Particle and Rigid Body Kinematics and Dynamics

Rigid body angular acceleration

A rigid body that rotates with angular acceleration \alpha obeys the rotational equation of motion, which relates the net moment to the product of the moment of inertia and the angular acceleration. For a free body, \sum M_G = I_G\alpha holds about the mass centre G, and the translational motion of G does not enter this equation. In a symmetric assembly the mass centre lies at the geometric centre, so the moment about that centre divided by I_G gives the angular acceleration.

The moment of inertia is the sum of mr^2 over all particles, taken about the chosen axis. The values to memorize are I_G = \frac{1}{2}mr^2 for a solid disc about its axis, I_G = \frac{1}{12}mL^2 for a uniform slender bar about its centre, and I = \frac{1}{3}mL^2 for the same bar about one end. The radius of gyration k is defined by I = mk^2.

For a disc rolling without slip, the contact point has zero velocity, so the kinematic condition is a_G = r\alpha. Taking moments about the contact point P gives \sum M_P = I_P\alpha, where I_P = I_G + mr^2 by the parallel-axis theorem. The friction force follows from the translational equation, and pure rolling holds only if this friction does not exceed \mu N.

Forces passing through a pin give no moment about that pin, so only the weight and the applied forces contribute. A uniform bar released from the horizontal while pinned at one end has \alpha = \frac{3g}{2L} immediately after release, and the pin reaction is then \frac{mg}{4} upward. Pinning the same bar at its midpoint removes the moment of its weight, so the two cases give different angular accelerations for the same applied force.

For a constant moment, \alpha is constant, and \omega = \omega_0 + \alpha t and \omega^2 = \omega_0^2 + 2\alpha\theta apply. Speeds given in rpm must be converted with \omega = 2\pi N/60 rad/s. The number of revolutions is the angular displacement \theta divided by 2\pi.

Coriolis acceleration

When a point moves relative to a frame that rotates with angular velocity \vec\omega, its absolute acceleration contains the Coriolis term \vec a_c = 2\vec\omega\times\vec v_{rel} in addition to the relative, Euler and centripetal parts. The Coriolis term depends only on the angular velocity and the relative velocity. The relative acceleration does not enter it.

Its magnitude is a_c = 2\omega v_{rel}\sin\theta, where \theta is the angle between \vec\omega and \vec v_{rel}, and its direction is perpendicular to both by the right-hand rule. With \hat k pointing out of the page, \hat k\times\hat j = -\hat i, so the order of the cross product fixes the sign. A clockwise rotation gives \vec\omega in the negative \hat k direction.

In polar coordinates the transverse acceleration is a_\theta = r\ddot\theta + 2\dot r\dot\theta, and the radial acceleration is a_r = \ddot r - r\dot\theta^2. When \dot\theta is constant and the relative acceleration is zero, the transverse part reduces to the Coriolis term 2\dot r\dot\theta and the radial part to the centripetal term -r\dot\theta^2. For example, with \omega = 3 rad/s and outward relative speed 2 m/s, the transverse acceleration is 12 m/s^2.

For a particle in a straight groove on a turntable rotating at constant \Omega, the radial component of absolute acceleration is -\Omega^2 r - 2\Omega v_\theta, where v_\theta is the transverse relative velocity. Setting this to zero requires |v_\theta| = \Omega r/2. A relative velocity inclined to the radial line must therefore be resolved into radial and transverse parts before this condition is applied.

Rigid body velocity constraint

For two points A and B on the same rigid body, \vec v_A = \vec v_B + \vec\omega\times\vec r_{A/B}, where \vec r_{A/B} runs from B to A. The rotational term is perpendicular to the line AB, so it changes only the velocity component perpendicular to that line. The rigid-body constraint therefore requires the components of \vec v_A and \vec v_B along AB to be equal.

A point confined to a horizontal channel has a velocity that is purely horizontal, and a point confined to a vertical channel has a purely vertical velocity. Combined with the equal-component rule, a known velocity at one point fixes the angular velocity. For example, if the lower end of a bar of length \ell, inclined at \theta from the vertical, moves horizontally at u while the upper end is held in a vertical channel, then \omega = u/(\ell\cos\theta).