Published by Patrick Mutisya · 14 days ago
Understand that centripetal acceleration is the cause of circular motion when an object moves with a constant angular speed.
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.\$
| 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 | \$ac\$ | \$ac = \dfrac{v^{2}}{r}= \omega^{2} r\$ | m·s\(^{-2}\) |
| Centripetal force | \$Fc\$ | \$Fc = m a_c = \dfrac{m v^{2}}{r}= m\omega^{2} r\$ | N |
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.
\$a_c = \frac{v^{2}}{r} = \frac{(4.0)^2}{0.80} = 20\ \text{m·s}^{-2}.\$
\$F_c = (0.50)(20) = 10\ \text{N}.\$
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.