PHYS 2211 Module 6.3

Centripetal Force

Recommended Reading

6.3 Centripetal Force

Learning Objectives

By the end of this section, you will be able to:

  • Explain the equation for centripetal acceleration
  • Apply Newton’s second law to develop the equation for centripetal force
  • Use circular motion concepts in solving problems involving Newton’s laws of motion

Centripetal Force

In module 4, we discussed centripetal acceleration — the acceleration that arises when a velocity vector changes direction. The magnitude of the centripetal acceleration can be expressed as:

where v is the speed and r is the radius of the circular path.

Every acceleration is produced by a net force. So the centripetal acceleration is produced by a net centripetal force. Any net force causing uniform circular motion is called a centripetal force, Fc. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration.

Discuss!

Consider how you would answer these questions. Then bring this to class for a group discussion.

A new package moving system in the new, improved post office consists of a large circular disc (i.e. turntable) which rotates once every 3.0 seconds at a constant speed in the horizontal plane. Packages are put on the outer edge of the turntable on one side of the room and taken off on the opposite side. The coefficient of static friction between the disc surface and the package is 0.80 while the coefficient of kinetic friction is 0.60. If this system is to work, what is the maximum possible radius of the turntable?

A car of mass m is driving with constant speed v through a valley that has a circular shape with radius R. Consider the situation at the instant the car is at the bottom of the curve. Is the car accelerating? Is there a net force? What forces act on the car? Which force is bigger? Solve for the normal force of the road on the car in terms of m, v, g, and R.

A car of mass m is driving with constant speed v over a hill that has a circular shape with radius R. Consider the situation at the instant the car is at the top of the hill. Is the car accelerating? Is there a net force? What forces act on the car? Which force is bigger? Solve for the normal force of the road on the car in terms of m, v, g, and R.

Keeping Mars in Orbit

Although the planet Mars orbits the Sun in a Kepler ellipse with an eccentricity of 0.09, we can approximate its orbit by a circle. If you have faith in Newton’s laws then you must conclude that there is an invisible centripetal force holding Mars in orbit. The data on the orbit of Mars about the Sun are shown below:

  • msun = 2.00 × 1030 kg
  • mMars = 6.42 × 1023 kg
  • dMars = 2.28 × 1011 m = distance from the sun (= radius of circular orbit)
  • v = 24.13 km/s (mean orbital speed of Mars)

What is the centripetal force needed to hold Mars in orbit? What is the direction of that force? What is the most likely source of this force? Is there anything in common with the force that attracts objects to Earth?

Practice!

Practice 6.3.1

A pendulum consists of a small object called a bob hanging from a light string of fixed length, with the top end of the string fixed, as represented in the figure. The bob moves without friction, swinging equally high on both sides. It moves from its turning point A through point B and reaches its maximum speed at point C.  Of these points, is there a point where the bob has nonzero centripetal acceleration and zero tangential acceleration?
(a) Yes, point A
(b) Yes, point B
(c) Yes, point C
(d) No
Practice 6.3.2

Of these points, is there a point where the bob has nonzero tangential acceleration and zero centripetal acceleration?
(a) Yes, point A
(b) Yes, point B
(c) Yes, point C
(d) No
Practice 6.3.3

Is there a point where the bob has no acceleration? If so, which point?
(a) Yes, point A
(b) Yes, point B
(c) Yes, point C
(d) No