Unit 5 | unit 5 kinetics of circular motion

Unit 5

Kinetics of circular motion

5.1 Introduction to Kinematics of Circular motion (no numerical on kinematics)

Introduction

The motion of a body along a circular path is known as circular motion. In circular motion the centre of rotation remains fixed. The examples of body is moving in a circular path are shafts, flywheel, pulleys etc. Rotating about their geometric axis.

5.2 Rotation with constant and variable angular acceleration

Angular velocity

It is defined as rate of change of angular displacement of body. It is denoted by .

Angular displacement is always measured in terms of angle covered by body from initial position.

Consider a body moving along a circular path as shown in figure 1

Consider a body initially at position A and after time t body moves to position B.

Let

Angular displacement=

Time taken=t

Angular velocity=

Mathematically

Angular velocity=

Angular velocity is expressed in terms of revolutions describe in one minutes.

Example if at an instant angular velocity of rotating body in N r.p.m. (i.e. revolutions per minute) the corresponding angular velocity (in radians) may found out as discussed below

One revolution/min=2πrad/min

N revolution/min=2πN rad/min

Angular velocity= =2πN rad/min

Relation between linear velocity and angular velocity

Consider a body moving in circle as shown in figure 1

Initial position of a body is at A. And after time t body moves to position B

Let

Angular velocity=

Let

v=linear velocity=

Linear displacement=arc AB

Angular acceleration:-

It is defined as the rate of change of angular velocity. It is measured as radians per second hand written as

It is denoted as

Mathematically,

=Rate of change of angular velocity

Also,

angular acceleration

Relation between linear acceleration and angular acceleration

We know that

Differentiating above equation with respect to t

But =Linear acceleration=a

=angular acceleration=

Linear acceleration is equal to r times the angular acceleration

Equations of motion along circular path consider a body which is moving along a circular path

Let

=Initial angular velocity of body in radians per second

=Final angular velocity of body in radians per second

t=time in second during which angular velocity changes from

=Angular displacement in radians

=Angular acceleration in

We know that

=Rate of change of angular velocity

We know that

Average angular velocity=

Angular displacement= =average angular velocity×time

Final angular velocity is given by

Angular displacement

Relation between linear motion and angular motion

SR.No.	Particulars	Linear Motion	Angular Motion
1	Initial velocity	u
2	Final velocity	v
3	Constant acceleration	a
4	Total distance traversed	s
5	Formula for final velocity	v=u+at
6	Formula for distance traversed
7	Formula for final velocity
8	Differential formula for velocity
9	Differential formula for acceleration

Motion of rotation of a body under variable angular acceleration

We discussed the motion under constant acceleration but in actual practice it is seldom possible that body move with constant acceleration i.e. rate of change of velocity is constant at t times.

A body which does not move with constant acceleration is said to be moving with non uniform or variable acceleration.

Angular velocity and angular acceleration at any instant

Consider a body moving in circular motion from position A to position B in time t second

Let

=Angular displacement in radians

t=Time taken by body to move from porition A to B

=Angular velocity

=Angular acceleration

We know that,

Angular velocity is the rate of change of angular displacement of a body.

Mathematically,

Similarly,

Angular acceleration is rate of change of angular velocity of a body.

Mathematically

Also

Methods for angular velocity, angular acceleration and angular displacement from mathematical expression

Angular velocity and angular acceleration and angular displacement from mathematical expression may be found out by either of following two methods

1) By differentiation

2) By integration

Angular velocity and angular acceleration by differentiation

Sometimes the given equation of motion is in terms of angular displacement and time t.

Example:-

Mow, differentiating both sides of equations w.r.t. t

The equations so obtained by differentiation give angular velocity of the body.

Again differentiating both sides of above equations

The equations so obtained by second differentiation gives angular acceleration of the body.

By integration:- by integration the given angular acceleration equation gives angular velocity and again integrating gives angular displacement.

Centripetal acceleration

Consider a particle moving along the circumference of a circle with as a centre as shown in figure 3(a)

Let

r=radius of a circular path in m

v=linear velocity of particle in m/s

Now consider and instant when A the position of particle. After a small interval of time. Let its portion is changed from A to B. When body is had a its linear velocity will be along dotted line AD. i.e tangent at A. Similarly when body is at B its linear velocity (again equal to v) will be along the dotted line BE i.e. tangent at B as shown in figure 3 a

First of all, take any suitable point O draw on parallel to AD and let it represent the velocity v at A in a direction and magnitude to same scale. Similarly draw ob parallel to BE let it represent the velocity v at B in a direction and magnitude to scale join ab as shown in figure to 2 b

Now in triangle oab, oa represents initial velocity, ob represent final velocity and AB represent change in velocity

We know,

Acceleration=a=

Since is considered to be very small

Therefore chord AB may be considered to be equal to arc AB. Now consider two triangles AOB and aob which are similar.

Substituting values of AB= , oa=v and OA=r in above equation

Note:- The same formula holds good for centrifugal acceleration also.

5.3 Centripetal and centrifugal force

Centripetal force:-

A body moving in circle or along a circular path with constant velocity suffers a continuous change in its direction at every point of its motion but magnitude of its speed remains constant since velocity is vector quantity i.e. magnitude and direction both are considered and velocity of body is continuously changing due to change in direction; therefore according to Newton’s first law of motion an external force must act continuously upon body to produce a change in direction of moving body.

Due to inertia body tends to move along the tangent at every point of its motion with constant velocity. Therefore some force must act at right angles to direction of motion at every point, which should be change the direction of motion of body leaving speed uniform.

Thus the force which act along radius of circle at every point and is always directed towards centre of circle along which body moves is known as centripetal force.

