<< Chapter < Page Chapter >> Page >
By the end of the 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

In Motion in Two and Three Dimensions , we examined the basic concepts of circular motion. An object undergoing circular motion, like one of the race cars shown at the beginning of this chapter, must be accelerating because it is changing the direction of its velocity. We proved that this centrally directed acceleration, called centripetal acceleration    , is given by the formula

a c = v 2 r

where v is the velocity of the object, directed along a tangent line to the curve at any instant. If we know the angular velocity ω , then we can use

a c = r ω 2 .

Angular velocity gives the rate at which the object is turning through the curve, in units of rad/s. This acceleration acts along the radius of the curved path and is thus also referred to as a radial acceleration.

An acceleration must be produced by a force. Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge. Any net force causing uniform circular motion is called a centripetal force    . The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: F net = m a . For uniform circular motion, the acceleration is the centripetal acceleration: . a = a c . Thus, the magnitude of centripetal force F c is

F c = m a c .

By substituting the expressions for centripetal acceleration a c ( a c = v 2 r ; a c = r ω 2 ) , we get two expressions for the centripetal force F c in terms of mass, velocity, angular velocity, and radius of curvature:

F c = m v 2 r ; F c = m r ω 2 .

You may use whichever expression for centripetal force is more convenient. Centripetal force F c is always perpendicular to the path and points to the center of curvature, because a c is perpendicular to the velocity and points to the center of curvature. Note that if you solve the first expression for r , you get

r = m v 2 F c .

This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve, as in [link] .

The figure consists of two semicircles. The semicircle on the left has radius r and bigger than the one on the right, which has radius r prime. In both the figures, the direction of the motion is given as counter-clockwise along the semicircles. A point is shown on the path, where the radius is shown with an arrow pointing out from the center of the semicircle. At the same point, the centripetal force, F sub c, is shown pointing inward, in the opposite direction to that of radius arrow. The velocity, v, is shown at this point as well, and it is tangent to the semicircle, pointing left and up, perpendicular to the forces. In both the figures, the velocity is same, but the radius prime is smaller and centripetal force is larger in the figure on the right. It is noted that vector F sub c is parallel to vector a sub c since vector F sub c equals m times vector a sub c.
The frictional force supplies the centripetal force and is numerically equal to it. Centripetal force is perpendicular to velocity and causes uniform circular motion. The larger the F c , the smaller the radius of curvature r and the sharper the curve. The second curve has the same v , but a larger F c produces a smaller r ′.

What coefficient of friction do cars need on a flat curve?

(a) Calculate the centripetal force exerted on a 900.0-kg car that negotiates a 500.0-m radius curve at 25.00 m/s. (b) Assuming an unbanked curve, find the minimum static coefficient of friction between the tires and the road, static friction being the reason that keeps the car from slipping ( [link] ).

Questions & Answers

a length of copper wire was measured to be 50m with an uncertainty of 1cm, the thickness of the wire was measured to be 1mm with an uncertainty of 0.01mm, using a micrometer screw gauge, calculate the of copper wire used
Nicole Reply
What is the answer please
Mustapha
If centripetal force is directed towards the center,why do you feel that you're thrown away from the center as a car goes around a curve? Explain
Maira Reply
Which kind of wave does wind form
Matthias Reply
calculate the distance you will travel if you mantain an average speed of 10N m/s for 40 second
Abdulai Reply
hw to calculate the momentum of the 2000.0 elephant change hunter at a speed of 7.50 m/s
Kingsley Reply
how many cm makes 1 inches
Hassan Reply
2.5
omwoyo
2.54cm=1inche
omwoyo
how do we convert from m/s to km/hr
Toni Reply
When paddling a canoe upstream, it is wisest to travel as near to the shore as possible. When canoeing downstream, it may be best to stay near the middle. Explain why?
SANA Reply
Explain why polarization does not occur in sound
Nuradeen
one ship sailing east with a speed of 7.5m/s passes a certain point at 8am and a second ship sailing north at the same speed passed the same point at 9.30am at what distance are they closet together and what is the distance between them then
Kuber Reply
density of a subtance is given as 360g/cm,put it in it s.i unit form
Linda Reply
if m2 is twice of m1. find the ration of kinetic energy in COM system to lab system of elastic collision
Raman Reply
What is a volt equal to?
Clifton Reply
list and explain the 3 ways of charging a conductor
Chidimma Reply
conduction convention rubbing
Asdesaw
formula of magnetic field
Yonas Reply
why polarization does not occur in sound
Nuradeen
Integral of a vector
Rahat Reply
define surface integral of a vector?
Rahat
Practice Key Terms 6

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'University physics volume 1' conversation and receive update notifications?

Ask