More Centripetal Force Examples Answers 1. A car of mass 1800 kg is traveling at 15 m/s around a curve of radius 35 m. The coefficient of friction between the tires and road is 1.0. Determine if the car successfully rounds the curve. m = 1800 kg v = 15 m/s r = 35 m μ = 1.0 For the car to round the curve, a centripetal force is required ; friction provides the force necessary for cars to change direction Is there enough friction in this case?

m = 1800 kg v = 15 m/s r = 35 m μ = 1.0 Centripetal force needed: m v2 ( 1800 kg )( 15 m/s )2 Fc = = r 35 m Fc = 11 571 N Friction force present: Ff = μ FN = μ m g = ( 1.0 )( 1800 kg )( 9.8 m/s2 ) Ff = 17 640 N

Ff = 17 640 N Fc = 11 571 N Ff > Fc Friction force is greater than the centripetal force There is more than enough friction force available for centripetal force, so car successfully rounds the curve

2. A motorcycle and rider have a combined mass of 400 kg and wish to traverse a curve of radius 25 m. If the coefficient of friction between the tires and road is 0.80, what is the maximum speed the motorcycle can go and still successfully negotiate the curve? Centripetal force provided by friction Friction present: Ff = μ FN = μ m g = ( 0.80 )( 400 kg )( 9.8 m/s2 ) Ff = 3136 N So 3136 N available for centripetal force Fc = 3136 N

Fc = 3136 N m = 400 kg r = 25 m m v2 r r Fc = r m v2 = r Fc m m r Fc v2 = m r Fc ( 25 m )( 3136 N ) v = = m 400 kg v = 14 m/s

3. Show that the maximum speed to get around a curve is independent of mass—that is, a dump truck has the same maximum speed as a motorcycle, assuming the coefficient of friction is the same for both vehicles. Set centripetal force = friction force Fc = Ff m v2 Mass term cancels out = μ m g r r v2 = μ g r r v2 = μ g r Max speed to negotiate curve of radius r Independent of mass v = μ g r

Apply to Example 2 : m = 400 kg r = 25 m μ = 0.80 v = μ g r = ( 0.80 )( 9.8 m/s2 )( 25 m ) v = 14 m/s Same result as earlier

4. A box of mass 12 kg is on the bed of a flatbed truck. It is unsecured. The coefficient of friction between the box and truck bed is 0.40. The truck turns a corner of radius 40 m at a speed of 25 mph. Does the box stay on the truck, or does the box fly off? m = 12 kg r = 40 m μ = 0.40 v = 25 mph v = μ g r = ( 0.40 )( 9.8 m/s2 )( 40 m ) v = 12.5 m/s This is the max speed the truck can have and still have the box stay on the truck Is 12.5 m/s greater than or less than 25 mph? Need to convert 25 mph to m/s

Conversion factor: 1 mi/hr = 0.447 m/s look it up! 25 mi/hr 0.447 m/s = 11.2 m/s 1 mi/hr So truck (and box) are moving at 11.2 m/s ; max speed for the box to stay on truck is 12.5 m/s Box stays on truck