This is the example of uniform circular motion in horizontal plane.
A bob of mass m attached to a light and in-extensible string rotates in a horizontal circle of radius r with constant angular speed \[\omega \] about the vertical. The string makes angle \[\theta \] with vertical and appears tracing the surface of a cone. So this arrangement is called conical pendulum.
The force acting on the bob are tension and weight of the bob.
From the figure \[T=\sin \theta =\frac{m{{\upsilon }^{2}}}{r}\] ...(i)
and \[T\cos \theta =mg\] ...(ii)
(1) Tension in the string : \[T=mg\sqrt{1+{{\left( \frac{{{\upsilon }^{2}}}{rg} \right)}^{2}}}\] ...(i)
\[T=\frac{mg}{\cos \theta }=\frac{mgl}{\sqrt{{{l}^{2}}-{{r}^{2}}}}\]
[As \[\cos \theta =\frac{h}{l}=\frac{\sqrt{{{l}^{2}}-{{r}^{2}}}}{l}\]]
(2) Angle of string from the vertical :
\[\tan \theta =\frac{{{\upsilon }^{2}}}{rg}\] ...(ii)
(3) Linear velocity of the bob :
\[\upsilon =\sqrt{gr\tan \theta }\]
(4) Angular velocity of the bob :
\[\omega =\sqrt{\frac{g}{r}\tan \theta }=\sqrt{\frac{g}{h}}=\sqrt{\frac{g}{l\cos \theta }}\]
(5) Time period of revolution :
\[{{T}_{p}}=2\pi \sqrt{\frac{l\cos \theta }{g}}=2\pi \sqrt{\frac{h}{g}}\]
\[=2\pi \sqrt{\frac{{{l}^{2}}-{{r}^{2}}}{g}}=2\pi \sqrt{\frac{r}{g\tan \theta }}\] ...(iii)
If \[\theta ={{90}^{o}}\], then pendulum becomes horizontal & from equations (i), (ii) and (iii) we get \[\upsilon =\infty ,\,\,T=\infty \] and \[{{T}_{p}}=0\] which is practically not possible.
This is an example of non-uniform circular motion. In this motion body is under the influence of gravity of earth. When body moves from lowest point to highest point. Its speed decrease and becomes minimum at highest point. Total mechanical energy of the body remains conserved and KE converts into PE and vice versa.
(1) Velocity at any point on vertical loop : If u is the initial velocity imparted to body at lowest point then velocity of body at height h is given by
\[v=\sqrt{{{u}^{2}}-2gh}=\sqrt{{{u}^{2}}-2gl(1-\cos \theta )}\]
[As \[h=l-l\cos \theta =l(1-\cos \theta )\]]
where l is the length of the string
(2) Tension at any point on vertical loop : Tension at general point P, According to Newton?s second law of motion.
Net force towards centre = centripetal force
\[T-mg\cos \theta =\frac{m{{v}^{2}}}{l}\]
or \[T=mg\cos \theta +\frac{m{{v}^{2}}}{l}\]
\[T=\frac{m}{l}[{{u}^{2}}-gl(2-3\cos \theta )]\]
[As \[v=\sqrt{{{u}^{2}}-2gl(1-\cos \theta )}\]]
Velocity and tension in a vertical loop
Where
\[{{\omega }_{1}}=\] Initial angular velocity of particle
\[{{\omega }_{2}}=\] Final angular velocity of particle
\[\alpha =\] Angular acceleration of particle
\[\theta =\] Angle covered by the particle in time t
\[{{\theta }_{n}}=\] Angle covered by the particle in nth second
If the speed of the particle in a horizontal circular motion changes with respect to time, then its motion is said to be non-uniform circular motion.
Consider a particle describing a circular path of radius r with centre at O. Let at an instant the particle be at P and \[\overrightarrow{\upsilon }\] be its linear velocity and \[\overrightarrow{\omega }\] be its angular velocity.
