Published by Patrick Mutisya · 14 days ago
Understand that a force of constant magnitude that is always perpendicular to the direction of motion produces a centripetal (radial) acceleration.
Consider an object moving with constant speed \$v\$ around a circle of radius \$r\$. After a short time \$\Delta t\$ the velocity vector has rotated through an angle \$\Delta\theta\$.
The change in velocity magnitude is zero; only its direction changes. The magnitude of the change in velocity is
\$\Delta v = 2v\sin\left(\frac{\Delta\theta}{2}\right) \approx v\Delta\theta\quad(\text{for small }\Delta\theta).\$
The average acceleration during \$\Delta t\$ is
\$\mathbf{a}_{\text{avg}} = \frac{\Delta\mathbf{v}}{\Delta t} \approx \frac{v\Delta\theta}{\Delta t}.\$
Since \$\displaystyle \frac{\Delta\theta}{\Delta t}\$ is the angular speed \$\omega\$, we have
\$a_c = v\omega.\$
Using the relationship \$v = \omega r\$, we obtain the familiar form
\$a_c = \frac{v^{2}}{r} = \omega^{2}r.\$
Newton’s second law states \$\mathbf{F}=m\mathbf{a}\$. For circular motion the acceleration is radial, so the net force must also be radial. If the force had any component parallel to the velocity, it would change the speed, violating the condition of uniform circular motion. Therefore a constant‑magnitude force that remains perpendicular to the instantaneous direction of motion supplies exactly the required centripetal force.
| Quantity | Symbol | Expression | Units |
|---|---|---|---|
| Centripetal acceleration | \$a_c\$ | \$\displaystyle a_c = \frac{v^{2}}{r} = \omega^{2}r\$ | m s⁻² |
| Centripetal force | \$F_c\$ | \$\displaystyle Fc = m ac = \frac{mv^{2}}{r} = m\omega^{2}r\$ | N |
| Angular speed | \$\omega\$ | \$\displaystyle \omega = \frac{v}{r}\$ | rad s⁻¹ |
Problem: A 0.50 kg stone is attached to a light 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.
Solution:
\$F_c = \frac{(0.50\ \text{kg})(4.0\ \text{m s}^{-1})^{2}}{0.80\ \text{m}} = \frac{0.50 \times 16}{0.80} = 10\ \text{N}.\$
A force of constant magnitude that remains perpendicular to the instantaneous direction of motion supplies a centripetal acceleration \$a_c = v^{2}/r\$. This acceleration is always directed towards the centre of the circular path, and the associated force is called the centripetal force. Understanding the geometric relationship between velocity, force, and acceleration is essential for analysing any situation involving uniform circular motion.