JEE Physics Circular Motion Notes Set 05

Angular Position, Displacement, Velocity and Acceleration

The angular position of a particle moving in a circle is the angle its position vector makes with a chosen reference line. The angle swept in a time interval is the angular displacement, measured in radians (\( 2\pi\ \text{rad}=360^\circ=1\) revolution). Only very small angular displacements add like vectors; finite ones do not.
Angular velocity is \( \omega=\frac{d\theta}{dt} \) and angular acceleration is \( \alpha=\frac{d\omega}{dt}=\omega\frac{d\omega}{d\theta} \). Their directions follow the right-hand thumb rule. For n turns in time t, \( \omega_{av}=\frac{2\pi n}{t} \), and for a period T, \( \omega=\frac{2\pi}{T}=2\pi f \).
For constant \( \alpha \) the equations mirror straight-line motion: \( \omega=\omega_0+\alpha t,\ \theta=\omega_0t+\tfrac12\alpha t^2,\ \omega^2=\omega_0^2+2\alpha\theta \).
Worked example: a fan turning at 100 rev/s is switched off and stops uniformly in 5 minutes. Total turns \( =\frac{100}{2}\times300=15000 \), and the average angular velocity is 50 rev/s.

Speed and Angular Velocity

The linear speed of a particle on a circle of radius r is \( v=\omega r \), or in vector form \( \vec v=\vec\omega\times\vec r \). All points of a rigid rotating body share the same \( \omega \) but have different speeds in proportion to their distance from the axis.
Worked example: the hour hand of a clock turns once in 12 hours and the Earth once in 24 hours, so the hour hand has twice the angular velocity of the Earth's rotation.

Relative Angular Velocity

Angular velocity depends on the point from which the position vector is drawn. The angular velocity of A with respect to B is the component of the relative velocity perpendicular to the line AB divided by the separation: \( \omega_{AB}=\frac{(v_{AB})_\perp}{r_{AB}} \).
For two particles circling the same centre O with angular speeds \( \omega_1 \) and \( \omega_2 \), the angle between OA and OB changes at the rate \( \omega_2-\omega_1 \). The time for one to gain a full turn on the other is \( T=\frac{2\pi}{\omega_{rel}}=\frac{T_1T_2}{T_1\sim T_2} \). Note that \( \omega_2-\omega_1 \) is not the angular velocity of B with respect to A.
Worked example: the minute hand and second hand coincide again after \( t \) given by \( (2\pi-\frac{2\pi}{60})t=2\pi \) in minutes, so \( t=\frac{60}{59} \) minute.

Tangential and Radial (Centripetal) Acceleration

The acceleration of a particle in circular motion has two perpendicular parts. The tangential acceleration \( a_t=\frac{d|\vec v|}{dt}=r\alpha \) changes the speed. The radial (centripetal) acceleration \( a_r=\frac{v^2}{r}=\omega^2r \) always points to the centre and changes only the direction of velocity.
The magnitude of the total acceleration is \( a=\sqrt{a_t^2+a_r^2} \) and its angle with the radius satisfies \( \tan\phi=\frac{a_t}{a_r} \). Remember that \( \left|\frac{d\vec v}{dt}\right| \) (magnitude of total acceleration) and \( \frac{d|\vec v|}{dt} \) (tangential acceleration) are different quantities. The formula \( v^2/r \) holds even when the speed is changing.
Worked example: a particle moves in a circle of radius 2 cm with speed \( v=4t \) cm/s. At t = 1 s, \( a_t=4 \) cm/s² and \( a_r=\frac{16}{2}=8 \) cm/s², so \( a=\sqrt{16+64}=4\sqrt5 \) cm/s².

Dynamics of Circular Motion

A body moving on a circle needs a net inward force. With constant speed the required centripetal force is \( F_c=\frac{mv^2}{r}=m\omega^2r \). If the speed changes there is also a tangential force \( F_t=m\,a_t \).
The quantity \( mv^2/r \) is not a new force. It is the required value of the resultant of all real forces (tension, friction, normal reaction, gravity, spring force) that act along the inward radial direction. So the method is always: identify the forces along the radius, add them towards the centre and equate to \( mv^2/r \).

