1. PHY 151: Lecture 6B 6.3 Extending Particle in Uniform Circular Motion Model (Continued)

2. PHY 151: Lecture 6B Circular Motion Other Applications of Newton’s Laws 6.3 Extending Particle in Uniform Circular Motion Model (Continued)

3. Conical Pendulum The object is in equilibrium in the vertical direction It undergoes uniform circular motion in the horizontal direction –∑F y = 0 → T cos θ = mg –∑F x = T sin θ = ma c v is independent of m

4. Motion in a Horizontal Circle The speed at which the object moves depends on the mass of the object and the tension in the cord The centripetal force is supplied by the tension The maximum speed corresponds to the maximum tension the string can withstand

5. Horizontal (Flat) Curve A car of mass 1.5 x 10 3 kg rounds a circular turn of radius 20 m The road is flat and the coefficient of static friction between the tires and the road is 0.50 How fast can the car travel without skidding?  Centripetal force is provided by static friction between the tires and the road  Car can make the turn without skidding if the centripetal force equals the maximum static friction  F c = mv 2 /r   s n =  s (mg) = mv 2 /r  v = sqrt(  s gr) = sqrt(0.50 x 9.8 x 20) = 9.9 m/s

6. Banked Curve F c = F N sin  = mv 2 /r F N cos  = mg Divide first equation by second equation tan  = v 2 /gr

7. Banked Curve - Example A car with a velocity of 30 m/s makes a turn of radius 150 m on a banked road of angle  Calculate   tan  = v 2 /gr = (30) 2 /(9.8 x 150) = 0.6122   = 31.47 0

8. Banked Curve - 2 The banking angle is independent of the mass of the vehicle If the car rounds the curve at less than the design speed, friction is necessary to keep it from sliding down the bank If the car rounds the curve at more than the design speed, friction is necessary to keep it from sliding up the bank

9. Road Hill A car goes over the top of a hill with a radius of 150 m Mass of the driver is 80 kg What normal force does driver feel?  F c = mv 2 /r  N - mg = -mv 2 /r  N = mg - mv 2 /r  N = 80(9.8 - 30 2 /180)=384 N At what speed is N = 0 0 = m(g-v 2 /r) v = sqrt(rg) = sqrt(150x9.8) v = 38.3 m/s At this speed the car leaves the road

10. Ferris Wheel - 1 The normal and gravitational forces act in opposite direction at the top and bottom of the path Categorize the problem as uniform circular motion with the addition of gravity –The child is the particle

11. Ferris Wheel - 2 At the bottom of the loop, the upward force (the normal) experienced by the object is greater than its weight

12. Ferris Wheel - 3 At the top of the circle, the force exerted on the object is less than its weight

13. Bicycle Loop - 1 Fighter plane dives towards ground and then pulls up (Position 1 in diagram) Velocity is 400 m/s Turning radius is 3600 m Mass of the pilot is 60 kg What is normal force on pilot?  F c = mv 2 /r  N – mg = mv 2 /r  N = m(g + v 2 /r)  N = 60(9.8 + 400 2 /3600) = 3254 N  Pilots weight is mg = 60(9.8) = 588 N  N = 5.53 times weight of pilot

14. Bicycle Loop - 2  What is the slowest velocity if bike is to stay on the track at position 3?  F c = mv 2 /r  -N – mg = -mv 2 /r  N + mg = mv 2 /r  v = sqrt[r(N/m + g)]  Just leave track when N = 0  v = sqrt(rg)

15. Airplane Turn - 1 Plane turns Main force is the lift of the wing Inclining plane in turn causes lift to act as normal force Concorde flies at 600 m/s Passenger can’t take normal force greater than 1.5 mg What is sharpest turn it can make?

16. Airplane Turn - 2 Find  L = 1.5 mg Lcos  = mg 1.5(mg)cos  = mg cos  = 1/1.5 = 2/3  = 48.19 0 Find r F c = mv 2 /r Lsin  = mv 2 /r 1.5(mg)sin  = mv 2 /r r = v 2 /1.5gsin  r=360000/[1.5(9.8)sin48.19]=32.9 km