Circular Motion

 Few important terms

Motion of particles in circular paths with constant or variable speed is called circular motion.

Inertial frame: Un-accelerated frames are inertial frames. An inertial frame is either rest or moving with uniform velocity.

Non inertial frame: An accelerated frame of reference is called non-inertial frames.

Radius Vector: Vector drawn from center to position of body performing circular motion.

Angular displacement: Angle traced by radius vector for a body in a circular motion. Infinite value is scalar but infinitesimal value is vector along the axis of rotation.

It can be measured by using a simple formula. The formula is:

$θ=\dfrac{s}{r}$ where,

θ is the angular displacement,

s is the distance travelled by the body, and

r is the radius of the circle along which it is moving.

Angular velocity: Rate of change of angular displacement of a body in a circular motion. SI unit is rad/s.

Formula is $v=r \omega$; Where v is the linear velocity and r is the radius.

Angular Acceleration: Rate of change of angular velocity of a body in a circular motion. SI unit is rad/s^2. It is a vector quantity and denoted by α.

$α=ωt=\dfrac{θ}{t^2}$

Pseudo Force: The force on a body due to acceleration of non-inertial frames.

Below figure is an explanation of Pseudo force or fictitious force.

Centrifugal Force: The force on a body is due to acceleration of the rotating frame. It is radially outward.

Centripetal force: Net force that acts on an object to keep it moving along a circular path. It is directed towards the center.

It is calculated by

$F=\dfrac{mv^2}{r} = mr \omega^2$

Uniform Circular Motion

Motion of particles in circular paths with constant speed but variable acceleration is called uniform circular motion.

Ucircular motion can occur only in a horizontal plane under gravity.

Speed, angular velocity, angular momentum, kinetic energy and distance from center remain constant.

Velocity, linear momentum, force, acceleration changes in direction only.

Work done on a circular path is 0.

Non uniform Circular motion:

The velocity changes both in magnitude as well as in direction.

The velocity vector is always tangential to the path.

The acceleration vector is not perpendicular to the velocity vector.

The acceleration vector has two components.

K.E, momentum, angular momentum , angular velocity get changed.

$\textbf{i. Tangential acceleration}$ $a_t$

$\to$ $a_t$ changes the magnitude of velocity vector and is defined as, $a_t = \dfrac{dv}{dt}$

$\textbf{ii. Normal acceleration or centripetal acceleration}$ $a_c$

$a_c$ changes the direction of the velocity vector and is defined as, $a_c = \dfrac{v^2}{r}.$

• iii. The total acceleration is the vector sum of the tangential and centripetal acceleration.

So, $a = \sqrt{a_t^2+a_c^2}$

Conical Pendulum

When a mass m is suspended from a fixed support by a string of length l is rotated in a horizontal circle around the vertical with angular velocity ω, then a formed pendulum is called a conical pendulum.

Let θ be the angle of string with vertical, then

● $T \sin θ = mrω^2 = m\dfrac{v^2}{R}$

● $T \cos θ =mg$

Dividing (i) by (ii)

● $\tan θ=\dfrac{v^2}{rg}$

● $V=\sqrt{gr \tan θ}$

● Time Period $(t)=2\pi \sqrt{\dfrac{R}{g\tan θ}}= 2\pi \sqrt{\dfrac{l \cos θ}{g}}$

● From the equation of time period we can say as θ increases v increases.

● For string should be completely horizontal speed should be infinite. This explained that the body can never be whirled in a horizontal circle.

Motion of vehicle on banked circular track:

When a vehicle goes round a curve on the road with excessive speed, then there is a tendency for the vehicle to overturn outwards. To avoid this road is given a slope rising outwards. The outer wheel of the vehicle is now raised. This is known as banking.

● Let θ be the inclination of the road, then

● $R \cos θ = mg$ ……(i)

● $R \sin θ = m\dfrac{V^2}{R}$…….(ii)

● From (i) and (ii)

$\tan θ =\dfrac{V^2}{Rg}$…………(ii)

$V= \sqrt{rg \tan θ}$

$V= \sqrt{\dfrac{grH}{d}}$

This equation (ii) gives the critical speed of the vehicle.

Also if centripetal force is provided by frictional force between tires then $V=\sqrt{µmg}$

Since, $V_{max}$ is inversely proportional to d , so sports car are designed of small heights to provide greater safe velocity.

Bending of Cyclist:

When a cyclist turns around he needs centripetal force which is provided by the horizontal component of reaction R or frictional force.

So, $m\dfrac{v^2}{r}= µmg$

$v=\sqrt{ µrg}$

Also, if θ is inclination with vertical we can write $V= \sqrt{rg\tan θ}$

For less bending of the cyclist, his speed v should be smaller and radius r of the circular path should be greater.

If μ is the coefficient of friction, then for no overturning of cyclist

$v\leq \sqrt{µrg}$

Motion in Vertical Circle

● Motion in a vertical circle is a case of non-uniform circular motion because speed keeps changing due to change in Kinetic energy.

