CIRCULAR MOTION

Basic Rotational Quantities

The angular displancment is defined by:

3600 = 2rad1 rad = 57.30

1rev = 3600

Rev to Rad to Degree

(2)

Circular Motion

For circular motion at a constant speed v, the centripetal acceleration of the motion can be derived. Since in radian measure, (2)

Notice that the velocity vector (in blue) is constantly changing direction. Even though the magnitude (amount) of the velocity stays constant, the direction is changing. This shows that the object is accelerating. Remember that acceleration is defined as the rate of change of velocity. So any change, even if it is just the direction, is an acceleration. Another way to think about this is to consider that to change your direction of motion requires a net force, and a net force causes acceleration.  (1)

Centripetal Acceleration The centripetal acceleration expression is obtained from analysis of constant speed circular motion by the use of similar triangles. From the ratio of the sides of the triangles (2)

Centripetal Acceleration

ac = Centripetal acceleration SI: m/s2

vT = Tangential velocity or speed SI: m/s

r = Radius of object's path SI: m

= Angular velocity SI: rad/s

In order for an object to execute circular motion - even at a constant speed - the object must be accelerating towards the center of rotation

Centripetal Force

Any motion in a curved path represents accelerated motion, and requires a force directed toward the center of curvature of the path. This force is called the centripetal force which means "center seeking" force. The force has the magnitude (2)

The centripetal acceleration can be derived for the case of circular motion since the curved path at any point can be extended to a circle

Note that the centripetal force is proportional to the square of the velocity, implying that a doubling of speed will require four times the centripetal force to keep the motion in a circle. If the centripetal force must be provided by friction alone on a curve, an increase in speed could lead to an unexpected skid if friction is insufficient (2)

Centripetal Force Calculation (2)

Centripetal force = mass x velocity2 = mac

radius

The smaller the mass, the smaller the centripetal force (shown by the red vector labeled as the force of tension in the rope, FT) you will have to apply to the rope (1)

The smaller the velocity of the object, the less centripetal force you will

have to apply. (1)

The smaller the length of rope (radius), the more centripetal force you will have to apply to the rope.

Notice that the centripetal force and the centripetal acceleration are always pointing in the same direction

If you let go of the rope (or the rope breaks) the object will no longer be kept in that circular path and it will be free to fly off on a tangent.

(1)

It is conceptually better to think about the Centripetal force that is calculated from the formula as a requirement.  If you meet the requirement, then you have circular motion at the radius and speed used in the formula.  If you do not meet the requirement, then the object moves into a larger curve (which requires less force) or defaults into straight line motion (going off on a Tangent). (1)

In this animation, the "sticky" or adhesive forces from the mud to the tire tread are large enough to be the centripetal force required to keep the mud in a circular path as the tire spins. (1)

In this animation, the tire is spinning faster which means a larger centripetal force is required to keep the mud in the circular path of the tire.  The adhesive forces of the mud to the tire are not large enough to meet the requirement.   The mud begins to move into a larger circular path but as soon as it is not touching the tread then there is no force (other than gravity) and so the mud continues with the velocity it had at the instant it was no longer touching the tread.  (It went off on a tangent).  It followed Newton's first law! (1)

Centrifugal Force - The False ForceCentrifugal force does not exist...there is no such thing...it is a ghost we tend to blame odd behavior on.

Take for example this common situation.  You are riding in a car going around a curve.  Sitting on your dashboard is a cassette tape.  As you go around the curve, the tape moves to outside edge of the car.

The car tires on the road have a enough static friction to act as centripetal force which forces the car to go around the curve.  The tape on the slippery dashboard does not have enough friction to act as a centripetal force, so in the absence of a centripetal force the tape follows straight line motion. (1)

Centripetal Force Example

The string must provide the necessary centripetal force to move the ball in a circle. If the string breaks, the ball will move off in a straight line. The straight line motion in the absence of the constraining force is an example of Newton's first law. The example here presumes that no other net forces are acting, such as horizontal motion on a frictionless surface. The vertical circle is more involved. (2)

Problem 1:A 150g ball at the end of a string is swinging in a horizontal circleOf radius 0.60m. The ball makes exactly 2.00 revolutions in a secondWhat is its centripetal acceleration?

Problem 2:The moon’s nearly circular orbit about the earth has a radius of about385,000km and a period of 27.3 days. Determine the acceleration Of the moon toward the earth.

V = 2r T

Problem 3:A 1000kg car rounds a curve on a flat road of radius 50m at a speedOf 50km/h. Will the car make the turn is (a) the pavement is dry and The coefficient of static friction is 0.60 (b) the pavement is icy andThe coefficient of friction is 0.20

On a bank: FNsin = m v2

r

1. http://regentsprep.org/Regents/physics/phys06/bcentrif/default.htmCopyright © 1999-2004 Oswego City School District Regents

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2. http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html