Circular Motion

uniform circular motion

an object rotating at a steady rate is said to have uniform circular motion. consider a point on the perimetre of a wheel or radius r with uniform circular motion. the circumference of the wheel = 2πr. the frequency of rotation = 1 / T where T is the time period for one rotation.

the speed ( v ) therefore at one point on the perimeter is circumference / time for one rotation = 2πr / T OR 2πrf or the angular speed can also be written as w = 2π / T or 2πf

for any object in uniform circular motion the object turns through an angle of 2π / T radians per second and therefore the angular displacement of the object per second is 2π / T. the angular displacement then at time t is classed as 2πt / T or 2πft. this can also be known as the ‘angle it turns through’

the angular speed ( w ) at this point is the angular displacement / time taken or 2π / T as displacement per second is equal to speed and is measure in rads^-1

centripetal acceleration

when an object is moving in a circle at a constant speed the velocity of the object is constantly changing direction. this changing velocity means that the object is accelerating. the path of the velocity at any point in the circular motion, is along the tangent to the circle at that point. the change in velocty is towards the centre of the circle therefore the acceleration of the object is towards the centre of the circle. this is the centripetal acceleration.

for an object moving at constant speed ( v ) in a circle of radius ( r ) its centripetal acceleration = v² / r or centripetal acceleration = w²r

centripetal force

to make an object move around on a circular path it must be acted upon by a resultant force that changes its direction of motion this resultant force is called the centripetal force as it acts in the same direction as centripetal acceleration which is towards the centre of the circle.

examples of this is an object moving in a circle by a string, most the centripetal force acting would be the tension of the string. as well as this a satellite moving around earth’s centripetal force would be gravity.

for an object moving at constant speed and where is acceleration = v² / r or w²r ( where w = v / r ) therefore applying newtons second law f = mv² / r or f = mw²r

on the road

over the top of a hill

to make any object move on a circular path it must be acted upon by a resultant force acting towards the centre of curvature of the path.

consider a vehicle of mass m moving at speed v along a road that passes the top of a hill of radius r. the support force from the road onto the vechile ( S ) will be acting upwards in the opposite direction of the weight ( mg ). the resultant force on the vechile will be the difference between these two forces. the difference will normally act towards the centre of the curvature of the hill therefore we can say:

mg - S = mv² / r

in this situation the car will loose contact with the road when the velocity passes a certain threshold eg: v0 when this happens the support force is zero therefore mg = mv₀² / r crossing out the masses we get v₀² = gr.

this speed should not be exceeded otherwhise the car will loose contact with the ground.

on a roundabout

consider a vehicle of mass m running at speed v on a roundabout with a radius r. the centripetal force in this situation is provided by friction between the cars tires and the road, in other words:

force of friction F = mv2 / r

for no skidding or sliding the force of friction must be below a limiting value F₀ that is proportional to the vehicles weight in addition the speed must also be below a certain value of v₀ therefore we can say that f₀ = mv₀² / r

at the fairground

the big dipper

on a ride that takes you at high speed through a big dip the difference between the support force and the force of weight provides the centripetal force. at the bottom of the dip the support force acts upwards and weight acts downwards. the support force here acts as the centripetal force it is the bottom of a circle. for a cart of speed v with a ride of radius r we can therefore state that

s - mg = mv² / r

this can be rearranged to s = mg + mv² / r. therefore we can say that the extra force added via circular motion is mv² / r.

the very long swing

when the swing is at the bottom all kinetic energy will have be tranfered to gravitational potential energy ( if its a perfecr system ) and therefore we can say 1/2mv² = mgh and therefore v² = 2gh

when there a rope of radius ( r ), the support force ( s ) ats towards this rop upwards and in opposite direction to weight ( mg ). s - mg therefore is the resultant centripetal force. therefore s - mg = mv² / r using subsitition this is s - mg = m2gh / r

the big wheel

at the maximum height on a large wheel the reaction force ( R ) and the weight ( mg ) act in the same direction and therefore the centripetal force is the sum of these two. therefore

R + mg = mv² / r