Circular Motion

definition

Non Uniform Circular Motion

In non uniform circular motion magnitude of velocity changes with time.
Direction of the particle changes at every point of time in circular motion.

definition

Axial Vector

Axial vector is a vector which does not change its sign on changing the coordinate system to a new system by a reflection in the origin.
An example of an axial vector is the vector product of two polar vectors, such as A = x $\times$

m, where A is the angular momentum of a particle, x is its position vector, and m is its momentum vector.

definition

Angular Displacement

The angular displacement is defined as the angle

θ

, in radians, through which an object moves on a circular path between the initial and final positions.

θ = θ_{f} - θ_{i}

definition

Angular Displacement Units

The unit of angular displacement is measured is radians.

2 π r a d i a n s = 360^{o}

⟹ 1 r a d i a n = {57.3}^{o}

diagram

Relation Between Angular Displacement and Path Length

A r c L e n g t h = A n g l e R o t a t e d \times R a d i u s o f c u r v a t u r e

definition

Angular Speed

The rate of change of angular position is called angular speed. Angular speed is a scalar quantity.
Angular speed = (final angle) - (initial angle) / time = change in position/time = angular speed in radians/sec.

definition

Angular Velocity

The rate of change of angular position is called the angular velocity. It is a vector quantity.

Angular Velocity

ω = \frac{d θ}{d t}

definition

Relation Between Angular Velocity and Linear Velocity

Relation Between Angular Velocity and Linear Velocity:

v = w r

definition

Relation between frequency, time period and angular velocity

w = \frac{2 π}{T} = 2 π ν

rad/s

definition

Tangential Velocity

Tangential Velocity is defined as:

v = \frac{d x}{d t}

unit is

m / s^{2}

definition

Tangential Acceleration

Tangential acceleration

a_{t}

is the component of acceleration along the tangent in a circular motion. It is a measure of the rate of change of speed along the tangential direction.

a = \frac{Δ v}{Δ t}

definition

Angular Acceleration

Angular acceleration is the rate of change of angular velocity. In SI units, it is measured in

r a d / s^{2}

, and is usually denoted by the Greek letter

α

α = \frac{d ω}{d t}

definition

Relationship between angular acceleration and linear acceleration

Linear acceleration (

a

) and angular acceleration (

α

) are related as:

\vec{a} = \vec{α} \times \vec{r}

definition

Position, Velocity and Acceleration in cartesian coordinate system for uniform circular motion

Position Vector:

\vec{r} = x \hat{i} + y \hat{j}

Velocity Vector:

\vec{v} = v_{1} \hat{i} + v_{2} \hat{j}

Acceleration Vector:

\vec{a} = a_{1} \hat{i} + a_{2} \hat{j}

definition

Position, Velocity and Acceleration Vector

The direction of velocity vector is tangential to the path and its expression is given by the value given in the figure.
The direction of acceleration vector is normal to the path and its expression is given by the value given in the figure.
Here

r

is the position vector.

definition

Radius of Curvature

The radius of a circle which touches a curve at a given point and has the same tangent and curvature at that point. Given above is the formula for radius of curvature for any trajectory

formula

Equations of motion for uniform angular acceleration in circular motion

Equations of motion for uniform angular acceleration are:

ω_{f} = ω_{i} + α t

θ = ω_{i} t + \frac{1}{2} α t^{2}

ω_{f}^{2} = ω_{i}^{2} + 2 α θ

Example:
A stationary wheel starts rotating about its own axis at uniform angular acceleration

8 r a d / s^{2}

. Find the time taken by it to complete

77

rotations.Solution:
From the equations of circular motion, we get:

θ = ω_{0} t + \frac{1}{2} α t^{2}

From the given conditions,

θ = 77 \times 2 π r a d

α = 8 r a d / s^{2}

and

ω_{0} = 0

as the wheels starts rotating from rest.
So, we get:

77 \times 2 π = \frac{1}{2} \times 8 \times t^{2}

or,

t = 11 s

Note:
Particle is at same positions at angular displacement of

θ

and

θ + 2 n π

definition

Centripetal Force

A body moving in a circle of constant radius with a constant speed has a non-zero force acting on it. This force is known as Centripetal force. It is directed towards the center of the circle. Its value is given by the formula:

