JEE Physics Motion In A Plane Notes Set 04

Describing Circular Motion

1. Angular displacement \(\theta\), angular velocity \(\omega=\frac{d\theta}{dt}\) and angular acceleration \(\alpha=\frac{d\omega}{dt}\) are related to the linear quantities by \(s=r\theta\), \(v=r\omega\), \(a_t=r\alpha\).
2. For constant \(\alpha\): \(\omega=\omega_0+\alpha t\), \(\theta=\omega_0t+\frac12\alpha t^2\), \(\omega^2=\omega_0^2+2\alpha\theta\). Starting from rest, the angles turned in successive equal intervals are in the ratio 1 : 3 : 5.
3. Average angular velocity is \(\frac{\Delta\theta}{\Delta t}\). The frequency f in revolutions per second gives \(\omega=2\pi f\); rpm values are divided by 60 first.
4. The displacement of a particle in uniform circular motion of radius a, measured from its starting point, is the chord \(2a\sin\frac{\omega t}{2}\).

Acceleration in Circular Motion

1. Radial (centripetal) acceleration \(a_r=\frac{v^2}{r}=\omega^2r\) points to the centre and changes only the direction of the velocity.
2. Tangential acceleration \(a_t=\frac{dv}{dt}\) changes the speed. The total acceleration is \(a=\sqrt{a_r^2+a_t^2}\), and its angle with the velocity satisfies \(\tan\alpha=\frac{a_r}{a_t}\).
3. In uniform circular motion the speed is constant but the velocity and the acceleration both change direction continuously; the acceleration is perpendicular to the velocity.
4. If the speed depends on the distance as \(v=k\sqrt s\), then \(a_t=\frac{k^2}{2}\); to find \(a_t\) in such problems write \(a_t=v\frac{dv}{ds}\).

Dynamics of Circular Motion

Vertical Circle

1. Mass on a string: tension \(T=\frac{mv^2}{r}+mg\cos\theta\) at angle \(\theta\) from the lowest point. At the top, \(T+mg=\frac{mv^2}{r}\), and the minimum speed there is \(\sqrt{rg}\).
2. To complete the circle on a string the speed at the lowest point must be at least \(\sqrt{5rg}\). The speed at the level of the centre is then \(\sqrt{3rg}\), so the centripetal acceleration there is 3g.
3. With a rigid rod or in a tube the minimum speed at the top is zero, so the minimum speed at the bottom is \(\sqrt{4rg}\).
4. To rise only to the level of the support, the speed needed at the lowest point is \(\sqrt{2gl}\); below that level the string stays taut.
5. Pendulum released from angle \(\theta_0\): at the lowest point \(T=mg(3-2\cos\theta_0)\) and \(v=\sqrt{2gl(1-\cos\theta_0)}\).
6. For a block leaving a track, the radius of curvature of its path is \(\frac{v^2}{g\cos\beta}\), where \(\beta\) is the angle of its velocity with the horizontal.
7. On a sharper valley the normal reaction \(mg+\frac{mv^2}{R}\) is greater; on a crest it is \(mg-\frac{mv^2}{R}\).

Centrifugal Force

1. In a frame rotating at angular speed \(\omega\) a particle at distance r from the axis experiences a pseudo force \(m\omega^2r\) directed away from the axis. It depends on the angular speed of the frame, not on how the particle itself moves.
2. A rider or vehicle on a curve tends to overturn or skid in proportion to \(v^2\); doubling the speed makes the tendency four times larger.