JEE Physics System Of Particles and Rotational Motion Notes Set 05

Rotational Quantities and Equations

A rigid body turning about a fixed axis is described by angular displacement \(\theta\), angular velocity \(\omega = \frac{d\theta}{dt}\) and angular acceleration \(\alpha = \frac{d\omega}{dt}\). Every particle of the body has the same \(\omega\) and \(\alpha\), while its linear speed is \(v = \omega r\), tangential acceleration \(a_t = \alpha r\) and centripetal acceleration \(a_c = \omega^2 r\). The vector form is \(\vec v = \vec\omega\times\vec r\).

For constant \(\alpha\) the equations mirror straight-line motion: \(\omega = \omega_0 + \alpha t\), \(\theta = \omega_0 t + \frac{1}{2}\alpha t^2\), \(\omega^2 = \omega_0^2 + 2\alpha\theta\). Frequency in rps is converted with \(\omega = 2\pi f\), and rpm with \(\omega = \frac{2\pi N}{60}\).

Moment of Inertia

Moment of inertia \(I = \sum m_i r_i^2\) measures how hard it is to change the rotation about a given axis. It depends on the mass, on how the mass is spread, and on the position of the axis. Radius of gyration \(K\) is defined by \(I = MK^2\).

1. Parallel axis theorem: \(I = I_{cm} + Md^2\), where the new axis is parallel to the one through the centre of mass at distance \(d\).
2. Perpendicular axis theorem (for flat plates only): \(I_z = I_x + I_y\), where the x and y axes lie in the plane and meet at the point where the z axis passes.
3. For identical pieces that together make up a symmetric body, divide \(I\) equally. A quarter disc of mass \(M\) about the centre has \(\frac{1}{2}MR^2\), the same as a full disc of mass \(M\).

Torque and Angular Momentum

1. Torque \(\vec\tau = \vec r\times\vec F\), with magnitude \(rF\sin\phi\). Rotation law: \(\tau = I\alpha\).
2. Angular momentum of a particle \(\vec L = \vec r\times\vec p\); for a rigid body about a fixed axis \(L = I\omega\). Also \(\tau = \frac{dL}{dt}\).
3. If the net external torque is zero, \(L\) is constant. A skater pulling in the arms, a diver tucking and the earth changing size (period \(T\propto R^2\)) are examples.
4. Angular velocity of a moving particle about a point is \(\omega = \frac{\vec r\times\vec v}{r^2}\).

Equilibrium and Toppling

A body is in rotational equilibrium when the total torque about any point is zero. A block just begins to topple when the line of action of the normal reaction shifts to its edge, so take torques about that edge and equate the toppling torque to the restoring torque of the weight.

Rolling Motion

1. Rolling without slipping needs \(v_{cm} = \omega R\). The contact point is instantaneously at rest and the top point moves at \(2v_{cm}\).
2. Total kinetic energy \(= \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2\left(1 + \frac{K^2}{R^2}\right)\). Rotational to translational KE ratio \(= \frac{K^2}{R^2}\).
3. On an incline of angle \(\theta\): \(a = \frac{g\sin\theta}{1 + K^2/R^2}\). The body with smaller \(\frac{K^2}{R^2}\) reaches the bottom first: solid sphere, then disc, then hollow sphere, then ring.
4. Friction needed on the incline is \(f = \frac{Mg\sin\theta}{1 + R^2/K^2}\), so \(\mu_{min} = \frac{\tan\theta}{1 + R^2/K^2}\).
5. In pure rolling static friction does no work, so mechanical energy is conserved. If the body slides first, kinetic friction acts, energy is lost, and angular momentum about the contact point (or any point on the surface line) is conserved.
6. Rolling bodies on an incline: a force applied above the centre can make friction act forward, so write the two equations \(F_{net} = Ma\) and \(\tau_{cm} = I\alpha\) with the rolling condition.

Worked Examples

1. A disc of radius \(0.2\ \text{m}\) and mass \(2\ \text{kg}\) has a string wound on its rim and is pulled with \(F = 4\ \text{N}\) while fixed on its axis. \(I = \frac{1}{2}(2)(0.04) = 0.04\ \text{kg m}^2\), torque \(= 4\times 0.2 = 0.8\ \text{N m}\), so \(\alpha = \frac{0.8}{0.04} = 20\ \text{rad s}^{-2}\).
2. A ring and a disc of the same mass and radius are released together on a rough incline. The disc has the larger acceleration: \(a_d = \frac{2}{3}g\sin\theta\) against \(a_r = \frac{1}{2}g\sin\theta\). The disc reaches the bottom first, and the time ratio is \(\frac{t_d}{t_r} = \sqrt{\frac{3}{4}}\).
3. A skater spins at \(2\ \text{rev s}^{-1}\) with \(I = 3\ \text{kg m}^2\). When she pulls in her arms, \(I\) becomes \(2\ \text{kg m}^2\). Then \(I\omega\) is conserved, so \(\omega' = 3\ \text{rev s}^{-1}\), and the kinetic energy rises from \(\frac{1}{2}(3)(4\pi)^2\) to \(\frac{1}{2}(2)(6\pi)^2\), the extra energy being supplied by her muscles.
4. A uniform rod of length \(L\) hinged at one end is released from the horizontal position. Its angular acceleration is \(\alpha = \frac{MgL/2}{ML^2/3} = \frac{3g}{2L}\). At the vertical position, \(\frac{1}{2}\cdot\frac{ML^2}{3}\omega^2 = Mg\frac{L}{2}\), so \(\omega = \sqrt{\frac{3g}{L}}\).

Common Mistakes

1. Using the perpendicular axis theorem for a three-dimensional body such as a sphere or cylinder. It is valid only for plates.
2. Using \(\frac{2}{5}MR^2\) for a hollow shell. The shell has \(\frac{2}{3}MR^2\).
3. Applying the parallel axis theorem between two axes, neither of which passes through the centre of mass.
4. Assuming friction always acts backwards in rolling. Its direction follows from the two equations, and it can be zero or forward.