angular kinematics and total acceleration

For a rigid body rotating about a fixed axis, the position of a point is conveniently described by the rotation angle $\theta$ rather than directly by the radius vector $\vec r$. The angular velocity $\omega = \frac{d\theta}{dt}$ and angular acceleration $\alpha = \frac{d\omega}{dt}$ then play the same role as linear velocity and acceleration in straight-line motion: for $\alpha = \text{const}$ the formulas

$$\omega = \omega_0 + \alpha t, \qquad \theta = \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha t^2, \qquad \omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$$

hold. The link to linear quantities follows by differentiating the position of a point on a circle: $s = r\theta$, $v = \omega r$, $a_t = \alpha r$, and $a_n = \omega^2 r = \frac{v^2}{r}$. The radius vector, displacement, velocity, and acceleration from the previous topic are here applied to a point moving along a circle. The total acceleration of a point is the sum of the tangential component $\vec a_t$ (along the tangent, parallel to velocity) and the centripetal component $\vec a_n$ (toward the center): $\vec a = \vec a_t + \vec a_n$, hence $a = \sqrt{a_t^2 + a_n^2}$. The direction of the total acceleration is given by the angle $\varphi$ between it and the radius: $\tan\varphi = \frac{a_t}{a_n}$.

A wheel rotates with constant angular acceleration. At $t = 0$ its angular velocity is $\omega_0 = 2.0$ rad/s, the angular acceleration is $\alpha = 3.0$ rad/s², and the initial angle is $\theta_0 = 0$. Find the angular velocity at $t = 4.0$ s and the angular displacement during this time.

A point lies on the rim of a disk of radius $r = 0.25$ m. At a certain instant the angular velocity of the disk is $\omega = 6.0$ rad/s and the angular acceleration is $\alpha = 4.0$ rad/s². Find the magnitude of the total acceleration of this point and the angle between the total acceleration and the radius.

A point on the rim of a rotating disk of radius $r = 0.50$ m has, at a given instant, total acceleration magnitude $a = 5.0$ m/s² and tangential acceleration magnitude $a_t = 3.0$ m/s². The angular acceleration is constant. Find: (a) the centripetal acceleration; (b) the angular velocity at that instant; (c) the angular velocity after $1.0$ s; (d) the angle between the total acceleration and the radius.

Safe self-study plan (simulation). Do not use real rotating machinery — drills, drilling machines, or fans with guards removed: that is hazardous. Work with an interactive rotating-disk simulation (for example, PhET "Ladybug Revolution" or an equivalent).

1. Set the radius to $r = 0.50$ m, the initial angular velocity to $\omega_0 = 2.0$ rad/s, and the angular acceleration to $\alpha = 1.5$ rad/s².
2. At $t = 0, 1, 2, 3$ s, record $\omega$ and compute $a_t = \alpha r$, $a_n = \omega^2 r$, $a = \sqrt{a_t^2 + a_n^2}$.
3. Compare the computed values with the acceleration vectors shown by the simulation and with your table.
4. Double the radius while keeping $\omega$ and $\alpha$ fixed, and check that $a_t$ and $a_n$ grow in proportion to $r$.
5. Draw a conclusion: which component of the total acceleration determines its direction at small $\omega$ and at large $\omega$?

Show that the total acceleration of a point on a rotating body is the vector sum of the tangential and centripetal accelerations, and how their directions are related to the radius vector and velocity.

Diagram: a circle of radius $r$ with center $O$; point $P$ on the rim; the radius vector $\vec r$ from $O$ to $P$; the velocity vector $\vec v$ directed along the tangent; the vector $\vec a_t$ along $\vec v$ (or opposite to it during braking); the vector $\vec a_n$ directed from $P$ toward the center; the resultant vector $\vec a$ constructed by the parallelogram rule from $\vec a_t$ and $\vec a_n$. The labels give $a_t = \alpha r$, $a_n = \omega^2 r$, $a = \sqrt{a_t^2 + a_n^2}$, and the angle $\varphi$ between $\vec a$ and the radius.

A car enters a roundabout of radius $R = 25$ m with speed $v_0 = 8.0$ m/s and speeds up uniformly to $v = 14$ m/s in $t = 4.0$ s while moving along the circular lane. At the moment it reaches $14$ m/s: tangential acceleration $a_t = (14 - 8)/4 = 1.5$ m/s$^2$, centripetal acceleration $a_c = v^2/R = 14^2/25 = 7.84$ m/s$^2$, angular speed $\omega = v/R = 0.56$ rad/s, and the total acceleration magnitude $a = \sqrt{a_t^2 + a_c^2} = \sqrt{1.5^2 + 7.84^2} \approx 7.98$ m/s$^2$.

This is a direct application of non-uniform circular motion: the tangential component $a_t = \mathrm{d}v/\mathrm{d}t$ changes the magnitude of the velocity, while the centripetal component $a_c = v^2/R = \omega^2 R$ changes its direction. The total acceleration is the vector sum $\vec a = \vec a_t + \vec a_c$ with the given magnitudes, and $v$ and $\omega$ are linked by $v = \omega R$.

A blender blade of radius $R = 4.0$ cm starts from rest and reaches $3000$ rpm in $t = 0.80$ s. Convert: $\omega = 3000 \times 2\pi/60 = 314$ rad/s. Angular acceleration $\alpha = \omega/t = 314/0.80 \approx 393$ rad/s$^2$. At the tip: tangential acceleration $a_t = \alpha R = 393 \times 0.040 \approx 15.7$ m/s$^2$, centripetal acceleration $a_c = \omega^2 R = 314^2 \times 0.040 \approx 3.9 \times 10^{3}$ m/s$^2$. Total acceleration magnitude $a = \sqrt{a_t^2 + a_c^2} \approx 3.9 \times 10^{3}$ m/s$^2$ — centripetal dominates.

This exercises the angular kinematic chain $\omega = \omega_0 + \alpha t$, $v = \omega R$, and $a_t = \alpha R$, together with $a_c = \omega^2 R$; the total acceleration at the blade tip is $\vec a = \vec a_t + \vec a_c$, and its magnitude is dominated by the centripetal term once the blade spins fast.

A hammer thrower swings a $m = 7.26$ kg ball on a cable of radius $R = 1.20$ m. During the final acceleration phase, the ball's speed increases from $15$ m/s to $28$ m/s in $0.50$ s. Tangential acceleration $a_t = (28 - 15)/0.50 = 26$ m/s$^2$. At release, centripetal acceleration $a_c = v^2/R = 28^2/1.20 \approx 653$ m/s$^2$, angular speed $\omega = v/R = 28/1.20 \approx 23.3$ rad/s. Total acceleration magnitude $a = \sqrt{26^2 + 653^2} \approx 654$ m/s$^2$.

The ball follows a horizontal circle: the centripetal acceleration $a_c = v^2/R = \omega^2 R$ is supplied by the cable tension, while the tangential acceleration $a_t = \mathrm{d}v/\mathrm{d}t = \alpha R$ comes from the thrower's pull along the path. The total acceleration $\vec a = \vec a_t + \vec a_c$ determines the force the thrower must exert.