A conical pendulum of a string AB (Figure) whose upper end is fixed at A and other end B is tied with a bob. When the bob is drawn aside and is given a calculated velocity in a perpendicular (horizontal) direction, it describe a horizontal circle with constant angular speed in such a way that AB makes a constant angle with the vertical. As the string traces the surface of a cone, it is known as conical pendulum.

Let `l` be the length of string `AB` . The forces acting on the bob are
(i) Weight mg acting downwards,
(ii) Tension `T` along the string (horizontal component is `T sin theta` and vertical components is `T cos theta`).

The bob is undergoing uniform circular motion. It has only centripetal acceleration towards the centre of circle.

`T cos theta = mg` (No acceleration along vertical) ...(1)


The horizontal component provides the centripetal force,

i.e., `T sin theta = m omega^2 r.....................(2)`

From eq. (1) and eq. (2) `(cos theta)/(sin theta) = g/(omega^2 r)`

`omega^2 =(g sin theta)/(r cos theta).....................(3)`

From Figure, `r= l sin theta`

`omega^2 = (g sin theta)/(l sin theta cos theta) = g/(lcos theta)...................(4)`

`t = (2 pi)/omega = 2 pi sqrt((l cos theta)/g)` [Time period].................(5)

But `l cos theta = h => t= 2 pi sqrt(h/g)..........................(6)`

`=>` The time period is independent of the mass of the body and depends only on `h`, i.e., distance between point of suspension A and the centre of the circle `O` .

In a death well, a person drives a bicycle or a motor cycle on the vertical surface of a large wooden well. The walls of death well are at rest. Here, the friction balances the weight of the person while reaction of the wall provides the centripetal force necessary for circular motion. The forces are shown in the figure.

Therefore,
`N=mvxxv/r`;
`f = mg` [No acceleration along vertical]

`=> (m v^2)/r =(mg)/mu; v_(min) = sqrt((rg)/mu)`

When vehicles go through turnings, they travel along a nearly circular arc. There must be some force which will produce the required centripetal acceleration. If the vehicles travel in a horizontal circular path, this resultant force is also horizontal. The necessary centripetal force is being provided to the vehicles in the fol lowing three ways.

a. By friction
b. By banking of roads
c. By both friction and banking of roads.

Now let us consider equations of motion in each of the three cases separately.

`text(By Friction Only :)`

Suppose a car of mass m is moving at a speed v in a horizontal circular arc of radius r. In this case, the necessary centripetal force to the car will be provided by force of friction f acting towards centre.

Thus , `f = (mv^2)/r`

Further, limiting value of `f` is `mu N`

or `f_L = mu N = mu m g`

Therefore, for a safe turn without sliding `(mv^2)/r` `<=` `f_L`

or `(mv^2)/r` `<=` `mu m g ` or `mu` `>=` `v^2/(rg)` or `v` `<=` `sqrt(mu r g)`


`text(By Banking of Roads :)`

To avoid dependence on friction, the roads are banked at the turn so that the outer part of the road is lifted compared to the inner part.

Applying Newton's second law along the radius and the first law in the vertical direction.

`N sin theta = (mv^2)/r`

and `N cos theta = mg`

From these two equations. we get(FIG-a)

`tan theta = v^2/(rg)...................(1)`

or `v = sqrt(rg tan theta)..................(2)`

`text(Note :)` This is the speed at which car does not slide down even if track is smooth. If track is smooth and speed is less than `sqrt(r g tan theta)`, vehicle will move down so that `r` decreases and if speed is more than this, the vehicle will move up.


`text(By Friction and Banking of Road)`

If a vehicle is moving on a circular road which is rough and banked, then th ree forces act on the vehicle, of these the first force is weight `(mg)` , the second force, is normal reaction `N` and is perpendicular to road, the third force, is friction `f` which can be either inwards or outwards while its magnitude can be varied up to a maximum limit `(f_L = mu N)` .

i) Friction f is outwards if the vehicle is at rest or `v =0`. Because in this case the component of weight `mg sin theta` is balanced by `f`.

ii) Friction f is inwards if `v > sqrt(r g tan theta)`

iii) Friction fis outwards if `v < sqrt(r g tan theta)` and

iv) Friction `f` is zero if , `v = sqrt(r g tan theta)`

Let us now see how the force of friction and normal reaction change as speed is gradually increased. (fig-c,fig-d)

In figure (c) : When the car is at rest force of friction is upwards. We can resolve the forces in two mutually perpendicular directions. Let us resolve them in horizontal and vertical directions

`sumF_H = 0` `therefore` `N sin theta - f cos theta =0 ...................(1)`

`sum F_V = 0` `therefore` `N cos theta + f sin theta = mg .................(2)`

In figure (d) : Now the car is given a small speed `v`, so that a centripetal force `(mv^2)/r` is now required in `r`
horizontal direction towards centre. So equation (1) will now become,

`N sin theta - f cos theta = (mv^2)/r`

or we can say, in case (c) N sin and f cos were equal while in case (d) their difference is `(mv^2)/r`

N will increase and `f` will decrease. This is because equation number (2). `N cos theta + f sin theta = mg` is still to be valid, as there is no acceleration along vertical. So to keep `N cos + f sin` to be constant `(= mg) N` should increase and `f` should decrease `(as theta =` constant)

Now as speed goes on increasing, force of friction first decreases, becomes zero at `v= sqrt( r g tan theta)` and then reverses its direction.


`text(Maximum velocity with which a vehicle can take a turn on a rough banked road)`

`N sin theta +f cos theta = (mv_(max)^2)/R ; N cos theta = mg +f sin theta `

`f = mu N` (Velocity is maximum)

`N sin theta + mu N cos theta =(mv_(max)^2)/R.....................(1) `

`N cos theta = mg + mu N sin theta ...................(2)`

Solving for (1) & (2) , `V_(max) = sqrt((gR(sin theta +mu cos theta))/(cos theta +mu sin theta))`


`text(Radius of curvature)`

Any curved path can be assumed to be made of infinite circular arcs. Radius of curvature at a point is the radius of the circular arc at a particular point which fits the curve at that point.

`F_c = (mv^2)/R` or `R = (mv^2)/(f_c) = (mv^2)/F_(bot)`

`F_(bot) = ` Force perpendicular to velocity (centripetal force)

If the equation of trajectory of a particle is given we can find the radius of curvature of the instantaneous circle by using the formula,

`R =([1+ ((dy)/(dx))^2]^(3/2))/|(d^2y)/(dx^2)|`
fig-e

Centrifugal force is a fictitious force which has to be employed in a travelling frame of reference attached to a body moving a circular path, in order to apply Newton's Laws of motion from that frame.

Its magnitude is equal to `(mv^2)/r`

When a body is moving in a circular path and the centripetal force vanishes, the body would leave the circular path. To an observer who is not sharing the motion along the circular path, the body appears to fly off tangentially at the point of release. To another observer, who is sharing the motion along the circular path (i.e., the observer is also moving with the body) it appears as if the object has been thrown off along the radius away from the centre by some force. This force is centrifugal force.

Analyzing the dynamics of a body moving in a circle, from a frame of reference that translates with the body, we have to assume a fictitious force directed radially outward, of magnitude `momega^2r` or `(mv^2)/r` apart from any real forces that may be acting on the body.