# Circular Motion Class 11 Physics | Notes

## Circular motion:

The motion of a body is said to be circular if the body is always at a fixed distance from a fixed point. The fixed distance is called radius and the fixed point is called the centre of the circular path.

## Angular displacement:

Angular displacement in circular motion is defined as the angle covered by the object from its initial position to final position in a given interval of time.

In figure,

$\angle$AOB = $\theta$ is angular displacement.

## Angular velocity:

The rate of change of angular displacement w.r.t. is called angular velocity. It is denoted by ‘$\omega$’ and given by,  angular velocity ($\omega$) =  $\frac{angular\text{ }displacement}{time}$

$\therefore \text{ }\omega$ =  $\frac{\theta }{t}$

## Angular acceleration ($\alpha$):

The rate of change of angular velocity with respect to time is called angular acceleration.

Angular acceleration, $\alpha =\frac{\Delta \omega }{t}$

## Time period (T):

Time period is defined as the time taken by the object to complete one revolution in a circular path. It is denoted by ‘T’.

From angular velocity, we have,

$\omega$ =  $\frac{\theta }{t}$

For one complete revolution, $\theta$ = 2$\pi$, t =T

Or, $\omega$ =  $\frac{2\pi }{T}$

$\therefore$   T =  $\frac{2\pi }{\omega }$

## Frequency (f):

Frequency is defined as the number of complete revolutions made in one second. It is denoted by ‘f ’ or ‘n’ and it is the reciprocal of time period. i.e,

f = $\frac{1}{T}$

Or, f = $\frac{1}{2\pi /\omega }$

∴ f = $\frac{\omega }{2\pi }$

It’s unit is (sec-1) or hertz (Hz) or revolution per second (rps). Another unit is revolution per minute (rpm).

## Types of circular motion:

### 1) Uniform circular motion:-

A circular motion is said to be uniform if an object covers equal angular displacement in equal interval of time. The work done in uniform circular motion is zero because the angle between force and displacement is 90ο.

## 2) Non uniform circular motion:-

A circular motion is said to be non-uniform if the object covers unequal angular displacement in equal interval of time.

Relation between linear velocity and angular velocity:

Let us consider an object revolving in a circular path of radius ‘r’ and centre ‘O’. Let ‘$\theta$’ be the small angular displacement when the object moves from A to B in a given time ‘t’. Let ‘v’ and ‘$\omega$’ be the linear velocity and angular velocity respectively.

From figure,

$\theta$ = $\frac{l}{r}$=  $\frac{AB}{r}$………(1)

For a small angle $\theta$, arc length AB= chord AB, i.e.

$\overset\frown{AB}$= AB = S  (say)

Now, equation (i) becomes

$\theta$ = $\frac{s}{r}$

Or,  s = r$\theta$

Differentiating above equation with respect to time,

$\frac{ds}{dt}$ = $\frac{d}{dt}$ (r$\theta$)

Or, $\frac{ds}{dt}$ = r $\left( \frac{d\theta }{dt} \right)$ [r is constant for circular path]

∴ v = r$\omega$ which is the required relation between ‘v’ and $\omega$

Again, differentiating above equation with respect to time we get

$\frac{dv}{dt}$ = $\frac{d(r\omega )}{dt}$

$\frac{dv}{dt}$ = r $\frac{d(\omega )}{dt}$

$\therefore$ a = r$\alpha$

This is the required relation between linear acceleration and angular acceleration.

Centripetal force and its expression:

Centripetal force is the force required to keep a body moving in a circular path with uniform speed. It is directed towards the center of the circular path.

Let us consider an object of mass ‘m’ revolving in a circular path of radius ‘r’ and centre ‘O’.

Let ‘$\theta$’ be the small angular displacement when the object moves from A to B.

Also, let $\overrightarrow{{{V}_{A}}}$ and $\overrightarrow{{{V}_{B}}}$ be the velocities of the object at position A and B respectively.

Then,

$\theta$= $\frac{l}{r}$ = $\frac{\overset\frown{AB}}{r}$

For a small angle$\theta$, arc length AB= chord AB, i.e.

