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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

state Hooke's law of elasticity
Aarti Reply
Hooke's law states that the extension produced is directly proportional to the applied force provided that the elastic limit is not exceeded. F=ke;
Shaibu
thanks
Aarti
You are welcome
Shaibu
thnx
Junaid
what is drag force
Junaid
A backward acting force that tends to resist thrust
Ian
solve:A person who weighs 720N in air is lowered in to tank of water to about chin level .He sits in a harness of negligible mass suspended from a scale that reads his apparent weight .He then dumps himself under water submerging his body .If his weight while submerged is 34.3N. find his density
Ian Reply
please help me solve this 👆👆👆
Ian
The weight inside the tank is lesser due to the buoyancy force by the water displaced. Weight of water displaced = His weight outside - his weight inside tank = 720 - 34.3 = 685.7N Now, the density of water = 997kg/m³ (this is a known value) Volume of water displaced = Mass/Density (next com)
Sharath
density or relative density
Shaibu
density
Ian
Upthrust =720-34.3=685.7N mass of water displayed = 685.7/g vol of water displayed = 685.7/g/997 hence, density of man = 720/g / (685.7/g/997) =1046.6 kg/m3
1046.8
R.d=weight in air/upthrust in water =720/34.3=20.99 R.d=density of substance/density of water 20.99=x/1 x=20.99g/cm^3
Shaibu
Kg /cubic meters
how please
Shaibu
Upthrust = 720-34.3=685.7N vol of water = 685.7/g/density of water = 685.7/g/997 so density of man = 720/g /(685.7/g/997) =1046.8 kg/m3
is there anyway i can see your calculations
Ian
Upthrust =720-34.3=685.7
Upthrust 720-34.3
=685.7N
Vol of water = 685.7/g/997
Hence density of man = 720/g / (685.7/g/997)
=1046.8 kg/m3
so the density of water is 997
Shaibu
Yes
Okay, thanks
Shaibu
try finding the volume then
Ian
Vol of man = vol of water displayed
I've done that; I got 0.0687m^3
Shaibu
okay i got it thanks
Ian
u welcome
Shaibu
HELLO kindly assist me on this...(MATHS) show that the function f(x)=[0 for xor=0]is continuous from the right of x->0 but not from the left of x->0
Duncan Reply
I do not get the question can you make it clearer
Ark
Same here, the function looks very ambiguous. please restate the question properly.
Sharath
please help me solve this problem.a hiker begins a trip by first walking 25kmSE from her car.she stops and sets her tent for the night . on the second day, she walks 40km in a direction 60°NorthofEast,at which she discovers a forest ranger's tower.find components of hiker's displacement for each day
Liteboho Reply
Take a paper. put a point (name is A), now draw a line in the South east direction from A. Assume the line is 25 km long. that is the first stop (name the second point B) From B, turn 60 degrees to the north of East and draw another line, name that C. that line is 40 km long. (contd.)
Sharath
Now, you know how to calculate displacements, I hope? the displacement between two points is the shortest distance between the two points. go ahead and do the calculations necessary. Good luck!
Sharath
thank you so much Sharath Kumar
Liteboho
thank you, have also learned alot
Duncan
No issues at all. I love the subject and teaching it is fun. Cheers!
Sharath
cheers!
Liteboho
cheers too
Duncan
what is the definition of model
matthew Reply
please is there any way that i can understand physics very well i know am not support to ask this kind of question....
matthew
yes
Duncan
prove using vector algebra that the diagonals of a rhombus perpendicular to each other.
Baijnath Reply
A projectile is thrown with a speed of v at an angle of theta has a range of R on the surface of the earth. For same v and theta,it's range on the surface of moon will be
Roshani Reply
0
Keshav
what is soln..
Keshav
o
Duncan
Using some kinematics, time taken for the projectile to reach ground is (2*v*g*Sin (∆)) (here, g is gravity on Earth and ∆ is theta) therefore, on Earth, R = 2*v²*g*Sin(∆)*Cos(∆) on moon, the only difference is the gravity. Gravity on moon = 0.166*g substituting that value in R, we get the new R
Sharath
Some corrections to my old post. Time taken to reach ground = 2*v*Sin (∆)/g R = (2*v²*Sin(∆)*Cos(∆))/g I put the g in the numerator by mistake in my old post. apologies for that. R on moon = (R on Earth)/(0.166)
Sharath
state Newton's first law of motion
Awal Reply
Every body will continue in it's state of rest or of uniform motion in a straight line, unless it is compelled to change that state by an external force.
Kumaga
if you want this to become intuitive to you then you should state it
Shii
changing the state of rest or uniform motion of a body
koffi
if a body is in rest or motion it is always rest or motion, upto external force appied on it. it explains inertia
Omsai
what is a vector
smith
a ship move due north at 100kmhr----1 on a River flowing be due east on at 25kmperhr. cal the magnitude of the resultant velocity of the ship.
Emmanuel Reply
The result is a simple vector addition. The angle between the vectors is 90 degrees, so we can use Pythagoras theorem to get the result. V magnitude = sqrt(100*100 + 25*25) = 103.077 km/hr. the direction of the resultant vector can be found using trigonometry. Tan (theta) = 25/100.
Kumar
103.077640640442km/h
Peter
state Newton's first law of motion
Kansiime Reply
An object continues to be in its state of rest or motion unless compelled by some external force
Alem
First law (law of inertia)- If a body is at rest, it would remain at rest and if the body is in the motion, it would be moving with the same velocity until or unless no external force is applied on it. If force F^=0 acceleration a^=0 or v^=0 or constant.
Govindsingh
how would you measure displacement in your car?
Grace Reply
what is constellation
Charles Reply
The product of a. (vector b× vector a)
Umesh Reply
I want to join the conversation
Kumaga Reply
ok
Kamal
Ok
Bishal
ok
Rohit
Two charges 1uc and 3uc are separated 4m apart. find the point on the line connecting them at which their electric field intensity balances each other
Chukwurah
hmmm
JMPSCL
a particle projected from origion moving on x-y plain passing through p&q point (9,7)(18,4)find the equation of trajectory
ali Reply
what is the equation of trajectory
Kumaga
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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