JEE Physics Circular Motion Notes Set 01

Describing Circular Motion

A particle is in circular motion when its distance from a fixed point (the centre) stays constant. The line from the centre to the particle is the radius vector; it has constant length but a changing direction.

Angular Quantities

Angular displacement \( \Delta\theta = \dfrac{\text{arc}}{\text{radius}} \) (radian). Angular velocity \( \omega = \dfrac{d\theta}{dt} \) in rad/s; for one revolution in time T, \( \omega = \dfrac{2\pi}{T} = 2\pi f \). Angular acceleration \( \alpha = \dfrac{d\omega}{dt} \). Both \( \vec\omega \) and \( \vec\alpha \) are axial vectors along the axis of rotation (right-hand rule). Finite angular displacements do not add as vectors, though very small ones do. 1 rps = 60 rpm.

Linear and Angular Variables

The speed of the particle is \( v = \omega r \), or in vector form \( \vec v = \vec\omega\times\vec r \). Velocity, angular velocity and radius vector are mutually perpendicular. The tangential acceleration is \( a_t = \alpha r = \dfrac{dv}{dt} \), and the centripetal (radial) acceleration is \( a_c = \omega^2 r = \dfrac{v^2}{r} = \omega v \). The total acceleration has magnitude \( \sqrt{a_t^2 + a_c^2} \) and makes an angle with the radius given by \( \tan\phi = a_t/a_c \).

Equations for Constant Angular Acceleration

These mirror the straight-line equations: \( \omega = \omega_0 + \alpha t \), \( \theta = \omega_0 t + \tfrac{1}{2}\alpha t^2 \), \( \omega^2 = \omega_0^2 + 2\alpha\theta \), \( \theta_{n^{th}} = \omega_0 + \tfrac{\alpha}{2}(2n-1) \). If α is not constant, use \( \omega = \dfrac{d\theta}{dt} \), \( \alpha = \omega\dfrac{d\omega}{d\theta} \) and integrate.

Uniform and Non-Uniform Circular Motion

Uniform circular motion: speed constant, \( a_t = 0 \), \( \alpha = 0 \), kinetic energy constant. The velocity changes in direction, so the acceleration \( v^2/r \) (towards the centre) is non-zero, constant in magnitude but not in direction. The net force \( mv^2/r \) does no work and delivers no power.
Non-uniform circular motion: the speed changes, so \( a_t \neq 0 \). The tangential force does work, and the power is \( F_t v \). The angle between velocity and acceleration satisfies \( \tan\phi = a_c/a_t \).

Relative Angular Velocity

The angular velocity of A with respect to B is \( \omega_{AB} = \dfrac{(v_{AB})_\perp}{r_{AB}} \), the component of relative velocity perpendicular to the line joining them, divided by their separation. For a particle moving parallel to the x-axis at height b, the angular velocity about the origin is \( \dfrac{v}{b}\sin^2\theta \).

Worked Examples

Worked Example 1. A wheel starts from rest with an angular acceleration of 4 rad/s². Find its angular velocity and the number of revolutions after 5 s.
Answer:
\( \omega = \alpha t = 20 \) rad/s.
\( \theta = \tfrac{1}{2}\alpha t^2 = 50 \) rad \( = 50/2\pi \approx 7.96 \) revolutions.

Worked Example 2. A particle moves on a circle of radius 2 m with speed \( v = 3t \) m/s. Find its total acceleration at t = 2 s.
Answer:
\( a_t = 3 \) m/s²; \( v = 6 \) m/s so \( a_c = 36/2 = 18 \) m/s².
\( a = \sqrt{9 + 324} \approx 18.2 \) m/s².

Worked Example 3. A particle's angular position is \( \theta = 4t^2 - t^3 \). At what time does its angular acceleration vanish?
Answer:
\( \alpha = \dfrac{d^2\theta}{dt^2} = 8 - 6t = 0 \Rightarrow t = \tfrac{4}{3} \) s.