Circular Motion in a Horizontal Plane

Conical pendulum: a bob of mass m on a string of length l, making an angle \( \theta \) with the vertical, moves in a horizontal circle. Vertically \( T\cos\theta=mg \) and horizontally \( T\sin\theta=m\omega^2r \), giving \( \tan\theta=\frac{\omega^2r}{g} \) and a period \( T=2\pi\sqrt{\frac{l\cos\theta}{g}} \).
Wall of death: a rider on the vertical wall of a cylinder is held up by friction, so \( \mu N\ge mg \) with \( N=\frac{mv^2}{r} \). The minimum speed is \( v=\sqrt{\frac{rg}{\mu}} \) (equivalently the minimum friction coefficient is \( \frac{rg}{v^2} \)).
A coin on a turntable stays at rest while \( \mu mg\ge m\omega^2r \), so it slips at \( r_{max}=\frac{\mu g}{\omega^2} \). For a particle whirled on a horizontal surface, the tension is \( T=m\omega^2r \), so doubling both r and \( \omega \) multiplies it by 8.
For a uniform rod rotating about one end, the tension at distance x from the axis falls steadily outward because each section supplies the centripetal force for the part beyond it.

Radius of Curvature

Any smooth path can be treated at a point as part of a circle of radius R, called the radius of curvature. Splitting the acceleration into components along and normal to the velocity, \( R=\frac{v^2}{a_n} \), where \( a_n \) is the component of acceleration perpendicular to the velocity.
For a projectile launched at speed u at angle \( \theta \) to the horizontal, \( a_n=g\cos\theta \) at launch, so \( R=\frac{u^2}{g\cos\theta} \). At the highest point the speed is \( u\cos\theta \) and \( a_n=g \), so \( R=\frac{u^2\cos^2\theta}{g} \). In general \( R\propto v^3 \) for a projectile. For a constant force F perpendicular to the initial velocity v of a mass m, \( R=\frac{mv^2}{F} \).

Motion in a Vertical Circle

For a mass m on a string of length l (or on the inside of a track) the speed changes with height, so both radial and tangential equations are needed. At an angle \( \theta \) from the lowest point, \( T-mg\cos\theta=\frac{mv^2}{l} \), and energy conservation gives \( v^2=u^2-2gl(1-\cos\theta) \). Hence \( T=\frac{mu^2}{l}+mg(3\cos\theta-2) \), which decreases as the body rises.
Looping the loop with a string: the tension at the top must not be negative, so \( v_{top}\ge\sqrt{gl} \) and the minimum speed at the bottom is \( \sqrt{5gl} \). The bottom tension is then \( 6mg \), the top tension is zero, and the difference between bottom and top tensions is always \( 6mg \).
If the bottom speed is between \( \sqrt{2gl} \) and \( \sqrt{5gl} \) the string goes slack somewhere above the horizontal and the body leaves the circle. The slack position satisfies \( v^2=gl\cos\phi \) with \( \phi \) measured from the upward vertical. If the bottom speed does not exceed \( \sqrt{2gl} \) the body only oscillates, and the tension is never zero.
Worked example: a bob whose tension at the mean position is 3mg rises to a height l, that is, to the horizontal (90 degrees), because \( \frac{mv^2}{l}=2mg \) gives \( v^2=2gl \).

Turning on Roads: Friction and Banking

On a level road friction supplies the centripetal force, so the safe speed is \( v_{max}=\sqrt{\mu rg} \). A perfectly banked road (angle \( \theta \)) needs no friction at the design speed: \( \tan\theta=\frac{v^2}{rg} \).
With both banking and friction the limits are \( v_{max}=\sqrt{rg\,\frac{\tan\theta+\mu}{1-\mu\tan\theta}} \) and \( v_{min}=\sqrt{rg\,\frac{\tan\theta-\mu}{1+\mu\tan\theta}} \). Below \( v_{min} \) the vehicle slips down the bank and above \( v_{max} \) it skids outwards.
For a railway track the outer rail is raised by h over a gauge width l such that \( \frac{h}{l}\approx\tan\theta=\frac{v^2}{rg} \).
On a bridge or hump the normal reaction is \( N=mg\cos\theta-\frac{mv^2}{r} \), so a vehicle on a convex bridge feels a lighter push from the road, and in a dip \( N=mg+\frac{mv^2}{r} \) is larger.

Centrifugal Force and the Effect of the Earth's Rotation

In a frame rotating with the body, a pseudo force \( m\omega^2r \) acts radially outward, called the centrifugal force. It is not a force from any object and does not appear in an inertial frame; there the only requirement is that the real forces add up to \( mv^2/r \) inwards.
Because the Earth rotates, a body at latitude \( \lambda \) has an apparent weight \( mg'=mg-m\omega^2R\cos^2\lambda \) (to a good approximation). At the equator the reading is smaller than the true weight by \( m\omega^2R \), about 0.34 percent, and at the poles there is no reduction. The apparent weight at the equator would become zero if \( \omega=\sqrt{g/R} \), which corresponds to a day of about 84 minutes.
A train running west to east along the equator moves faster in the inertial frame than one running east to west, so it presses less on the track.