$\overset\frown{AB}=AB=\Delta S$   (say)

So,    $\theta$ = $\frac{\Delta S}{r}$     ……..(i)

Let $\left| \overrightarrow{{{V}_{A}}} \right|=\left| \overrightarrow{{{V}_{B}}} \right|=V$(say)

From figure (b) change in velocity is given as

And$\left| \overrightarrow{{{V}_{AB}}} \right|=\left| \overrightarrow{{{V}_{B}}}-\overrightarrow{{{V}_{A}}} \right|=\Delta V$(say)

Also, $\theta$ = $\frac{\Delta V}{V}$ …………..(ii)

From equations (i) and (ii)

$\frac{\Delta V}{V}$ = $\frac{\Delta S}{r}$

r $\Delta$V = V $\Delta$S ………..(iii)

Dividing above equation by$\Delta t$and taking$\underset{\Delta t\to 0}{\mathop{\lim }}\,$on both sides we get,

$\text{ }\underset{\Delta t\to 0}{\mathop{\lim }}\,\text{ r }\frac{\Delta V}{\Delta t}\text{= }\underset{\Delta t\to 0}{\mathop{\lim }}\,\text{ V}\frac{\Delta S}{\Delta t}$

$r\text{ }\underset{\Delta t\to 0}{\mathop{\lim }}\,\text{ }\frac{\Delta V}{\Delta t}\text{=V }\underset{\Delta t\to 0}{\mathop{\lim }}\,\text{ }\frac{\Delta S}{\Delta t}$

Or, r a = V V

Or, a = $\frac{{{V}^{2}}}{r}$ Also, V = r $\omega$ $\therefore$a = $\omega$2r

This is the required expression for centripetal acceleration.

Now, centripetal force,

Fc = ma

Fc = $\frac{m{{V}^{2}}}{r}$

OR,

Fc = m$\omega$2r

Which is the required expression for centripetal force.

Motion of a body in horizontal circle (conical pendulum):

A conical pendulum is attached to a string and whirls in a horizontal circle with uniform speed and makes a constant angle with the vertical.

Let us consider an object of mass ‘m’ attached to a string length ‘l’ is whirled in a horizontal circle of radius ‘r’ with uniform speed ‘v’. The string makes an angle ‘$\theta$’ with the vertical. Resolving tension ‘T’ into its constituent components, we get Tcos$\theta$ and Tsin$\theta$. Tsin$\theta$ provides necessary centripetal force$\left( \frac{m{{v}^{2}}}{r} \right)$. The weight (mg) acts vertically downward which is balanced by Tcos$\theta$ as shown in figure above.

i.e,

Tsin$\theta$ = $\frac{m{{v}^{2}}}{r}$………………… (i)

Tcos$\theta$ = mg ………………. (ii)

Dividing (i) by (ii),

tan$\theta$ = $\frac{{{v}^{2}}}{rg}$ …………….. (iii)

Or, tan$\theta$ = $\frac{r}{h}$ ………………. (iv)

From (iii) and (iv),

$\frac{r}{h}$ = $\frac{{{v}^{2}}}{rg}$ …………………………………. (v)

Or, $\frac{g}{h}$ = $\frac{{{v}^{2}}}{{{r}^{2}}}$

Or,   $\frac{g}{h}$ = $\frac{{{r}^{2}}{{\omega }^{2}}}{{{r}^{2}}}$     [$\because$v = r$\omega$]

Or, $\frac{g}{h}$ = $\omega$2

Or, $\frac{g}{h}$ = ${{\left( \frac{2\pi }{T’} \right)}^{2}}$, where T’ is the time period.

Or, ${{\left( \frac{T’}{2\pi } \right)}^{2}}$ = $\frac{h}{g}$

Or, $T’=2\pi \sqrt{\frac{h}{g}}$

$\therefore$  $T’=2\pi \sqrt{\frac{l\cos \theta }{g}}$

This is the required expression for the time period of conical pendulum.

Now, tension in the string of conical pendulum,

Squaring and adding (i) and (ii)

T2 (sin2$\theta$ + cos2$\theta$) =  ${{\left( \frac{m{{v}^{2}}}{r} \right)}^{2}}$+ (mg)2

T2  =  m2g2 + $\frac{{{m}^{2}}{{v}^{4}}}{{{r}^{2}}}$

T2  =  m2g2  $\left( 1+\frac{{{v}^{4}}}{{{g}^{2}}{{r}^{2}}} \right)$

T = mg $\sqrt{1+\frac{{{v}^{4}}}{{{g}^{2}}{{r}^{2}}}}$

Using eqn (V)

Or,     T = mg $\sqrt{1+\frac{r{}^{2}}{{{h}^{2}}}}$

This is the required expression for tension in conical pendulum.

Motion of a body attached to a string in a vertical circle:

Let us consider an object of mass ‘m’ attached to a string is whirled in a vertical circle of radius ‘r’ with speed ‘v’. Let at any instant of time, the position of the object is at P, where the string makes an angle ‘$\theta$’ with the vertical (OC). The weight (mg) acts vertically downward and resolving it, we get (mgcos$\theta$) and (mgsin$\theta$). The resultant of tension (T) and mgcos$\theta$ provides the necessary centripetal force $\frac{m{{v}^{2}}}{r}$ .

i.e, T– mg cos$\theta$ =  $\frac{m{{v}^{2}}}{r}$

T=  $\frac{m{{v}^{2}}}{r}$ + mg cos$\theta$

Which is the required expression for tension.