Then, \[\vec{\upsilon }=\vec{\omega }\times \vec{r}\] ...(i)
Differentiating both sides of w.r.t. time t we have
\[\frac{\overset{\to }{\mathop{d\upsilon }}\,}{dt}=\frac{\overset{\to }{\mathop{d\omega }}\,}{dt}\,\times \vec{r}\,+\,\vec{\omega }\,\times \,\frac{\overset{\to }{\mathop{dr}}\,}{dt}\] ...(ii)
Here, \[\frac{\overset{\to }{\mathop{dv}}\,}{dt}=\vec{a},\,\,\](Resultant acceleration)
\[\vec{a}=\vec{\alpha }\,\times \,\vec{r}\,\,\,\,\,+\,\,\,\,\vec{\omega }\,\times \,\vec{\upsilon }\]
\[\frac{\overset{\to }{\mathop{d\omega }}\,}{dt}=\vec{\alpha }\,\,\](Angular acceleration)
\[\vec{a}=\,\,\,\,\,\,\,{{\vec{a}}_{t}}\,\,\,\,\,+\,\,\,\,\,{{\vec{a}}_{c}}\] ...(iii)
\[\frac{\overset{\to }{\mathop{dr}}\,}{dt}=\vec{\upsilon }\,\] (Linear velocity)
Thus the resultant acceleration of the particle at P has two component accelerations
(1) Tangential acceleration : \[\overrightarrow{{{a}_{t}}}=\overrightarrow{\alpha }\times \overrightarrow{\,r}\]
It acts along the tangent to the circular path at P in the plane of circular path.
According to right hand rule since \[\vec{\alpha }\] and \[\vec{r}\] are perpendicular to each other, therefore, the magnitude of tangential acceleration is given by
\[|{{\overrightarrow{a}}_{t}}|\,=\,|\overrightarrow{\alpha }\,\times \,\overrightarrow{r}|\,=\,\alpha \,r\,\sin \,{{90}^{o}}\,=\alpha \,r.\]
(2) Centripetal (Radial) acceleration : \[\overrightarrow{{{a}_{c}}}=\overrightarrow{\omega }\times \overrightarrow{v}\]
It is also called centripetal acceleration of the particle at P. It acts along the radius of the particle at P.
According to right hand rule since \[\overrightarrow{\omega }\] and \[\overrightarrow{\upsilon }\] are perpendicular to each other, therefore, the magnitude of centripetal acceleration is given by
\[|{{\vec{a}}_{c}}|\,=\,|\vec{\omega }\,\times \,\vec{\upsilon }|\,=\,\omega \,\upsilon \,\sin \,{{90}^{o}}\]= \[\omega \,\upsilon \,=\,\omega (\omega \,r)\,=\,{{\omega }^{2}}r={{\upsilon }^{2}}/r\]
Tangential and centripetal acceleration
(1) When car moves on a concave bridge then
Centripetal force = \[R-mg\cos \theta \] \[=\frac{m{{v}^{2}}}{r}\]
and reaction \[R=mg\cos \theta +\frac{m{{v}^{2}}}{r}\]
(2) When car moves on a convex bridge
Centripetal force = \[mg\cos \theta -R\] \[=\frac{m{{v}^{2}}}{r}\]
and reaction \[R=mg\cos \theta -\frac{m{{v}^{2}}}{r}\]
When a charged particle having mass m, charge q enters perpendicularly in a magnetic field B with velocity \[\upsilon \] then it describes a circular path. Because magnetic force (q\[\upsilon \]B) works in the perpendicular direction of v and it provides required centripetal force
Magnetic force = Centripetal force
q\[\upsilon \]B = \[\frac{m{{v}^{2}}}{r}\]
\[\therefore \] radius of the circular path
\[r=\frac{mv}{qB}\]
When a car moves in a circular path with speed more than a certain maximum speed then it overturns even if friction is sufficient to avoid skidding and its inner wheel leaves the ground first
Weight of the car = mg
Speed of the car = \[\upsilon \]
Radius of the circular path = r
Distance between the centre of wheels of the car = 2a
Height of the centre of gravity (G) of the car from the road level = h
Reaction on the inner wheel of the car by the ground \[={{R}_{1}}\]
Reaction on the outer wheel of the car by the ground \[={{R}_{2}}\]
When a car move in a circular path, horizontal friction force F provides the required centripetal force
i.e., \[F=\frac{m{{v}^{2}}}{R}\] ...(i)
For rotational equilibrium, by taking the moment of forces \[{{R}_{1}},\,\,{{R}_{2}}\] and F about G
\[Fh+{{R}_{1}}a={{R}_{2}}a\] ...(ii)
As there is no vertical motion so \[{{R}_{1}}+\,{{R}_{2}}=mg\] ...(iii)
By solving (i), (ii) and (iii)
\[{{R}_{1}}=\frac{1}{2}M\left[ g-\frac{{{v}^{2}}h}{ra} \right]\] ...(iv)
and \[{{R}_{2}}=\frac{1}{2}M\left[ g+\frac{{{v}^{2}}h}{ra} \right]\] ...(v)
It is clear from equation (iv) that if \[\upsilon \] increases value of \[{{R}_{1}}\] decreases and for \[{{R}_{1}}=0\]
\[\frac{{{v}^{2}}h}{ra}=g\] or \[v=\sqrt{\frac{gra}{h}}\]
i.e. the maximum speed of a car without overturning on a flat road is given by \[v=\sqrt{\frac{gra}{h}}\]
For getting a centripetal force, cyclist bend towards the centre of circular path but it is not possible in case of four wheelers.
