 m/s

v =  {r g/ ( cos  +  sin  )}{ (sin  -  cos  )}

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

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