Special cases:

1. At position A (highest point), $\theta$ = 180°

T=  $\frac{m{{v}^{2}}}{r}$ + mg cos180°

Tmin =  $\frac{m{{v}^{2}}}{r}$ – mg which is the minimum tension.

2. At position C (lowest point), $\theta$ = 0°

T=  $\frac{m{{v}^{2}}}{r}$ + mg cos0°

Tmax  =  $\frac{m{{v}^{2}}}{r}$ + mg which is the maximum tension.

3. At position D, $\theta$ = 90°

T =  $\frac{m{{v}^{2}}}{r}$ + mg cos90°

T =  $\frac{m{{v}^{2}}}{r}$

Critical velocity:

The minimum velocity needed to keep the object moving in the vertical circle is called critical velocity.

1. At position A: (highest point), for minimum velocity tension can be assumed zero.

T =  $\frac{m{{v}^{2}}}{r}$ – mg

0 =  $\frac{m{{v}^{2}}}{r}$ – mg

$\therefore$VA= $\sqrt{rg}$

This is the critical velocity at point A.

2. At position C:

Mechanical energy at A = Mechanical energy at C

Or, (M.E)C = (M.E)A

or, KEC + PEC = KEA + PEA

or, KEC = KEA + (PEA –PEC)

or, $\frac{1}{2}$ m Vc2 = $\frac{1}{2}$ m V2A + mg.2r

or, Vc2 =  VA2 + 4rg

or, Vc = $\sqrt{5rg}$ ,

This is the critical velocity at lowest point.

3. At point B:

Mechanical energy at C = Mechanical energy at A

Or, (M.E)B = (M.E)A

or, KEB + PEB = KEA + PEA

or, KEB = KEA + (PEA –PEB)

or, $\frac{1}{2}$ m V2B = $\frac{1}{2}$ m V2A + mg.r

or, V2B =  V2A + 2rg

or, VB = $\sqrt{3rg}$

This is the critical velocity at point B.

Motion of a car in a level curved path:

Let us consider a car of mass ‘m’ moving with speed ‘v’ in a level circular path of radius ‘r’. The weight of the car acts vertically downward, which is balanced by the normal reaction (R = R1 + R2). The frictional force (Ff = F1 + F2) provides necessary centripetal force $\frac{m{{v}^{2}}}{r}$, i.e.

Ff = FC

Or, $\mu$R = $\frac{m{{v}^{2}}}{r}$

Or, $\mu$mg = $\frac{m{{v}^{2}}}{r}$ ($\because$R = mg)

$\therefore$ v = $\sqrt{\mu rg}$, Which is the required maximum speed of car with which it can take turn in a circular path

Motion of a car in a banked curve path:

Let us consider a car of mass ‘m’ moving with velocity ‘v’ in a banked curve path of radius ‘r’. Let the surface of the road make an angle$\theta$ with the horizontal. The weight of the car acts vertically downward and normal reaction acts perpendicularly outward to the surface of the road. Resolving normal reaction (R = R1+R2) into its constituent components, we get Rsin$\theta$ and Rcos$\theta$. The component Rcos$\theta$ is balanced by weight (mg) and the component Rsin$\theta$ provides necessary centripetal force $\frac{m{{v}^{2}}}{r}$ i.e.

R sin$\theta$ = $\frac{m{{v}^{2}}}{r}$ …(i)

R cos$\theta$ = mg …(ii)

Dividing (i) by (ii)

tan$\theta$ = $\frac{{{v}^{2}}}{rg}$

$\therefore$ $\theta$ = tan–1$\left( \frac{{{v}^{2}}}{rg} \right)$  which is the expression for banking angle of a road.

And, v2 = tan$\theta$. rg

$\therefore$v =$\sqrt{rg\text{ }tan\theta }$,which is the required expression for maximum velocity in a banked curve path

Motion of a cyclist in a curved path / Bending of a cyclist in a curved path:

Let us consider a cyclist of mass ‘m’ moving in a circular path of radius ‘r’ with velocity ‘v’. Let the cyclist make an angle ‘$\theta$’ with the vertical. The weight of the cyclist acts vertically downward. Resolving normal reaction into its constituent components, we get Rsin$\theta$ and Rcos$\theta$. The component Rcos$\theta$  is balanced by mg and the component Rsin$\theta$ provides necessary centripetal force $\frac{m{{v}^{2}}}{r}$ i.e.