Therefore, outer bed of the road is raised so that a vehicle moving on it gets automatically inclined towards the centre.
In the figure (A) shown reaction R is resolved into two components, the component R cos \[\theta \] balances weight of vehicle
\[\therefore \] \[R\,\,\cos \,\,\theta =mg\] ...(i)
and the horizontal component R sin \[\theta \] provides necessary centripetal force as it is directed towards centre of desired circle
Thus \[R\,\,\sin \,\,\theta =\frac{m{{v}^{2}}}{r}\] ...(ii)
Dividing (ii) by (i), we have
\[\tan \,\theta =\frac{{{v}^{2}}}{r\,g}\] ...(iii)
or \[\tan \theta =\frac{{{\omega }^{2}}r}{g}=\frac{v\omega }{g}\] ...(iv) \[[As\,\,\upsilon =r\omega ]\]
If l = width of the road, h = height of the outer edge from the ground level then from the figure (B)
\[\tan \theta =\frac{h}{x}=\frac{h}{l}\] ...(v) [since \[\theta \] is very small]
From equation (iii), (iv) and (v)
\[\tan \theta =\frac{{{v}^{2}}}{rg}\]\[=\frac{{{\omega }^{2}}r}{g}=\frac{v\omega }{g}=\frac{h}{l}\]
Note :
If friction is also present between the tyres and road then \[\frac{{{v}^{2}}}{rg}=\frac{\mu +\tan \theta }{1-\mu \tan \theta }\]
Maximum safe speed on a banked frictional road \[v=\sqrt{\frac{rg(\mu +\tan \theta )}{1-\mu \tan \theta }}\]
A cyclist provides himself the necessary centripetal force by leaning inward on a horizontal track, while going round a curve. Consider a cyclist of weight mg taking a turn of radius r with velocity \[\upsilon \]. In order to provide the necessary centripetal force, the cyclist leans through angle \[\theta \] inwards as shown in figure.
The cyclist is under the action of the following forces :
The weight mg acting vertically downward at the centre of gravity of cycle and the cyclist.
The reaction R of the ground on cyclist. It will act along a line-making angle \[\theta \] with the vertical.
The vertical component R cos \[\theta \] of the normal reaction R will balance the weight of the cyclist, while the horizontal component R sin \[\theta \] will provide the necessary centripetal force to the cyclist.
\[R\sin \theta =\frac{m{{v}^{2}}}{r}\] ...(i)
and R cos q = mg ...(ii)
Dividing equation (i) by (ii), we have
\[\frac{R\sin \theta }{R\cos \theta }=\frac{m{{{v}^{2}}}/{r}\;}{mg}\]
or \[\tan \theta =\frac{{{v}^{2}}}{rg}\] ...(iii)
Therefore, the cyclist should bend through an angle \[\theta ={{\tan }^{-1}}\left( \frac{{{v}^{2}}}{rg} \right)\]
It follows that the angle through which cyclist should bend will be greater, if
(i) The radius of the curve is small i.e. the curve is sharper
(ii) The velocity of the cyclist is large.
Note :
For the same reasons, an ice skater or an aeroplane has to bend inwards, while taking a turn.
On a rotating platform, to avoid the skidding of an object (mass m) placed at a distance r from axis of rotation, the centripetal force should be provided by force of friction.
