Rotational Dynamics — Formula Sheet

$v$: Linear speed (m s⁻¹) · $\omega$: Angular speed (rad s⁻¹) · $r$: Radius of the circular path (m) · $T$: Period of revolution (s) · $n$: Frequency of revolution (Hz = rev s⁻¹)

Valid for any point at distance r from the axis. In vector form v⃗ = ω⃗ × r⃗. ω is in rad/s; n is the frequency in rev/s (Hz).

$a_r$: Centripetal acceleration (m s⁻²) · $v$: Linear speed (m s⁻¹) · $\omega$: Angular speed (rad s⁻¹) · $r$: Radius (m)

Directed towards the centre. Present even in uniform circular motion because the direction of velocity keeps changing. Unit: m s⁻².

$a_t$: Tangential acceleration (m s⁻²) · $\alpha$: Angular acceleration (rad s⁻²) · $a_r$: Radial (centripetal) acceleration (m s⁻²)

When the speed changes, a tangential component a_t appears in addition to a_r. The two are perpendicular, so they add as a right-angled triangle.

$F$: Centripetal force (N) · $m$: Mass of the body (kg) · $v$: Linear speed (m s⁻¹) · $r$: Radius (m)

The net force towards the centre required to keep a body on a circular path. It is provided by tension, friction, gravity or a component of the normal reaction. Unit: N.

${\mu }_s$: Coefficient of static friction (no unit) · $r$: Radius of the curve (m) · $g$: Acceleration due to gravity (m s⁻²)

On an unbanked road the only horizontal force is static friction, so μ_s·mg ≥ mv²/r. The safe speed does not depend on the mass of the vehicle.

$\theta$: Angle of banking (with the horizontal) · $v$: Most safe speed (m s⁻¹) · $r$: Radius of the curve (m)

At this speed no friction is needed. The horizontal component of the normal reaction alone provides the centripetal force. Independent of the vehicle's mass.

${\mu }_s$: Coefficient of static friction · $\theta$: Angle of banking

At v_max the friction acts down the slope. The minimum speed (friction up the slope) is v_min = √[rg(tan θ − μ_s)/(1 + μ_s tan θ)]. Putting μ_s = 0 gives back tan θ = v²/rg.

$T$: Period of revolution (s) · $L$: Length of the string (m) · $\theta$: Angle made by the string with the vertical · $r$: Radius of the horizontal circle = L sin θ (m)

The bob moves in a horizontal circle of radius r = L sin θ. L cos θ is the vertical height of the point of suspension above the circle. Period does not depend on the mass of the bob.

$r$: Radius of the vertical circle (m) · $g$: Acceleration due to gravity (m s⁻²)

These are the minimum speeds for a body on a string (or a vehicle in a loop) to just complete the vertical circle. v_mid is at the horizontal position, level with the centre.

$T_{bottom}$: Tension at the lowest point (N) · $T_{top}$: Tension at the highest point (N) · $m$: Mass of the body (kg)

Holds for any speed, as long as the string stays taut over the full circle. Here T means tension, not period.

$I$: Moment of inertia about the given axis (kg m²) · $m_i$: Mass of the i-th particle (kg) · $r_i$: Perpendicular distance of the i-th particle from the axis (m) · $M$: Total mass of the body (kg) · $K$: Radius of gyration (m)

Moment of inertia is the rotational analogue of mass. The radius of gyration K is the distance from the axis at which the whole mass could be placed to give the same I. SI unit of I: kg m².

$M$: Mass of the body (kg) · $R$: Radius of ring, disc or sphere (m) · $L$: Length of the rod (m)

Axes: ring and disc about their own axis (through the centre, perpendicular to the plane); thin rod about a perpendicular axis through its centre; solid sphere about a diameter. Thin hollow sphere about a diameter: ⅔MR².

$I_O$: MI about the given axis through O (kg m²) · $I_C$: MI about the parallel axis through the centre of mass (kg m²) · $h$: Distance between the two parallel axes (m)

One axis MUST pass through the centre of mass C; the other is parallel to it at a distance h. I_C is the minimum MI among all parallel axes. Example: rod about one end = ML²/12 + M(L/2)² = ML²/3.

$I_Z$: MI about the axis perpendicular to the plane (kg m²) · $I_X$: MI about the X axis in the plane (kg m²) · $I_Y$: MI about the Y axis in the plane (kg m²)

Valid only for a plane lamina (ring, disc, plate). X and Y lie in the plane of the lamina, Z is perpendicular to it, and all three meet at one point. Example: disc about a diameter = ½ × ½MR² = ¼MR².

$E_{rot}$: Rotational kinetic energy (J) · $I$: Moment of inertia (kg m²) · $\omega$: Angular speed (rad s⁻¹)

The rotational analogue of ½mv². Doubling ω makes the rotational energy four times larger. Unit: J.

$\vec{L}$: Angular momentum (kg m² s⁻¹) · $\vec{r}$: Position vector from the axis (m) · $\vec{p}$: Linear momentum (kg m s⁻¹)

L = Iω applies to a rigid body rotating about a fixed axis. SI unit: kg m² s⁻¹ (equivalently J s).

$\tau$: Torque (N m) · $I$: Moment of inertia (kg m²) · $\alpha$: Angular acceleration (rad s⁻²)

Rotational analogue of F = ma. τ = Iα holds when I is constant. Unit of torque: N m.

$I_1,I_2$: Initial and final moment of inertia (kg m²) · ${\omega }_1,{\omega }_2$: Initial and final angular speed (rad s⁻¹ or rpm, same unit on both sides)

If no external torque acts, L stays constant. A spinning dancer or diver who pulls the arms in reduces I and so spins faster. Rotational KE is not conserved in this process.

$m$: Mass (kg) · $v$: Speed of the centre of mass (m s⁻¹) · $K$: Radius of gyration about the axis through the centre of mass (m) · $R$: Radius of the rolling body (m)

For rolling without slipping, v = Rω. K²/R² = 1 (ring), ½ (disc or solid cylinder), ⅖ (solid sphere), ⅔ (hollow sphere).

$a$: Linear acceleration along the incline (m s⁻²) · $\theta$: Angle of inclination · $h$: Vertical height descended (m)

Smaller K²/R² means larger acceleration. So a solid sphere reaches the bottom first, then a disc, and a ring last, whatever their masses and radii.