Discussion Problems Thursday Feb 19 Circular motion

Circular motion relations T=period (time to complete one revolution)

Amusement park fun During the summer, you decide to visit the an amusement park. You want to ride the Ferris wheel, which has seats on the rim of a circle with a radius of 25 m. The Ferris wheel rotates at a constant speed and makes one complete revolution every 20 seconds. While waiting, you decide to calculate the acceleration and forces (both magnitude and direction) on a rider at different positions on the ride. What is the acceleration direction at each position? What must the forces (normal and weight) look like at each position?

Ferris wheel physics n w w n n w n w a aa a C DB A

A rider’s “apparent weight” is the magnitude of the normal force on the rider. It would be least –A) at the top –B) halfway down –C) at the bottom –D) halfway up –E) back on the ground “Apparent Weight”

Forces on a Ferris Wheel rider If the radius of the wheel is 25 meters and the period is 20 seconds, what would the normal force be on a rider whose weight is 600 N at: A) the highest point? B) the lowest point?

Forces on a Ferris Wheel rider If the radius of the wheel is 25 meters and the period is 20 seconds, what would the normal force be on a rider whose weight is 600 N at: A) the highest point? ma c =w-n n=w-ma c =mg-ma c a c =(2  /T) 2 r=2.5 m/s 2 m=w/g=(600 N)/(9.8 m/s 2 )=61 kg n=(61 kg)( ) m/s 2 =450 N B) the lowest point? ma c =n-w n=w+ma c =m(g+a c )= n=(61 kg)( ) m/s 2 =750 N n w a a w n

Slippery road On a trip, you find yourself driving in your 3000-lb car along a flat level road at 55 mph. The road makes a turn which you take without changing your speed. The curve is approximately an arc of a circle with a radius of 0.20 miles. You notice that the curve is flat and level with no sign of banking. There are no warning signs but you wonder if it would be safe to try to go 55 mph around the curve in the rain when the wet surface has a lower coefficient of friction. What is the minimum coefficient of static friction between the road and your car's tires which will allow your car to make the turn? Note: 1 mile=1.6x10 3 meter, 55 mph=25 meter/second On a dry road µ s is about 1.

Staying on the road x w n y fsfs n-w=0: n=w=mg f s =ma c f s,max =µ s n=µ s mg≥ma c µ s ≥a c /g a c =v 2 /r =(25 m/s) 2 /(320 m) =1.95 m/s 2 µ s ≥a c /g=1.95/9.8=0.20 acac

Slippery road 2 Curves are often “banked” to make them easier to negotiate in poor driving conditions. What banking angle would be ideal (friction not needed!) for a speed of 55 mph for the same curve (0.2 mile radius)? Assume that the car rounds the curve on a horizontal path.

Slippery road 2 Without friction, the combination of normal force and weight must produce the necessary centripetal force: x acac  w n y  n w