Centripetal force \[\le \] Force of friction
\[m{{\omega }^{2}}r\le \mu mg\]
\[\therefore \,\,{{\omega }_{\max }}=\sqrt{(\mu g/r)},\]
Hence maximum angular velocity of rotation of the platform is \[\sqrt{(\mu g/r)}\,,\] so that object will not skid on it.
(1) Tension in the string : \[T=mg\sqrt{1+{{\left( \frac{{{\upsilon }^{2}}}{rg} \right)}^{2}}}\] ...(i)
\[T=\frac{mg}{\cos \theta }=\frac{mgl}{\sqrt{{{l}^{2}}-{{r}^{2}}}}\]
[As \[\cos \theta =\frac{h}{l}=\frac{\sqrt{{{l}^{2}}-{{r}^{2}}}}{l}\]]
(2) Angle of string from the vertical :
\[\tan \theta =\frac{{{\upsilon }^{2}}}{rg}\] ...(ii)
(3) Linear velocity of the bob :
\[\upsilon =\sqrt{gr\tan \theta }\]
(4) Angular velocity of the bob :
\[\omega =\sqrt{\frac{g}{r}\tan \theta }=\sqrt{\frac{g}{h}}=\sqrt{\frac{g}{l\cos \theta }}\]
(5) Time period of revolution :
\[{{T}_{p}}=2\pi \sqrt{\frac{l\cos \theta }{g}}=2\pi \sqrt{\frac{h}{g}}\]
\[=2\pi \sqrt{\frac{{{l}^{2}}-{{r}^{2}}}{g}}=2\pi \sqrt{\frac{r}{g\tan \theta }}\] ...(iii)
If \[\theta ={{90}^{o}}\], then pendulum becomes horizontal & from equations (i), (ii) and (iii) we get \[\upsilon =\infty ,\,\,T=\infty \] and \[{{T}_{p}}=0\] which is practically not possible. 
(2) Tension at any point on vertical loop : Tension at general point P, According to Newton?s second law of motion.
Net force towards centre = centripetal force
\[T-mg\cos \theta =\frac{m{{v}^{2}}}{l}\]
or \[T=mg\cos \theta +\frac{m{{v}^{2}}}{l}\]
\[T=\frac{m}{l}[{{u}^{2}}-gl(2-3\cos \theta )]\]
[As \[v=\sqrt{{{u}^{2}}-2gl(1-\cos \theta )}\]]
Velocity and tension in a vertical loop
\[\frac{\overset{\to }{\mathop{d\omega }}\,}{dt}=\vec{\alpha }\,\,\](Angular acceleration)
\[\vec{a}=\,\,\,\,\,\,\,{{\vec{a}}_{t}}\,\,\,\,\,+\,\,\,\,\,{{\vec{a}}_{c}}\] ...(iii)
\[\frac{\overset{\to }{\mathop{dr}}\,}{dt}=\vec{\upsilon }\,\] (Linear velocity)
Thus the resultant acceleration of the particle at P has two component accelerations
(1) Tangential acceleration : \[\overrightarrow{{{a}_{t}}}=\overrightarrow{\alpha }\times \overrightarrow{\,r}\]
It acts along the tangent to the circular path at P in the plane of circular path.
According to right hand rule since \[\vec{\alpha }\] and \[\vec{r}\] are perpendicular to each other, therefore, the magnitude of tangential acceleration is given by
\[|{{\overrightarrow{a}}_{t}}|\,=\,|\overrightarrow{\alpha }\,\times \,\overrightarrow{r}|\,=\,\alpha \,r\,\sin \,{{90}^{o}}\,=\alpha \,r.\]
(2) Centripetal (Radial) acceleration : \[\overrightarrow{{{a}_{c}}}=\overrightarrow{\omega }\times \overrightarrow{v}\]
It is also called centripetal acceleration of the particle at P. It acts along the radius of the particle at P.
