Circular Motion

An object moving in a circle, either with uniform or non-uniform speed, is accelerating.  Since it is accelerating, it must be acted on by a force. 

Under different circumstances, what is the force?

Let us solve some problems investigating this question.

Problems:

A 3 kg mass attached to a light string rotates on a horizontal frictionless table.  The radius of the circle is 0.8m and the string can support a mass of 25kg before breaking.  What range of speeds can the mass have before the string breaks?
Solution:

A mass attached to a string rotates on a horizontal, frictionless table.

We assume that the mass rotates with uniform speed.  It is accelerating.  The direction of the acceleration is towards the center of the circle, and its magnitude is v2/r.  There is tension T in the string.  The string pulls on the mass with a force T directed towards the center of the circle.  This force T is responsible for the centripetal acceleration, T = mv2/r. 
The string can support a mass of 25kg before breaking, i.e. we can let a mass of up to 25kg hang from the string near the surface of the earth.  The maximum tension in the string therefore is 25kg´9.8m/s2 = 245N.

Given Tmax = 245N and T = mv2/r, we can find vmax.

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A coin placed 30 cm from the center of a rotating, horizontal turntable slips when its speed is 50cm/s.
(a) What force provides the centripetal acceleration when coin is stationary relative to the turntable?
(b) What is the coefficient of static friction between coin and turntable?
Solution:
(a) When the coin is at rest relative to the rotating turntable, the force of static friction between the coin and the turntable provides the centripetal acceleration.
(b) The magnitude of the maximum force of static friction is fs = msN.  This maximum force of static friction is equal to mv2/r when v = 0.5m/s.  We have msN = msmg = mv2/r,
or ms = v2/(rg) = (0.5m/s)2/(0.3m 9.8m/s2) = 0.085.
Consider a conical pendulum with a 80kg bob on a 10m wire making an angle of q = 5o with the vertical.  Determine
(a) the horizontal and vertical component of the force exerted by the wire on the pendulum and
(b) the centripetal acceleration of the bob.

Solution:
(a) A free-body diagram of the bob is shown.

The bob does not change its vertical position, y = constant, vy = ay = 0.  The vertical component of T must have magnitude mg.
Tcos(5o) = mg,  T = (80kg9.8m/s2)/cos(5o) = 787N
The magnitude of horizontal component of T is Tsin(5o) = 68.6N.  The horizontal component of the force points towards the center of the circle.

(b) The horizontal component of T provides the centripetal (radial) acceleration ar.
Tsin(5o) = mar, ar = 68.6N/80kg = 0.857m/s2.
The speed of the bob is found from ar = v2/r, . 
Since r = 10m´sin(5o), we have v = 0.86m/s.

An 1800kg car passes over a hump in a road that follows the arc of a circle of radius 42m.


(a) What force does the road exert on the car as the car passes the highest point of the hump if the car travels at 16m/s?
(b) What is the maximum speed the car can have as it passes this hump before losing contact with the road?
Solution:
(a) A Free body diagram of the car is shown.

The only forces acting on the car moving with constant speed are gravity and the normal force, which is exerted by the road.  If these forces are equal in magnitude, the car does not accelerate.  If the car is moving on a circular arc, then it is accelerating.  The acceleration is ar = v2/r.  The gravitational force must therefore have a larger magnitude than the normal force. We need

mg - n = mv2/r, or n = m(g - v2/r).

n = 1800kg(9.8m/s2 - (16m/s)2/42m) = 6669N

(b) The car looses contact with the road when n becomes zero.  Then the road does no longer support the car.  This happens when g - v2/r = 0, or v2 = gr = 411.6m2/s2, v = 20.3m/s.

Link:

Circular motion and the Centripetal force

Gravity

Newton's law of gravity states that any two objects with mass m1and m2, respectively, attract each other with a force proportional to the product of their masses and inversely proportional to the square of the distance R between them.

The proportional constant G = 6.67´ 10-11Nm2/kg2 is a constant of nature.  The point in an object from which the distance R is measured is its center of mass.  Mass m1 pulls on mass m2, and mass m2 pulls on mass m1.  The center of mass of each object is pulled towards the center of mass of the other object.

Weighing the earth

The force of gravity on an object of mass m on the surface of the earth is F = mg.  Using Newton's law of gravity we write GMm/R2 = mg, where M is the mass of the earth.  We therefore have M = gR2/G.

The radius of the earth is R = 6.4´106 m.  Therefore

M = (9.8 m/s2)(6.4´106m)2/(6.67´10-11Nm2/kg2) = 6´1024kg.

Orbiting

The gravitational attraction between an object and the earth pulls the object towards the center of the earth.  When an object circles the earth, the direction of the gravitational force on the object constantly changes.  The radius of the earth is so large, that the earth appears locally flat to an observer standing on the surface.  When a problem involves only distances which are much smaller than the radius of the earth we often neglect the curvature of the earth’s surface and assume that the gravitational force points in the same downward direction everywhere.

Link: 

The satellite as a projectile

Assume that near the surface of the earth an object is thrown in the x-direction as shown in the figure above.  Initially it accelerates only in the y-direction.  But as the object moves, the direction of the acceleration changes.  If the objects initial speed is high enough, we have to take the change in the direction of the force into account when calculating the objects trajectory.  An object in a circular orbit around the earth is constantly falling towards the center of the earth.  It is constantly accelerating.  But while it moves on a curved trajectory, the surface of the earth curves away from the object so that the distance between the earth and the object stays constant.

The force of gravity always points towards the center of the object's circular orbit and is responsible for the centripetal acceleration of the object.

F = mv2/r

For an object near the surface of the earth F = mg and r = 6.4´106 m.  The speed of the orbiting object is found from mg = mv2/r, v2 = gr = (9.8m/s2)(6.4´106m).  We have v = 7919m/s, or approximately 8000m/s.  It takes the object t = 2pr/v = 6.28´6.4´106m/(7919 m/s) = 5075s = 84min to complete an orbit.

If the same object moved in a circular orbit with a larger radius, the force of gravity on the object would be smaller.  As we double the distance from the center of the earth the force of gravity decreases by a factor of 1/4.  The centripetal acceleration v2/r decreases by a factor of 1/4.  This means that v2 must decrease by a factor of 1/2 .  We have v = 5600m/s and it take 14355s = 240min to complete an orbit.

Objects in geo-synchronous orbits complete an orbit in 24 hours or 86400s.  Their speed is therefore is v = 2pr/86400s.  Writing GMm/r2 = mv2/r = m(2pr/86400s)2/r, or r3 = GM(86400s)2/4p2, we can solve this equation for the radius of the geo-synchronous orbit.  With M = 6´1024kg we have r = 42260km.  A geo-synchronous satellite orbits about 42260km - 6400km = 35860km above the surface of the earth.  The radius of its orbit is 6.6 times the radius of the earth.

The moon orbits the earth once every 27.3 days.  We can find the distance to the moon in the same way we found the distance to a geo-synchronous satellite.  The earth-moon distance is 384400km.

Problem:

When a falling meteor is a distance above the earth's surface of 3 times the Earth's radius, what is its free-fall acceleration due to the gravitational force exerted on it?
Solution:

The force on the meteor is F = ma = GMm/r2.
Therefore a = GM/r2, a = (6.67´10-11Nm2/kg2)´(6´1024kg)/(4´6.4´106 m)2 = 0.61m/s2.
(The meteor is 3 earth radii above the earth surface, so it is 4 earth radii from its center.)

Links to other Web materials:

Circular Motion

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