Circular Motion

Centripetal force

The centripetal force on a body is defined as the external force which causes the body to move in a circular path with a constant speed and acts along the radius and towards the centre of the circular path.

                                                                      a=rω2$\mathrm{a}=\mathrm{r}{\omega }^2$

∴ Fc=mrω2${\mathrm{F}}_{\mathrm{c}}=\mathrm{m}\mathrm{r}{\omega }^2$

Centrifugal force

The outward forces acting on bodies when they move in circular paths are called centrifugal forces.

centrifugal force=mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$

It is not a real force but a fictitious force and is due to the inertial property of the body.

An expression for the centripetal force:

It is defined as the inward force acting on a body when it moves in a circular path. This force is directed towards the centre of the circular path. It is given by,

F = mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$

Let us consider a body of mass ‘m’ is moving in a circular path of radius ‘r’ with uniform speed ‘v’. The velocity of the body is changing due to direction. At any instant, the body is at the point A.

Let VA−→$\overrightarrow{{\mathrm{V}}_{\mathrm{A}}}$ be the velocity of the body at A

After time Δ$\mathrm{\Delta }$T , the body reaches the point B

Let VB−→$\overrightarrow{{\mathrm{V}}_{\mathrm{B}}}$ be the velocity of the body at B.

Change in velocity is,

ΔV⃗ $\mathrm{\Delta }\overrightarrow{\mathrm{V}}$ = VB−→$\overrightarrow{{\mathrm{V}}_{\mathrm{B}}}$ - VA−→$\overrightarrow{{\mathrm{V}}_{\mathrm{A}}}$ --------------i)

Angular velocity (w):

It is defined as the rate of change of angular displacement.

It is denoted by w

It is given by w = θt$\frac{\theta }{\mathrm{t}}$

Unit of W = rads-1

Average angular velocity Wav= ΔθΔt$\frac{\mathrm{\Delta }\theta }{\mathrm{\Delta }\mathrm{t}}$

Angular acceleration (α$\alpha$);

It is defined as the rate of change of angular velocity.

It is denoted by α$\alpha$

It is given by α$\alpha$ = wt$\frac{\mathrm{w}}{\mathrm{t}}$

Unit of W = rads-2

Average angular acceleration α$\alpha$av = ΔWΔt$\frac{\mathrm{\Delta }\mathrm{W}}{\mathrm{\Delta }\mathrm{t}}$

Relational between linear velocity (v) and angular velocity (ω$\omega$):

Let us consider a body is moving in a circular path of radius ‘r’. When the body reaches to B from A

Then, the angular displacement of the body is θ.

Let the arc length arc AB is denoted by S.

By trigonometry,

θ = arclengthradius$\frac{\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}}{\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{i}\mathrm{u}\mathrm{s}}$

θ = ABr$\frac{\mathrm{A}\mathrm{B}}{\mathrm{r}}$

θ$\theta$ = sr$\frac{\mathrm{s}}{\mathrm{r}}$

Or, s = r$$θ-----------i)

Differentiating eqn (i) w.r.t to t

d(S)dt$\frac{\mathrm{d}\left(\mathrm{S}\right)}{\mathrm{d}\mathrm{t}}$ = ddt$\frac{\mathrm{d}}{\mathrm{d}\mathrm{t}}$ (rθ$\theta$)

dSdt=r.dθdt$\frac{\mathrm{d}\mathrm{S}}{\mathrm{d}\mathrm{t}}=\mathrm{r}.\frac{\mathrm{d}\theta }{\mathrm{d}\mathrm{t}}$

V=rw (dSdt$\frac{\mathrm{d}\mathrm{S}}{\mathrm{d}\mathrm{t}}$ = velocity, dθdt$\frac{\mathrm{d}\theta }{\mathrm{d}\mathrm{t}}$= angular velocity)

Centripetal force in terms of angular velocity:

Let us draw a vector triangle PQR. So, that the triangles AOB and PQR are similar triangles because both are isosceles triangle with same vertex angle.

In similar triangle, the ration of corresponding side is equal.