Rsin$\theta$ = $\frac{m{{v}^{2}}}{r}$ …….(i)

Or,     Rcos$\theta$ = mg ……(ii)

Dividing (i) and (ii),

tan$\theta$ = $\frac{{{v}^{2}}}{rg}$

$\theta$ = tan-1$\left( \frac{{{v}^{2}}}{rg} \right)$

This is the required bending angle of the cyclist with the vertical in a circular path

Also, v2 = rg tan$\theta$

$\therefore$v = $\sqrt{rg\text{ }tan\theta }$

This is the required expression for the velocity of cyclist in a circular path.

Also Read: Work, Energy and Power Class 11 Notes

CIRCULAR MOTION

NUMERICAL PROBLEMS

1.         At what angle should a circular road be banked so that a car running at 50 km/hr be safe to go round the circular turn of 200 m radius?

Ans:    5.5°

2.         An object of mass 4 kg is whirled round a vertical circle radius 1 m with a constant speed of 3 ms-1. Calculate the maximum tension in the string.

Ans:    76 N

3.         An object of mass 8.0 kg is whirled round in a vertical circle of radius 2m with a constant speed of 6 ms-1. Calculate the maximum and the minimum tensions in the string.

Ans:    224 N, 64 N

4.         A mass of 0.2 kg is rotated by a string at a constant speed in a vertical circle of radius 1m. If the minimum tension in the string is 3 N, calculate the magnitude of the speed and the maximum tension in the string.

Ans:    5 m/sec: 7 N

5.         A body of mass 0.2 kg is whirled in a horizontal circle of radius 0.5m by a string inclined at 30° to the vertical. Calculate i. the tension in the string ii. The speed of the mass in the horizontal circle.

Ans:    2.3 N, 1.7 m/sec

6.         A bob of mass 200 gram is whirled in a horizontal circle of radius 50 cm by a string inclined at 30° to the vertical. Calculate the tension in the string and the speed of the bob in the horizontal circle.

Ans:    2.3 N. 1.7m/sec

7.         A mass of 1 kg is attached to the lower end of the string 1 m long whose upper end is fixed. The mass is made to rotate in a horizontal circle of radius 60 cm. If the circular speed of the mass is constant, find the tension in the string and the period of motion.

Ans:    12.5 N, 1.77sec

8.         An object of mass 0.5 kg is rotated in a horizontal circle by a string in 1 m long. The maximum tension in the string before it breaks is 50 N. What is the greatest number of revolutions per second of the object?

Ans:    1.6 rev/sec

9.         A stone with mass 0.8 kg is attached to one end of a string 0.9 m long. The string will break if its tension exceeds 600 N. The stone is whirled in a horizontal circle, the other end of the string remains fixed. Find the maximum speed, the stone can attain without breaking the string.

Ans:    0.85 rev/sec

10.       A certain string breaks when a weight of 25 N acts on it. A mass of 500 gram is attached to one end of the string of 1 m long and is rotated in a horizontal circle. Find the greatest number of revolutions per minute which can be made without breaking the string.

Ans:    67.5 rpm

11.       A coin placed on a disc rotates with speed of 100/3 rev. min-1 provided that the coin is not more than 10 cm. from the axis. Calculate the coefficient of static friction between the coin and the disc.

Ans:    0.122

12.       A stone of mass 0.5 kg is attached to a string of length 50 cm which will break if the tension in it exceeds 20 N. The stone is whirled in a vertical circle, the axis of rotation being at a height of 100 cm above the ground. The angular speed is very slowly increased until the string breaks. In what position is this break most likely to occur, and what angular speed? Where will the stone hit the ground?

Ans:    At lowest point, 7.74 rad/sec. 1.22 m

13.       A train has to take a circular turn of radius 500 m with a speed of 36 km/hr. By how much should the outer rail be raised above the inner rail so that there is no side pressure on the rails?  The distance between the rails is 1 m.

Ans:    2 cm

14.       A stone of mass 5 kg is attached to a string 8 m long and is whirled in a horizontal circle. The string can stand maximum tension 160 N. What is the greatest number of revolutions per seconds of the stone?

Ans:    0.318 rps.

15.       A bucket of water having total mass 3 kg is whirled around the vertical circle of radius 1.5 m with speed of 6 m/s. Find the maximum and minimum tension.

Ans:    102 N, 42 N

16.       Find the maximum speed of vehicle moving on a circular track without skidding if the radius of circular path is 80 m and the coefficient of friction between the road and the tyres of vehicle is 0.8

Ans:    25.3 m/s

17.       A bucket filled with water is revolved in vertical circle of radius 1m and the water does just not fall. Find period of revolution.

Ans:    2 sec