JEE Physics Circular Motion Notes Set 04

Kinematics of Circular Motion

For motion on a circle of radius r, the angular velocity is \( \omega=\frac{d\theta}{dt} \), the linear speed is \( v=\omega r \) (as a vector, \( \vec v=\vec\omega\times\vec r \)), and the angular acceleration is \( \alpha=\frac{d\omega}{dt} \). For uniform angular acceleration from rest, \( \theta=\frac12\alpha t^2 \), so successive equal intervals cover angles in the ratio 1 : 3 : 5.
The chord between two positions separated by angle \( \phi \) is \( 2r\sin\frac\phi2 \), and the change in velocity of a particle of speed v whose direction turns through \( \phi \) is \( 2v\sin\frac\phi2 \). A seconds hand has \( \omega=\frac{\pi}{30} \) rad/s.

Centripetal and Tangential Acceleration

The radial (centripetal) acceleration is \( a_c=\frac{v^2}{r}=\omega^2r \) towards the centre, and the tangential acceleration is \( a_t=\frac{dv}{dt} \). They are perpendicular, so the total acceleration is \( \sqrt{a_c^2+a_t^2} \). In uniform circular motion \( a_t=0 \) and the total acceleration is purely centripetal, with constant magnitude but changing direction.
The centripetal force is not a new force; it is the net inward force supplied by tension, friction, the normal reaction or gravity. If the sense of rotation is reversed at the same speed, the centripetal force is unchanged.

Turning on Level and Banked Roads

On a flat road friction supplies the centripetal force, so the maximum safe speed is \( v=\sqrt{\mu rg} \). A cyclist or motorcyclist leans inward at an angle with \( \tan\theta=\frac{v^2}{rg} \) so that the torques about the centre of gravity balance.
On a road banked at angle \( \theta \) with no reliance on friction, \( \tan\theta=\frac{v^2}{rg} \); with friction the safe speed range is \( v^2=rg\,\frac{\tan\theta\pm\mu}{1\mp\mu\tan\theta} \). For a road of width b, the height difference between its edges is \( h=b\sin\theta\approx\frac{v^2b}{rg} \). For a fixed banking angle, \( v^2\propto r \).

Vertical Circle

For a particle on a string, the tension at angle \( \theta \) from the lowest point is \( T=\frac{mv^2}{l}+mg\cos\theta \). With energy conservation \( v^2=u^2-2gl(1-\cos\theta) \). To complete the circle the speed at the bottom must be at least \( \sqrt{5gl} \), the speed at the top at least \( \sqrt{gl} \), and the tensions at bottom and top then differ by 6mg.
If the bottom speed is below \( \sqrt{2gl} \) the particle oscillates below the level of suspension, and between \( \sqrt{2gl} \) and \( \sqrt{5gl} \) the string goes slack above the horizontal. With a light rod instead of a string the minimum speed at the top is zero, so the bottom speed need only be \( \sqrt{4gl} \). A block sliding down a smooth track completes a loop of radius r if released from a height \( h\ge\frac{5r}{2} \). A pendulum released from the horizontal has tension 3mg at the lowest point. The total acceleration of a pendulum bob combines \( g\sin\theta \) (tangential) and \( \frac{v^2}{l} \) (radial).