OAOB$\frac{\mathrm{O}\mathrm{A}}{\mathrm{O}\mathrm{B}}$ = PQPR$\frac{\mathrm{P}\mathrm{Q}}{\mathrm{P}\mathrm{R}}$

or, rAB$\frac{\mathrm{r}}{\mathrm{A}\mathrm{B}}$ = VΔV$\frac{\mathrm{V}}{\mathrm{\Delta }\mathrm{V}}$ ---------ii)

We know,

Distance = speed ∗$\ast$ time

AB = v ∗$\ast$Δ$\mathrm{\Delta }$t ----------iii)

Putting the value of AB in eqn ii)

rVΔt$\frac{\mathrm{r}}{\mathrm{V}\mathrm{\Delta }\mathrm{t}}$ = VΔV$\frac{\mathrm{V}}{\mathrm{\Delta }\mathrm{V}}$

VΔV$\frac{\mathrm{V}}{\mathrm{\Delta }\mathrm{V}}$ = v2r$\frac{{\mathrm{v}}^2}{\mathrm{r}}$

a = v2r$\frac{{\mathrm{v}}^2}{\mathrm{r}}$  --------------iv)

It gives the centripetal acceleration.

By Newton’s 2nd law of motion,

F = ma

F = mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$

It gives centripetal force

As we have,

V = ω$\omega$r

Then,

F = mω2r2r$\frac{\mathrm{m}{\omega }^2{\mathrm{r}}^2}{\mathrm{r}}$

F = mrω2${\omega }^2$

It gives centripetal force in terms of angular velocity.

Motion of cyclist on a circular path:

Let us consider a cyclist of mass ‘m’ is moving on a circular path of radius ‘r’ with uniform speed ‘v’.

Let ‘mg’ be the weight of the cyclist and ‘R’ be the normal reaction on the cyclist.

To move on a circular path, he should bend through an angle θ$\theta$ from vertical . Let us resolve ‘r’ into two components. Rcosθ$\theta$ and Rsinθ$\theta$

The component Rcosθ$\theta$ balances weight of the cyclist.

∴ Rcosθ$\theta$ = mg ------------i)

The component Rsinθ$\theta$ provides necessary centripetal force to the cyclist to move on a circular path.

∴Rsinθ$\theta$ = mv2R$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{R}}$  ------------ii)

Dividing eqn ii) by i)

RsinθRcosθ$\frac{\mathrm{R}\mathrm{s}\mathrm{i}\mathrm{n}\theta }{\mathrm{R}\mathrm{c}\mathrm{o}\mathrm{s}\theta }$ = mv2r∗mg$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}\ast \mathrm{m}\mathrm{g}}$

Or, tanθ$\theta$ = v2rg$\frac{{\mathrm{v}}^2}{\mathrm{r}\mathrm{g}}$

Or, v2 = tanθ$\theta$rg

Or, v = tanθrg−−−−−√$\sqrt{\mathrm{t}\mathrm{a}\mathrm{n}\theta \mathrm{r}\mathrm{g}}$

This relation gives permissible speed of a cyclist to move on a circular path.

To provide centripetal force by himself and to allow maximum speed cyclist lean from vertical wheel moving in a circular path.

Banked track:

If the outer edge of road is slightly raised than inner edge, then it is banked road. To provide necessary centripetal force to the vehicle, roads are banked. The component of normal reaction R sinθ$\theta$ provides centripetal force to the vehicle. On the banked road the permissible speed is more, side slip of vehicle is prevented and the motion becomes independent of friction.

Let us consider a car of mass ‘m’ is moving in a circular banked track of radius ‘r’ with uniform speed ‘v’. Let ‘mg’ be the weight of the car and ‘R’ be the normal reaction on the car.

Let us resolve R in two components. Rcosθ$\theta$ and Rsinθ$\theta$, where θ$\theta$is the angle of banking.

The component ‘Rcosθ$\theta$’ balances weight of the ar.

∴ Rcosθ = mg -------- i)

The component Rsinθ$\theta$ provides necessary centripetal force to the car to move in a circular path.