Centrifugal force:-

According to Newton’s third law of motion, the force which acts opposite to centripetal force is known as centrifugal force.

Centrifugal force always acts away from centre of path i.e. centrifugal force always tends to throw the body away from centre of circular path.

Example 1:- A stone of mass 1 kg is revolving in circle of radius 1m with linear velocity of 10m/s what is value of centrifugal force acting on the stone.

Solution:-

Given

m=1Kg

v=10m/s

r=1m

We know

Centrifugal force acting on the stone

Example 2:- A ball of mass 0.25Kg is attached to end of 2m long string. The string will break if tension in string is more than 25N. What is the maximum angular velocity at which a ball can be rotated.

Solution:- given

m=0.25Kg

r=2m

F=25N

Centrifugal force

(tension in string)

Example 3:- a body of mass 0.5Kg tied to string is whirled in vertical circle working 2 rev/sec. If radius of circle is 1.2m, then find tension in string when body is at top.

Solution:- given

m=0.5Kg

Angular rotation (N)=2 rev/sec

r=1.2m

1) Tension in the string when body is at top

Angular velocity= =2πN

Tension in string

2) Tension in string when body is at bottom

Tension in string

Centrifugal force acting on a body moving along a circular path

Consider a body moving along a circular path with constant velocity

Let

m=mass of body

r=radius of circular path

v=linear velocity of body

We know,

Centrifugal acceleration of body=a=

And

Centrifugal force mass× centrifugal acceleration

Example 4:- an automobile of mass 1.5t, travelling at of 54Km/hr traverses a sag in the road. The sag is part of circle of radius 25m. Find the reaction between automobile and road while travelling at lowest part of the sag.

Solution:-

Given

Mass of automobile (m)=1.5t

Radius of sag (r)=25m

We know that reaction between automobile and road while travelling at lowest part of sag

5.4 Condition of skidding and overturning

Equilibrium of vehicle moving along a level circular path

We have seen that outer edge of road is always raised for circular turn. This is known as banking of road. Similarly, whenever alignment of railway takes circular turn outer rail is raised by an amount equal to super elevation then inner rail. The main idea of banking of road or super elevation in railway line is to distribute the load of vehicle equally on both wheels.

But if roads are not banked then a vehicle may have to face following miss happenings. This will also happen when vehicle moves with a velocity more than permissible velocity.

1) Vehicle may overtone

2) Vehicle may skid away.

Maximum velocity to avoid over it earning of vehicle moving along a level circular path

Consider a vehicle moving on a level circular path with O as a centre as shown in figure 4

Let

m=mass of vehicle

v=velocity of vehicle

r=radius of circular path in m.

h=height of C.G. Of vehicle from ground level in m.

2a=distance between reactions A and B

reaction at A

=reaction at B

We know that centrifugal force acting through C.G. Of vehicle

Also

Weight of vehicle=mg acting downward through its centre of gravity will exert reaction equal to on A and B (because of symmetry) taking moment about A and equating the same

Similarly by taking moment about B and equating the same we get

Little consideration will show that reaction at B. I.e. can never be negative. But reaction at A i.e. can be negative if value of becomes more than unity if we observe we can say that can be negative if value of v is increase, all other things are constant.

When this condition reaches, the vehicle will overturn at wheel B. Therefore in order to avoid overturning

In order to avoid overturning velocity of vehicle is less than that obtained from above equation the vehicle will not overturn if velocity is more vehicle is bound to overturn.

Therefore to avoid overturn maximum velocity is given

Note:- From above equation we say that maximum velocity of vehicle to avoid overturn is independent of its mass.

Example 5:- A vehicle weighing 1 tonnes is to turn on a circular curve of 40 m radius. The height of its C.G. Above road level is 75cm and distance between the centre lines of wheels is 120cm. Find speed at which vehicle should be run in order to avoid overturning.

Solution:-

Given

M=1 tonnes

h=75cm=0.75m

2a=120cm=1.2m

a=0.6m

We know maximum speed at which vehicle should to run in order to avoid overturning

=17.7m/s

63.72 Km/hr

Example 6:- Racing motor car is negotiating a curve of radius 50m. The distance between the centre to centre of wheel is 1.5m and height of C.G. Of car is 90cm above the ground level, find maximum speed of motor to avoid overturning.

Solution:- given

r=50m

2a=1.5m

a=0.75m

h=90cm

=0.9m

We know

=20.21m/s

=72.78 km/hr

Maximum velocity to avoid skidding away of vehicle moving along a level of circular path

Consider a vehicle moving on a level circular path with O as shown in figure 4

Let

m=mass of vehicle

v=velocity of vehicle

r=radius of circular path in in m.

=Coefficient of friction between the wheels of vehicle and ground

We know that,

Centrifugal force which tends to skid away the vehicle

And friction force between the wheels of vehicle and ground

The skidding away can only be avoiding if frictional force is more than centrifugal force. Therefore in order to avoid skidding away.

It is thus obvious that if velocity is less than that obtained from above equation, the vehicle will not skid away. But velocity is more than that of obtain from above equation, vehicle will skid away.

Therefore, in order to avoid skidding maximum velocity

Note:- From above equation we can say that maximum velocity of vehicle to avoid skidding is independent of its mass.

Example 7:- A car is travelling on a level track of radius 50m find maximum speed at which he can travel as curved truck. If coefficient of friction between tires and track is 0.45. Take g=

Solution:- Given

r=0.5m

We know

for avoid skidding

=14.85m/s=53.48 km/hr

Example 8:- A motorcyclist riding at 20km/HR has to turn a corner. Find least radius of curve he should follow for safe travelling. If coefficient of friction between tires and road is 0.2

Solution:- Given