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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 tire 0.5m in radius rotate at constant rate 200rev/min. find speed and acceleration of small lodged in tread of tire.
Tahira Reply
hmm
Ishaq
100
Noor
define the terms as used in gravitational mortion 1:earth' satellites and write two example 2:parking orbit 3:gravitation potential 4:gravitation potential energy 5:escping velocity 6:gravitation field and gravitation field strength
Malima Reply
what larminar flow
Rajab Reply
smooth or regular flow
Roha
Hii
Sadiq
scalar field define with example
Malik Reply
what is displacement
Isaac Reply
the change in the position of an object in a particular direction is called displacement
Noor
The physical quantity which have both magnitude and direction are known as vector.
Malik
good
Noor
Describe vector integral?
Malik
define line integral
Malik
Examples on how to solve terminal velocity
Louis Reply
what is Force?
Bibas Reply
ans:loading...
Lumai
the sideways pressure exerted by fluid is equal and canceled out.how and why?
Chaurasia
what is blackbody radiation
Syed Reply
Particles emitted by black holes
Lord
what is oscillation
Iwelomen Reply
Bernoulli's equation which applies a fluid flow states that P+h✓g+½PV²=k. where ✓=density. P=pressure. h=height. V= velocity. g=acceleration due to gravity. k= constant. Show that the equation is dimensionally constant and show the S.I unit for k.
amoni Reply
The nature of physics
Darshan Reply
what physics
Darshan
Physics deals with nature and its natural phenomenons
Sushant
physics explains more about mechanics and other natural phenomenon.
Ferdinand
define realitives motion
Zahid Reply
Relative motion is the calculation of the motion of an object with regard to some other moving object. Thus, the motion is not calculated with reference to the earth, but is the velocity of the object in reference to the other moving object as if it were in a static state.
Lansana
I am unable to access the mcq can someone help me with it?
Harkamal Reply
What is free fall?
Barham Reply
when an object is falling under the the influence of the earth gravitational force, the term is called free fall.
Bra
V=E½-P-½ where v; velocity, P; density and E; constant. Find dimension and it's units of E (constant)
michael Reply
ML-3
LAWAL
Practice Key Terms 6

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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