understand that centripetal acceleration causes circular motion with a constant angular speed

Centripetal Acceleration

Learning Objective

Understand that centripetal acceleration is the cause of circular motion when an object moves with a constant angular speed.

Key Concepts

• Uniform circular motion (UCM) – motion in a circle at constant angular speed $\omega$.
• Radial (centripetal) acceleration $a_c$ points towards the centre of the circle.
• Relationship between linear speed $v$, radius $r$, and angular speed $\omega$: $v = \omega r$.
• Force required to produce $a_c$: $F_c = m a_c$.

Derivation of Centripetal Acceleration

Consider an object moving in a circle of radius $r$ with constant angular speed $\omega$. After a short time $\Delta t$, its velocity vector changes direction by an angle $\Delta\theta = \omega\Delta t$ while its magnitude remains $v$.

The change in velocity magnitude is

$$\Delta v = 2v\sin\frac{\Delta\theta}{2} \approx v\Delta\theta \quad (\text{for small }\Delta\theta).$$

The average acceleration over the interval is

$$a_{\text{avg}} = \frac{\Delta v}{\Delta t} \approx \frac{v\Delta\theta}{\Delta t}=v\omega.$$

Taking the limit $\Delta t \to 0$ gives the centripetal (radial) acceleration

$$a_c = v\omega = \frac{v^2}{r} = \omega^2 r.$$

Important Formulas

Quantity	Symbol	Formula	Units
Linear speed	$v$	$v = \omega r$	m·s\(^{-1}\)
Angular speed	$\omega$	$\omega = \dfrac{v}{r}$	rad·s\(^{-1}\)
Centripetal acceleration	$a_c$	$a_c = \dfrac{v^{2}}{r}= \omega^{2} r$	m·s\(^{-2}\)
Centripetal force	$F_c$	$F_c = m a_c = \dfrac{m v^{2}}{r}= m\omega^{2} r$	N

Common Misconceptions

1. “Centripetal force is a new type of force.” It is simply the net radial force acting on the object (tension, gravity, normal reaction, etc.).
2. “If speed is constant, there is no acceleration.” Acceleration also includes a change in direction; in circular motion the direction changes continuously.
3. “Centrifugal force is a real force acting outward.” In an inertial frame there is no outward force; the apparent outward force appears only in a rotating (non‑inertial) reference frame.

Worked Example

Problem: A 0.50 kg ball is attached to a string and whirled in a horizontal circle of radius 0.80 m at a constant speed of 4.0 m·s\(^{-1}\). Find the tension in the string.

1. Calculate the centripetal acceleration: $$a_c = \frac{v^{2}}{r} = \frac{(4.0)^2}{0.80} = 20\ \text{m·s}^{-2}.$$
2. Apply $F_c = m a_c$: $$F_c = (0.50)(20) = 10\ \text{N}.$$
3. Since the only horizontal force providing the centripetal force is the tension, the tension $T = 10\ \text{N}$.

Experimental \cdot erification

One simple laboratory setup uses a low‑friction turntable, a small mass attached to a string, and a stopwatch. By measuring the period $T$ of one revolution, the angular speed is $\omega = 2\pi/T$, and the radius $r$ is measured directly. Using $a_c = \omega^2 r$ the predicted centripetal acceleration can be compared with the force measured by a spring balance.

Summary

• Centripetal acceleration $a_c$ is required for any object to follow a circular path at constant angular speed.
• It is given by $a_c = v^2/r = \omega^2 r$ and always points toward the centre of the circle.
• The necessary centripetal force is $F_c = m a_c$ and can be supplied by tension, gravity, normal reaction, or friction, depending on the situation.
• Even with constant speed, the continuous change in direction means the object is accelerating.