● Consider a body tied to one end of string be whirled in a vertical circle of radius r whose other end is fixed. Let at any time t, the body is at position p where the string makes an angle θ with vertical.

Speed, K.E, momentum , tension decreases as the body moves from lowest to highest point and vice versa.

Better You Know

A body moving with constant speed in a circular path is not in equilibrium.

Centrifugal force is an apparent force or pseudo force and is directed radially outward.

If a spirit level is placed with its center at the axis of a turn table , the air bubble will be at the center and it is placed at the edge , the air bubble will be at the inner edge .

When vehicle moves on horizontal circular track reaction of inner wheel is less than that of outer wheel .

When a car driver takes a circular turn and his body is pressed against the side of the car , the force acting on him is centrifugal force .

Water doesn’t fall from bucket when it is rotated because centrifugal force balances the force of gravity .

When speed of vehicle increases then reaction of inner wheel decreases and reaction of outer wheel increases .

A car sometimes overturns while taking a turn, the inner wheel leaves the ground first .

When the body moves in a uniform circular motion no works is done on it .

At critical speed , reaction at inner wheel becomes zero and it gets and it gets overturned .

In a curved bridge , the speed of vehicle as well as the force exerted by its maximum at concave side than at the convex side .

For a particle in circular motion , if its velocity becomes zero , then the particle moves towards the center along radius and if radial acceleration becomes zero , then the particles flies off tangentially .

A particle is tied to an end of a string which is whirled in a horizontal circle . If the string is cut , the particles flies off tangentially .

A sphere is suspended by a thread of length l. The minimum horizontal velocity which has to be imparted to the sphere for it to reach the highest of suspension is $\sqrt{2gl}$.

The angular velocity of a particle about the circumference and the center of a circle is in the ratio 1: 2.

A toy car moves in a horizontal circle of radius 2a with the time period T, kept on the track by tan elastic string of unstretched length ‘a’. If the car now moves in a circle of radius 3a, ( Hooke’s law being followed ), the new time period is $\sqrt{3}/2 T$.

If a bar with a bead set into angular acceleration $\alpha$,then the beads starts slipping after time \sqrt{\dfrac{\mu}{\alpha}} where μ is co-efficient of friction between bead and rod

When a tube of length l is filled completely with an incompressible fluid of mass , closed at both ends and then ratated in horizontal plane with uniform angular velocity ω , the force exerted on the other end by the liquid is $mω^2l/2$ .

The minimum rotational speed of a cylinder of radius r so as to enable an object of mass m to remain stuck to the wall is $ω_{min} = \sqrt{\dfrac{g}{\mu r}}$.

The minimum rotational speed so that a coin placed on a rotating table doesn’t slip ( i.e move with the table ) is $ω_{min} = \sqrt{\dfrac{\mu g}{r}}$.

[ HINT : $μmg \geq mω^2r$]

In a vertical circular motion , if speed at lowest point is less than critical velocity . i.e ( < $\sqrt{5rg}$. ) in two cases arises ;

i) if velocity becomes zero earlier than tensions, then the velocity at lower point is ≤ $\sqrt{2rg}$ and the object oscillates .

ii) If the tensions becomes zero earlier than velocity, then velocity at lowest point is between $\sqrt{2rg}$ and √$\sqrt{5rg}$ and the body leaves the circular path .

If the two vehicles moving two circular tracks of radius r1 and r2 at different constant speed i. e $v_1 \neq v_2$ complete revolution in equal time ($T_1= T_2$) then

i) $\dfrac{ω_1}{ω_2} =1:1 [ =\dfrac{2π}{T}$ ]

ii) $\dfrac{v_1}{v_2} = r_1 ∶ r_2 [$v = rω$ and ω is constant]

iii) $\dfrac{a_1}{a_2} = r_1:r_2$ [ $a= ω^2r$ and ω is constant ]

b. If they move at constant speed i.e $v_1 = v_2$ then

i) $\dfrac{ω_1}{ω_2} = \dfrac{ r_2}{r_1}$ [ v=rω , i.e $ω \propto 1/r$ ,v is constant]

ii) $\dfrac{a_1}{a_2} = \dfrac{r_2}{r_1}$ [ $a= v^2/r$ i.e $a \propto 1/r$ ,v is constant

When a car moves on a convex circular road

$mg \cos θ – N = mv^2/R$

At highest point ,

$N = mg- mv^2 /R <mg$

When the body/car moves on concave road

$N-mg \cos θ = mv^2/R$

$N = mg \cosθ + mv^2/R$

• At lowest point,

$N = mg + mv^2/R > mg$

[ Note : Normal reaction (N) first increases then decreases on convex bridge as well as in concave road .]

Height above which the ball should be released such that the ball just completes the circular arc of radius R

To just complete the circle $v = \sqrt{5gR}$

$v^2 = 5gR$

$KE = PE$

$\dfrac{1}{2} mv^2 = mg h$

$h =5R/2$