Grade 11 Physics – Unit 3.5
Uniform Circular Motion
👋 Hello grade 11! Today we explore uniform circular motion – moving in a circle at constant speed. Even though speed is constant, direction changes, so there is acceleration. This topic often appears in entrance exams. Let’s go step by step.
What is uniform circular motion? It’s motion along a circular path with constant speed. Examples: a point on a spinning fan, a car turning a curve at constant speed, or an athlete running on a circular track.
Even though speed is constant, the velocity changes because direction changes. So there is always acceleration – we call it centripetal acceleration (toward the center).
Angular and tangential displacement
When an object moves in a circle, we measure angular displacement \(\theta\) (in radians) where \(\theta = \frac{s}{r}\). Here \(s\) is the arc length (tangential displacement), \(r\) is radius. One full revolution = \(2\pi\) radians = 360°.
Angular velocity and tangential velocity
Angular velocity \(\omega = \frac{\Delta \theta}{\Delta t}\) (rad/s). Tangential velocity \(v_t = \frac{s}{t} = \omega r\). For one full period \(T\), \(v_t = \frac{2\pi r}{T}\).
Direction of angular velocity is given by right-hand rule (thumb along axis if fingers curl in motion direction).
Example 3.9: A wheel has angular velocity 40 rad/s, diameter 60 cm. Find tangential velocity.
\(r = 0.3\,\text{m}\), \(v_t = \omega r = 40 \times 0.3 = 12\,\text{m/s}\).
Centripetal Acceleration
In uniform circular motion, acceleration is toward the center, magnitude \(a_c = \frac{v^2}{r} = \omega^2 r\). This is not slowing you down, it’s changing your direction.
Example 3.10: Car moves at 50.4 km/h (14 m/s) on circular track diameter 40 m (r=20 m). Find angular speed and period.
\(\omega = v/r = 14/20 = 0.7\,\text{rad/s}\). Period \(T = 2\pi r / v = (2\pi \times 20)/14 \approx 8.98\,\text{s}\).
📘 Entrance tip: Centripetal acceleration increases if speed increases or radius decreases. They often ask: if speed doubles, how does \(a_c\) change? (\(a_c \propto v^2\) → four times larger).
Centripetal Force
Centripetal force is the net force toward the center that causes circular motion: \(F_c = m a_c = m\frac{v^2}{r} = m\omega^2 r\). Examples: tension (string), friction (car turning), gravity (satellite).
Example 3.11: 100 g ball whirled in horizontal circle radius 40 cm. Breaking strength 50 N. Find max speed.
\(F_c = m v^2 / r\) → \(50 = 0.1 \times v^2 / 0.4\) → \(v^2 = 200\) → \(v \approx 14.14\,\text{m/s}\).
Example 3.12 (centrifuge): radius 7.5 cm, spins at \(7.5\times10^4\) rev/min. Find centripetal acceleration and compare to g.
Convert: \(\omega = 2\pi f = 2\pi \times (75000/60) \approx 7854\,\text{rad/s}\). \(a_c = \omega^2 r = (7854)^2 \times 0.075 \approx 4.63\times10^6\,\text{m/s}^2\). Ratio \(a_c/g \approx 4.72\times10^5\).
Applications: Banked Roads and Conical Pendulum
Banked road (no friction): \(\tan\theta = \frac{v^2}{rg}\). The banking angle is designed so that the normal component provides centripetal force.
Conical pendulum: \(\tan\theta = \frac{v^2}{rg}\), where \(\theta\) is angle from vertical.
Example 3.14: Find max speed on curve radius 40 m, μ=0.7, g=10. \(v_{\max} = \sqrt{\mu g r} = \sqrt{0.7\times10\times40} = \sqrt{280} \approx 16.73\,\text{m/s}\).
📘 On entrance exam, they may combine circular motion with friction or banking. Remember: for flat curves friction provides centripetal force; for banked curves (ideal) it’s the horizontal component of normal force.
Review Questions 3.5 (selected)
Q5 (Earth’s rotation): Radius of Earth = 6400 km = 6.4×10⁶ m. Period T = 24×3600 = 86400 s.
Tangential velocity at equator: \(v = 2\pi r / T = 2\pi ×6.4e6 / 86400 ≈ 465\,\text{m/s}\).
Angular velocity \(\omega = 2\pi / T ≈ 7.27\times10^{-5}\,\text{rad/s}\).
Centripetal acceleration \(a_c = \omega^2 r ≈ (7.27e-5)^2 × 6.4e6 ≈ 0.034\,\text{m/s}^2\).
Q6: Bicycle wheel radius 40 cm = 0.4 m, speed 20 km/h = 5.556 m/s. \(T = 2\pi r / v = 2\pi×0.4 / 5.556 ≈ 0.452\,\text{s}\). \(\omega = 2\pi/T ≈ 13.9\,\text{rad/s}\).
Q7: Railway track radius 1500 m, speed 15 m/s. Banking angle \(\theta = \tan^{-1}(v^2/(rg)) = \tan^{-1}(225/(1500×9.8)) = \tan^{-1}(0.0153) ≈ 0.877°\). Elevation = \(1.8 × \sin\theta ≈ 1.8×0.0153 ≈ 0.0275\,\text{m} = 2.75\,\text{cm}\).
Key Takeaways
- Uniform circular motion: speed constant, velocity changing → centripetal acceleration \(a_c = v^2/r\).
- Angular velocity \(\omega = v/r = 2\pi/T\).
- Centripetal force \(F_c = m v^2/r\) is provided by tension, friction, normal force, etc.
- Banked road: \(\tan\theta = v^2/(rg)\).
🎯 Well done! You’ve covered all of Unit 3. Now practice the end-of-unit questions in your textbook. Circular motion is all about identifying the force that pulls toward the center. You’ve got this!