F = {m v}^{2} / R

Note: Centripetal force for uniform circular motion is constant in magnitude. However, its direction is continuously changing as it is always directed towards the center of the circle.

definition

Centrifugal Force

Centrifugal force is a fictitious force that appears in a rotating reference frame. Its direction is opposite to that of the centripetal force i.e. radially outwards from the center. Its magnitude is equal to the centripetal force on the object. Hence,

F = {m v}^{2} / R

Example: A person sitting in a car feels a force towards right when the car is turning left.
Note: Centrifugal force is a fictitious force. It is only valid inside the rotating frame of reference. For an observer outside, it is no longer valid and the interpretation is different.

definition

Difference between cetrifugal and centripetal force

Centripetal Force	Centrifugal Force
A centripetal force is a force that makes a body follow a curved path. Its direction is always orthogonal to the motion of the body and towards the fixed point of the instantaneous center of curvature of the path.	Centrifugal force is an outward force apparent in a rotating reference frame; it does not exist when measurements are made in an inertial frame of reference. All measurements of position and velocity must be made relative to some frame of reference.
Centripetal Force is observed from inertial frame of reference.	Centrifugal Force is observed from non- inertial frame of reference.

example

Variation of centripetal force with radius

Example: Two stones of masses

m

and

2 m

are whirled in horizontal circles, the heavier one in a radius,

\frac{r}{2}

, and the lighter one in radius,

r

. The tangential speed of lighter stone is

n

times that of the value of heavier stone when they experience same centripetal forces. The value of

n

is:
Solution:
Centripetal force is same on both as given.
So,

\frac{m v_{1}^{2}}{r_{1}} = \frac{2 m v_{2}^{2}}{r_{2}}

\Rightarrow \frac{m (n v_{2})^{2}}{r} = \frac{2 m v_{2}^{2}}{\frac{r}{2}}

\Rightarrow n^{2} = 4

\Rightarrow n = 2

example

Cars moving on smooth circular banked roads

Question: A circular road of radius r is banked for a speed

v = 40

km/hr. A car of mass m attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible. Then comment on the speed of car when it turns.
Solution: Applying Newton's laws in horizontal and vertical directions. we get

N \sin θ = \frac{m v^{2}}{r}

...

(I)

and

N \cos θ = m g

...

(I I)

from these two equations, we get

\tan θ = \frac{v^{2}}{r g}

v = \sqrt{r g \tan θ}

This is the speed at which the car doesn't slide down even if there is no
friction. Since the car is in horizontal and vertical equilibrium. So the option 'A' is wrong.
If the car's speed is less then the banking speed then It will slip down to reduce the

r

.

If the car turns at correct speed of

40

m/s. Then the force by the road on the car is given by

N = \frac{m v^{2}}{r \sin θ} = \frac{m g}{\cos θ}

....

(I I I)

[from

(I)

and

(I I)

]
By looking at equation

(I I I)

. we can say that, the force by the road on
the car

N

is equal to

\frac{m v^{2}}{r \sin θ}

, not

\frac{m v^{2}}{r}

,
Since

θ < \frac{π}{2}; \Rightarrow \sin θ < 1 & \cos θ < 1 \Rightarrow \frac{1}{\sin θ} > 1 & \frac{1}{\cos θ} > 1 \Rightarrow \frac{m v^{2}}{r \sin θ} > \frac{m v^{2}}{r} & \frac{m g}{\cos θ} > m g \Rightarrow N > \frac{m v^{2}}{r} & N > m g

example

Cars moving on circular banked roads with friction acting on tyres

Example: A circular track of radius

300

m is banked at an angle

\frac{π}{12}

radian. If the coefficient of friction between wheel of a car and road is

0.2

, the maximum safe speed of car is:

Solution: Maximum velocity,

v_{m a x} = {[(\frac{μ + \tan θ}{1 - μ \tan θ}) r g]}^{1 / 2}

= {[(\frac{0.2 + \tan 15^{\circ}}{1 - 0.2 \tan 15^{\circ}}) 300 \times 9.8]}^{1 / 2}

≃ 38 m s^{- 1}

example

Uniform Circular Motion in Horizontal Plane

Example: A simple pendulum of length l is set in motion such that the bob, of mass m, moves along a horizontal circular path, and the string makes a constant angle

θ

with the vertical. The time period rotation of the bob is t and then find the tension T in the thread.