∴ Rsinθ = mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$  -------------ii)

Dividing eqn ii) by i)

RsinθRcosθ$\frac{\mathrm{R}\mathrm{s}\mathrm{i}\mathrm{n}\theta }{\mathrm{R}\mathrm{c}\mathrm{o}\mathrm{s}\theta }$ = mv2r∗mg$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}\ast \mathrm{m}\mathrm{g}}$

Or, tanθ$\theta$ = v2rg$\frac{{\mathrm{v}}^2}{\mathrm{r}\mathrm{g}}$

Or, v2 = tanθ$\theta$rg

Or, v = tanθrg−−−−−√$\sqrt{\mathrm{t}\mathrm{a}\mathrm{n}\theta \mathrm{r}\mathrm{g}}$

It gives the permissible speed of car on a banked truck. Here it should be noted θ$\theta$ should be less than angle of repose.

Motion of a body in a vertical circle:

Let us consider a body of mass ‘m’ is connected at one end of a string of length ‘l’ and it is rotated in a vertical circle with uniform velocity ‘v’. Here, length of string = radius

At lowest point A,

Let ‘mg’ be the weight of the body T1 be the tension produced in the string then,

Net force towards centre = T1 – mg

mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ = T1 – mg

T1 = mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ + mg

Tmax = mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ + mg

At highest point B

Let ‘mg’ be the weight of the body and ‘T2’ be the tension produced in the string then,

Net force towards centre = T2 + mg

mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ = T2 + mg

T2 = mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ - mg

Tmin = mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ - mg

At pont C (string is horizontal)

Let ‘mg’ be the weight and T3 be the tension produced in the string then,

Net force towards centre = T3

mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ = T3

At point D

Let T1 be the tension produced in the string then net force towards centre = T – mgcosθ$\theta$

mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ = T- mgcosθ$\theta$

T = mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ + mgcosθ$\theta$

If θ$\theta$ = 0∞$\mathrm{\infty }$, then T = mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ + mgcos0∞$\mathrm{\infty }$ = mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ + mg (At A)

If θ$\theta$ = 90∞$\mathrm{\infty }$, then T = mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ + mgcos90∞$\mathrm{\infty }$ = mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ (At C)

If θ$\theta$ = 180∞$\mathrm{\infty }$, then T = mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ + mgcos180∞$\mathrm{\infty }$ = mv2r$\frac{\mathrm{m}{\mathrm{v}}^2}{\mathrm{r}}$ - mg (At B)

Horizontal plane:

Speed, kinetic energy and angular momentum physical quantities remain constant for a particle moving along a circular path in a horizontal plane.

A particle is executing circular motion with constant speed, is its acceleration also constant:

When a body moves in a circular motion with constant speed, it has acceleration because the direction of the velocity changes, while magnitude of velocity (speed) remains constant. When a particle is moving with uniform speed v the magnitude of acceleration v2/r always remains constant but its direction continuously changes which is directed towards the centre of the circular path and perpendicular to the velocity.

Cyclist lean inwards while rounding a curve:

When cyclist tends to move in curved path, sidewise frictional forces come into play between the tyres and the road.  The forces of friction act towards the centre of the curved path and hence provide necessary centripetal force and cyclist lean inwards while rounding a curve.

Curved railway tracks banked:

When a fast moving train takes a curved path, it tends to move away tangentially off the track. In order to prevent this, the curved tracks are banked on the outside to produce the necessary centripetal force required to keep the train moving in a curved path.

The passenger of a car rounding a curve does is thrown outward:

The passenger of a car rounding a curve are thrown outward by turning the tires, friction from the road exerts a force on the car which pushes the car round the curve. Hence provide necessary centrifugal force and car thrown outward while rounding a curve.

A smooth ball is placed on the circumference of a smooth disc. When the disc rotates, the ball falls down:

A smooth ball is placed on the circumference of a smooth disc. When the disc rotates, the ball falls down because The center of mass isn't stable and centrifugal force will throw it outwards.

An airplane tilts when it makes a curved flight;

An airplane tilts when it makes a curved flight because the weight of the air plane gets used in providing it the necessary centripetal force.

If a small can filled with water is rapidly swing in a vertical circle:

As a bucket is tied to a string and is rapidly spin in a vertical circle the tension force acting upon the bucket provides the centripetal force that is required for the circular motion and prevent the water from falling outside. As a bucket of water is tied to a string and spun in a circle, the tension force acting upon the bucket provides the centripetal force required for circular motion."