According to right hand rule since \[\overrightarrow{\omega }\] and \[\overrightarrow{\upsilon }\] are perpendicular to each other, therefore, the magnitude of centripetal acceleration is given by
\[|{{\vec{a}}_{c}}|\,=\,|\vec{\omega }\,\times \,\vec{\upsilon }|\,=\,\omega \,\upsilon \,\sin \,{{90}^{o}}\]= \[\omega \,\upsilon \,=\,\omega (\omega \,r)\,=\,{{\omega }^{2}}r={{\upsilon }^{2}}/r\]
Tangential and centripetal acceleration
Centripetal force = \[R-mg\cos \theta \] \[=\frac{m{{v}^{2}}}{r}\]
and reaction \[R=mg\cos \theta +\frac{m{{v}^{2}}}{r}\]
(2) When car moves on a convex bridge
Centripetal force = \[mg\cos \theta -R\] \[=\frac{m{{v}^{2}}}{r}\]
and reaction \[R=mg\cos \theta -\frac{m{{v}^{2}}}{r}\] 
Weight of the car = mg
Speed of the car = \[\upsilon \]
Radius of the circular path = r
Distance between the centre of wheels of the car = 2a
Height of the centre of gravity (G) of the car from the road level = h
Reaction on the inner wheel of the car by the ground \[={{R}_{1}}\]
Reaction on the outer wheel of the car by the ground \[={{R}_{2}}\]
When a car move in a circular path, horizontal friction force F provides the required centripetal force
i.e., \[F=\frac{m{{v}^{2}}}{R}\] ...(i)
For rotational equilibrium, by taking the moment of forces \[{{R}_{1}},\,\,{{R}_{2}}\] and F about G
\[Fh+{{R}_{1}}a={{R}_{2}}a\] ...(ii)
As there is no vertical motion so \[{{R}_{1}}+\,{{R}_{2}}=mg\] ...(iii)
By solving (i), (ii) and (iii)
\[{{R}_{1}}=\frac{1}{2}M\left[ g-\frac{{{v}^{2}}h}{ra} \right]\] ...(iv)
and \[{{R}_{2}}=\frac{1}{2}M\left[ g+\frac{{{v}^{2}}h}{ra} \right]\] ...(v)
It is clear from equation (iv) that if \[\upsilon \] increases value of \[{{R}_{1}}\] decreases and for \[{{R}_{1}}=0\]
\[\frac{{{v}^{2}}h}{ra}=g\] or \[v=\sqrt{\frac{gra}{h}}\]
i.e. the maximum speed of a car without overturning on a flat road is given by \[v=\sqrt{\frac{gra}{h}}\] 
In the figure (A) shown reaction R is resolved into two components, the component R cos \[\theta \] balances weight of vehicle
\[\therefore \] \[R\,\,\cos \,\,\theta =mg\] ...(i)
and the horizontal component R sin \[\theta \] provides necessary centripetal force as it is directed towards centre of desired circle
Thus \[R\,\,\sin \,\,\theta =\frac{m{{v}^{2}}}{r}\] ...(ii)
Dividing (ii) by (i), we have
\[\tan \,\theta =\frac{{{v}^{2}}}{r\,g}\] ...(iii)
or \[\tan \theta =\frac{{{\omega }^{2}}r}{g}=\frac{v\omega }{g}\] ...(iv) \[[As\,\,\upsilon =r\omega ]\]
If l = width of the road, h = height of the outer edge from the ground level then from the figure (B)
\[\tan \theta =\frac{h}{x}=\frac{h}{l}\] ...(v) [since \[\theta \] is very small]
From equation (iii), (iv) and (v)
\[\tan \theta =\frac{{{v}^{2}}}{rg}\]\[=\frac{{{\omega }^{2}}r}{g}=\frac{v\omega }{g}=\frac{h}{l}\]
Note :
\[R\sin \theta =\frac{m{{v}^{2}}}{r}\] ...(i)
and R cos q = mg ...(ii)
Dividing equation (i) by (ii), we have
\[\frac{R\sin \theta }{R\cos \theta }=\frac{m{{{v}^{2}}}/{r}\;}{mg}\]
or \[\tan \theta =\frac{{{v}^{2}}}{rg}\] ...(iii)
Therefore, the cyclist should bend through an angle \[\theta ={{\tan }^{-1}}\left( \frac{{{v}^{2}}}{rg} \right)\]
It follows that the angle through which cyclist should bend will be greater, if
(i) The radius of the curve is small i.e. the curve is sharper
(ii) The velocity of the cyclist is large.
Note :