Solution:
Balancing the forces in Horizontal and vertical direction:

T s i n θ = m ω^{2} l s i n θ o r T = m ω^{2} l T c o s θ = m g o r \frac{m ω^{2}}{c o s θ} = m g o r ω^{2} = \frac{g}{l c o s θ} t = \frac{2 π}{ω} = 2 π \sqrt{\frac{l c o s θ}{g}} T = m [\frac{4 π^{2}}{t^{2}}] l

example

Uniform Circular Motion in Horizontal Plane with Normal Reaction From a surface as the centripetal force

Example: A particle moving with constant speed

u

inside a fixed smooth spherical bowl of radius

a

describes a horizontal circle at a distance

\frac{a}{2}

below its centre. Then, find u.
Solution:
From the fig., we get:

r = a s i n 60^{0} = \frac{a \sqrt{3}}{2}

And the components of normal reaction:

N s i n

60^{0} = \frac{m u^{2}}{r}

and

N c o s

60^{0} = m g

From the above two equations we get

u = 3^{1 / 4} \sqrt{a g}

example

Problem on Uniform Circular Motion in Horizontal Plane

Example: A vehicle is travelling with uniform speed along a concave road of radius of curvature

19.6 m

. At lowest point of concave road if the normal reaction on the vehicle is three times its weight, the speed of vehicle is:

Solution: As normal reaction is three times weight, net force will balance the centripetal force

N - m g = \frac{m v^{2}}{R}

2 m g = \frac{m v^{2}}{R}

\Rightarrow v = \sqrt{2 g R} = \sqrt{2 \times 9.8 \times 19.6}

= 19.6 m s^{- 1}

example

Accelerated Circular motion in horizontal plane

Example: A small block of mass

1 k g

is released from rest at the top of a rough track. The track is a circular arc of radius

40 m

. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure, above, is

150 J

. (Take the acceleration due to gravity,

g = 10 m s^{- 2}

).

Solution: Using work energy theorem we get

m g R s i n 30^{o} + W = \frac{1}{2} m v^{2}

200 - 150 = \frac{v^{2}}{2}

Thus we get

v = 10 m / s

Now, the force equation gives:

N - m g \cos 60^{o} = \frac{m v^{2}}{R}

or,

N = 5 + \frac{5}{2} = 7.5 N

example

Problems on Uniform Circular Motion

Example: A body of mass

0.2 k g

is rotated along a circle of radius

0.5 m

in horizontal plane with uniform speed

3 m / s

. The centripetal force on that body is:
Solution: Given,

m = 0.2 k g

R = 0.5 m

and

v = 3 m / s

Therefore,

F_{c} = \frac{m v^{2}}{R}

= \frac{(0.2) (3^{2})}{(0.5)}

= 3.6 N

example

Problem on horizontal circle and projectile motion

Example: An open umbrella is held upright and rotated about the handle at a uniform rate of 21 revolution in 44 s44 s. If the rim of the umbrella is a circle of 100 cm100 cm diameter and the height of the rim above ground is 150 cm150 cm, then the drops of water spinning off the rim will hit the ground from center of umbrella, at a distance of:Solution:
Velocity of water drop,

v = r ω = 0.5 \times \frac{(2 π \times 21)}{44} = 1.5 m s^{- 1}

Time taken to reach ground,

t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 1.5}{9.8}} = 0.55 s

Range of drop,

x = v t = 1.5 \times 0.55 = 0.83 m

∴

distance:

R = \sqrt{r^{2} + x^{2}} = \sqrt{{0.5}^{2} + {0.83}^{2}} = 0.97 m

= 97 c m

example

Problems on Conical Pendulum

Vertical :

