Centripetal Force

Law of Action in Circles  Motion in a circle has a centripetal acceleration.  For every acceleration there is a net force.  There must be a centripetal force. Points to the center of the circlePoints to the center of the circle Magnitude is ma = mv 2 /rMagnitude is ma = mv 2 /r  The centrifugal force that we describe is just inertia. It points in the opposite direction – to the outsideIt points in the opposite direction – to the outside It isn’t a real forceIt isn’t a real force

Conical Pendulum  A 200. g mass hung is from a 50. cm string as a conical pendulum. The period of the pendulum in a perfect circle is 1.4 s. What is the angle of the pendulum? What is the tension on the string? FTFT 

Radial Net Force  The mass has a downward gravitational force, -mg.  There is tension in the string. The vertical component must cancel gravityThe vertical component must cancel gravity  F Ty = mg  F T = mg / cos   Tension: F T = mg / cos  = 2.0 N  Centripetal force:  F Tr = mg sin  / cos  = mg tan  mg FTFT F T cos  F T sin  

Acceleration to Velocity  The acceleration and velocity on a circular path are related. mg FTFT mg tan   r

Period of Revolution  The pendulum period is related to the speed and radius. FTFT mg tan   r L cos  =  = 13 °

Vertical Curve  A loop-the-loop is a popular rollercoaster feature.  There are only two forces acting on the moving car. GravityGravity Normal forceNormal force  There is a centripetal acceleration due to the loop. Not uniform circular motionNot uniform circular motion FgFg FNFN

Staying on Track  If the normal force becomes zero, the coaster will leave the track in a parabolic trajectory. Projectile motionProjectile motion  At any point there must be enough velocity to maintain pressure of the car on the track. FgFg

Force at the Top  The forces of gravity and the normal force are both directed down.  Together these must match the centripetal force.  The minimum occurs with almost no normal force.  The maximum is at the bottom: a = v 2 / r. FgFg FNFN

Horizontal Curve  A vehicle on a horizontal curve has a centripetal acceleration associated with the changing direction.  The curve doesn’t have to be a complete circle. There is still a radius (r) associated with the curveThere is still a radius (r) associated with the curve The force is still F c = mv 2 /r directed inwardThe force is still F c = mv 2 /r directed inward r FcFc

Curves and Friction  On a turn the force of static friction provides the centripetal acceleration.  In the force diagram there is no other force acting in the centripetal direction. r FcFc

Skidding  The limit of steering in a curve occurs when the centripetal acceleration equals the maximum static friction.  A curve on a dry road (  s = 1.0) is safe at a speed of 90 km/h.  What is the safe speed on the same curve with ice (  s = 0.2)? 90 km/h = 25 m/s r dry = v 2 /  s g = 64 m v 2 icy =  s g r = 120 m 2 /s 2 v icy = 11 m/s = 40 km/h

Banking  Curves intended for higher speeds are banked.  Without friction a curve banked at an angle  can supply a centripetal force F c = mg tan .  The car can turn without any friction. next