Uniform Circular Motion: Velocity and Acceleration

In uniform circular motion, the speed is constant but the direction is constantly changing. Two key facts:

• The velocity vector is always tangent to the circle.
• The acceleration vector always points toward the center (centripetal acceleration).

These two vectors are perpendicular at every instant — that's the geometric signature of circular motion at constant speed.

The Formulas

For an object moving in a circle of radius $r$ at speed $v$:

$\omega = \dfrac{v}{r} \qquad a_c = \dfrac{v^{2}}{r} = \omega^{2} r$

$T = \dfrac{2\pi r}{v} = \dfrac{2\pi}{\omega}$

where $\omega$ is the angular velocity (in rad/s), $a_c$ is the centripetal acceleration, and $T$ is the period (time for one full revolution).

Step-by-Step Solution

Given: $r = 2\;\text{m}$, $v = 3\;\text{m/s}$.

Find: Angular velocity $\omega$, centripetal acceleration $a_c$, and period $T$.

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Step 1 — Compute the angular velocity.

$\omega = \dfrac{v}{r} = \dfrac{3}{2} = 1.500\;\text{rad/s}$

Step 2 — Compute the centripetal acceleration.

$a_c = \dfrac{v^{2}}{r} = \dfrac{(3)^{2}}{2} = \dfrac{9}{2} = 4.500\;\text{m/s}^{2}$

(Equivalently $a_c = \omega^{2} r = (1.5)^{2}(2) = 4.5\;\text{m/s}^{2}$.)

Step 3 — Compute the period.

$T = \dfrac{2\pi r}{v} = \dfrac{2\pi(2)}{3} = \dfrac{4\pi}{3} \approx 4.189\;\text{s}$

Step 4 — Convert period to revolutions per minute (rpm).

$\text{rpm} = \dfrac{60}{T} = \dfrac{60}{4.189} \approx 14.32\;\text{rpm}$

That's about one revolution every 4.19 seconds, or 14.3 rpm — a very gentle rotation.

Step 5 — Centripetal force (if mass is given).

If the ball has mass $m$, the tension in the string provides the centripetal force:

$F_c = m\,a_c = 4.5\,m\;\text{N}$

For example, a 0.5 kg ball would need 2.25 N of string tension to maintain this motion.

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Answer:

• Angular velocity: $\boxed{\omega = 1.5\;\text{rad/s}}$
• Centripetal acceleration: $\boxed{a_c = 4.5\;\text{m/s}^{2}}$
• Period: $\boxed{T \approx 4.19\;\text{s}\;(\approx 14.3\;\text{rpm})}$

The velocity is always tangent to the circle (3 m/s in magnitude, direction always changing). The acceleration is always toward the center (4.5 m/s², direction always changing) — but its magnitude is constant because the speed is constant.

Try It

• Adjust radius and speed with the sliders.
• Watch the velocity vector (tangent) and acceleration vector (toward center) rotate around with the ball.
• Notice that doubling the speed quadruples the centripetal acceleration (since $a_c \propto v^{2}$).
• Doubling the radius (with constant speed) halves the centripetal acceleration.