F c o s θ = m g

Horizontal :

F s i n θ = \frac{m v^{2}}{r}

t a n θ = \frac{v^{2}}{r g}

t a n θ = \frac{r ω^{2}}{g}

\frac{s i n θ}{c o s θ} = \frac{λ s i n θ ω^{2}}{g}

ω^{2} = \frac{g}{λ c o s θ}

T = \frac{2 π}{ω} = 2 π \sqrt{\frac{λ c o s θ}{g}}

example

Problem on centrifugal force

Example:
A car is moving along a circular track of radius

10 \sqrt{3}

m with a constant speed of

36

kmph. A plumb bob is suspended from the roof of the car by a light rigid rod of length

1

m. The angle made by the rod with the track is

(g = 10 m s^{- 2})

Solution:Balancing the forces in the frame of car.

\Rightarrow \frac{m V^{2}}{r} \sin θ = m g \cos θ

\tan θ = \frac{r g}{V^{2}}

= \frac{10 \sqrt{3} \times 10}{10 \times 10} = \sqrt{3}

\Rightarrow θ = 60^{\circ}

example

Force equations for an object in uniform circular motion

For an object tied on a string rotating in a vertical circle equation for particle's motion will be given by:
1. At the bottom most position:

T = \frac{m v^{2}}{R} + m g

2. At the position 90 degree from vertical line:

T = \frac{m v^{2}}{R}

3. At the top most point of the circle:

T = \frac{m v^{2}}{R} - m g

definition

Conservation of Energy for an Object in a vertical Circle

Considering level of center of the circle as reference point:
1. Energy at the lowest point of the circle will be:

\frac{m v^{2}}{2} - m g R

2. Energy at

90^{0}

degree from the vertical line of the circle will be:

\frac{m v^{' 2}}{2}

3. Energy at the topmost point of the circle will be:

\frac{m u^{2}}{2} + m g R

Since there isn't any energy dissipative force all the energies will be same.

example

Motion in Vertical Circle in which tension becomes zero

Example: A test tube of mass

20 g

is filled with a gas and fitted with a stopper of

2 g

. It is suspended horizontally by means of a thread of

1 m

length and heated. When the stopper kicks out, the tube just completes a circle in vertical plane. Find the velocity with which the stopper is kicked out.

Solution: If tubes completes circle

v = \sqrt{5 g R}

= \sqrt{5 g}

Conserving momentum

0 = - 20 \times v^{'} + 2 \times v

\Rightarrow v^{'} = 10 \sqrt{5 g}

v^{'} = 70

m/s

example

Objects moving in vertical circle with tension as the centripetal force

Example: A mass of

2 k g

tied to a string of

1 m

length is rotating in a vertical circle with a uniform speed of

4 m s^{- 1}

.Find the position of mass when tension in the string will be

52 N

.[Take

g = 10 m s^{- 2}

]

Solution:

T - m g s i n θ = \frac{m v^{2}}{R}

(

θ

is the angle from the horizontal line
Thus we get:

52 - 2 \times 10 \times s i n θ = \frac{2 \times 16}{1}

Thus

s i n θ = 90^{o}

Thus the tension of 52 N would occur at the bottom most point.

example

Objects in vertical circle with normal reaction as centripetal force

Example: A block is freely sliding down from a vertical height

4 m

on a smooth inclined plane. The block reaches bottom of inclined plane then it describes vertical circle of radius

1 m

along smooth track. What is the ratio of normal reactions on the block while it is crossing lowest point and highest point of vertical circle is:

Solution:
Applying the law of conservation of energy at point A and B, we have

m g h = \frac{1}{2} m {v_{1}}^{2} \Rightarrow v_{1} = \sqrt{2 g h} \Rightarrow v_{1} = \sqrt{8 g}

Now, net force is towards the center is the centripetal force, we have at point B

\frac{m {v_{1}}^{2}}{r} = N_{1} - m g

\Rightarrow N_{1} = \frac{m {v_{1}}^{2}}{r} + m g = \frac{m (8 g)}{1} + m g = 9 m g

..........

(I)

applying the law of conservation of energy at point B and C, we have

\frac{1}{2} m {v_{1}}^{2} = m g (2 r) + \frac{1}{2} m {v_{2}}^{2} \Rightarrow \frac{1}{2} m (8 g) = m g (2) + \frac{1}{2} m {v_{2}}^{2} \Rightarrow v_{2} = \sqrt{4 g}

Now, net force is towards the center is the centripetal force, we have at point C

\frac{m {v_{2}}^{2}}{r} = N_{2} + m g

\Rightarrow N_{2} = \frac{m {v_{2}}^{2}}{r} - m g = \frac{m (4 g)}{1} - m g = 3 m g

..........

(I I)

So the ratio of normal reactions on the block while it is crossing lowest point, highest point of vertical circle is:

\frac{N_{1}}{N_{2}} = \frac{9 m g}{3 m g} = 3 : 1

..from

(I)

and

(I I)

result

Limiting condition to complete a vertical circle

The minimum velocity required at the bottom of the circle to complete the vertical circle is:

\frac{m v^{2}}{R} - m g = 0

v = \sqrt{R g}

Using energy conservation between topmost and bottommost point of the circle:

\frac{m v_{1}^{2}}{2} = \frac{m v^{2}}{2} + m g (2 R)

This gives:

v_{1} = \sqrt{5 g r}

This is the velocity the topmost point of the circle when

v = \sqrt{R g}

is velocity at the bottom.

example

Combination of Vertical Circular Motion and Projectile Motion

Example: A mass is attached to one end of a mass less string (length

l

), the other end of which is attached to a fixed support O. The mass swings around in a vertical circle as shown in fig. The mass has the minimum speed (u =

\sqrt{5 g l})

necessary at the lowermost point of the circle to keep the string from going slack. If you cut the string, when it is making angle

θ

(as shown in figure) with vertical line through centre O of the circle, the resulting projectile motion of the mass has its maximum height located directly above the center of the circle.
Find the value of

10 \cos θ

Solution:
By conserving energy,

m v^{2} / 2 = 0.5 m \times 5 g l - m g (l + l c o s θ)

\Rightarrow v = \sqrt{3 g l - 2 l c o s θ}

.
Since the highest point is above the O. So

t = l s i n θ / v c o s θ

.
At highest point, final velocity is zero.

v = g t

solving,

θ = 60^{0} \Rightarrow 10 c o s θ = 5

Monday, July 31, 2017

Circular Motion

Non Uniform Circular Motion

Axial Vector

Angular Displacement

Angular Displacement Units

Relation Between Angular Displacement and Path Length

Angular Speed

Angular Velocity

Relation Between Angular Velocity and Linear Velocity

Relation between frequency, time period and angular velocity

Tangential Velocity

Tangential Acceleration

Angular Acceleration

Relationship between angular acceleration and linear acceleration

Position, Velocity and Acceleration in cartesian coordinate system for uniform circular motion

Position, Velocity and Acceleration Vector

Radius of Curvature

Equations of motion for uniform angular acceleration in circular motion

Centripetal Force

Centrifugal Force

Difference between cetrifugal and centripetal force

Variation of centripetal force with radius

Cars moving on smooth circular banked roads

Cars moving on circular banked roads with friction acting on tyres

Uniform Circular Motion in Horizontal Plane

Uniform Circular Motion in Horizontal Plane with Normal Reaction From a surface as the centripetal force

Problem on Uniform Circular Motion in Horizontal Plane

Accelerated Circular motion in horizontal plane

Problems on Uniform Circular Motion

Problem on horizontal circle and projectile motion

Problems on Conical Pendulum

Problem on centrifugal force

Force equations for an object in uniform circular motion

Conservation of Energy for an Object in a vertical Circle

Motion in Vertical Circle in which tension becomes zero

Objects moving in vertical circle with tension as the centripetal force

Objects in vertical circle with normal reaction as centripetal force

Limiting condition to complete a vertical circle

Combination of Vertical Circular Motion and Projectile Motion

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