Circular Motion — Why Things Move in Circles

Why Does Circular Motion Require a Force?

Newton’s first law says an object moves in a straight line at constant speed unless a net force acts on it. So when an object moves in a circle, it must be experiencing a continuous force — because its direction keeps changing (even if its speed stays the same).

That force is centripetal force — directed always toward the centre of the circle. Remove it, and the object flies off in a straight line (tangent to the circle at that point). This is not intuitive until you feel it: hold a stone on a string and swing it in a circle. The string tension is the centripetal force. Let go — the stone flies straight.

Circular motion is crucial for JEE Main (typically 2–3 questions per year), NEET (1–2 questions), and CBSE Class 11. Understanding the WHY behind each formula separates students who can solve novel problems from those who can only apply familiar formulas.

Key Terms & Definitions

Uniform circular motion (UCM): Motion in a circle with constant speed. Speed is constant, but velocity is not (direction changes continuously).

Angular displacement ( $\theta$ ): Angle swept by the radius vector. Measured in radians (rad).

Angular velocity ( $\omega$ ): Rate of change of angular displacement. $\omega = d\theta/dt$ . Unit: rad/s.

Period ( $T$ ): Time for one complete revolution. $T = 2\pi/\omega$ .

Frequency ( $f$ ): Number of revolutions per second. $f = 1/T$ . Unit: Hz.

Linear (tangential) speed ( $v$ ): Speed of the object at any instant, directed along the tangent to the circle. $v = r\omega$ .

Centripetal acceleration ( $a_c$ ): Acceleration directed toward the centre of the circle. $a_c = v^2/r = r\omega^2$ .

Centripetal force ( $F_c$ ): The net inward force causing circular motion. $F_c = ma_c = mv^2/r = mr\omega^2$ .

Key Formulas

v = r\omega

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

F_c = \frac{mv^2}{r} = mr\omega^2

T = \frac{2\pi r}{v} = \frac{2\pi}{\omega}

\omega = 2\pi f = \frac{2\pi}{T}

Important: Centripetal force is not a new type of force. It is always provided by an existing force: tension (in circular motion of a pendulum/stone on string), gravity (satellite orbit), normal force (roller coaster loop), friction (car on curved road).

Uniform Circular Motion — Detailed Analysis

Why the Acceleration Points Inward

The velocity vector is always tangent to the circle. As the object moves from point A to point B, the velocity direction rotates by the same angle as the radius vector. The change in velocity ( $\Delta v$ ) points toward the centre (for a small arc, $\Delta v$ is perpendicular to the velocity, directed inward). Therefore, $a = \Delta v / \Delta t$ is centripetal.

Centrifugal Force — The Pseudo-Force

In a rotating reference frame (e.g., sitting in a car going around a bend), you feel thrown outward. This “feeling” is the centrifugal pseudo-force — an apparent outward force that arises because the rotating frame is non-inertial.

The centrifugal force is NOT a real force in an inertial frame. There is no real outward force on the car passenger — they tend to move in a straight line (due to inertia) while the car turns inward. From outside (inertial frame), only centripetal force (friction) acts.

JEE problems often involve switching between inertial and non-inertial frames. In an inertial frame, only centripetal force exists. In the rotating frame, you add centrifugal force (equal and opposite to centripetal) and solve static equilibrium. Both approaches give the same answer for the object’s position/normal force.

Vertical Circular Motion

When an object moves in a vertical circle (like a ball on a string, or a roller coaster loop), gravity plays a dual role: it provides part of the centripetal force at some positions and works against it at others. The speed is NOT constant (it changes as the object gains/loses height).

Critical Points

At the top of the loop: The centripetal force is provided by both gravity AND the normal force (or tension):

mg + N = \frac{mv^2}{r}

For the minimum speed at the top (where N = 0, the object barely maintains contact):

mg = \frac{mv^2_{min}}{r} \Rightarrow v_{min} = \sqrt{gr}

At the bottom of the loop: Gravity acts downward, centripetal force is upward, so:

N - mg = \frac{mv^2}{r} \Rightarrow N = mg + \frac{mv^2}{r}

$N > mg$ at the bottom — you feel heavier at the bottom of a roller coaster.

At the top (minimum speed to maintain tension/contact):

v_{top} = \sqrt{gr}

At the bottom (using energy conservation, given $v_{top} = \sqrt{gr}$ ):

v_{bottom} = \sqrt{5gr}

The ratio $v_{bottom}/v_{top} = \sqrt{5}$ . This appears in JEE frequently.

Banking of Roads

A car moving on a flat circular road relies on friction to provide centripetal force. On a banked road (inclined at angle $\theta$ ), the horizontal component of the normal force provides centripetal force — reducing the dependence on friction.

Optimum speed (zero friction needed):

v_{opt} = \sqrt{rg\tan\theta}

Maximum speed (friction acts inward, preventing outward sliding):

v_{max} = \sqrt{rg \cdot \frac{\tan\theta + \mu}{1 - \mu\tan\theta}}

Minimum speed (friction acts outward, preventing inward sliding):

v_{min} = \sqrt{rg \cdot \frac{\tan\theta - \mu}{1 + \mu\tan\theta}}

At the optimum speed, the road exerts no lateral friction on the tyre — reducing tyre wear and increasing safety.

Conical Pendulum

A pendulum whose bob traces a horizontal circle. The string makes angle $\theta$ with the vertical.

Forces: tension $T$ (along string) and weight $mg$ (downward).

Equations:

Vertical: $T\cos\theta = mg$
Horizontal (centripetal): $T\sin\theta = mr\omega^2$

Dividing: $\tan\theta = r\omega^2/g$ . Since $r = L\sin\theta$ : $\tan\theta = L\sin\theta \cdot \omega^2/g$ , giving:

$\omega = \sqrt{\frac{g}{L\cos\theta}}$ , $\quad T = 2\pi\sqrt{\frac{L\cos\theta}{g}}$$

Solved Examples

Easy \text{---} CBSE Level

Q: A stone of mass 0.1 kg is swung in a horizontal circle of radius 0.5 m with a speed of 4 m/s. Find the centripetal force.

Solution:$$ F_c = \frac{mv^2}{r} = \frac{0.1 \times 4^2}{0.5} = \frac{0.1 \times 16}{0.5} = \frac{1.6}{0.5} = 3.2 \text{ N}

### Medium — JEE Main Level **Q: A car of mass 1000 kg moves on a circular road of radius 100 m. The coefficient of friction is 0.3. Find the maximum safe speed.** **Solution:** On a flat road (unbanked), friction provides centripetal force: $f_{max} = \mu mg = 0.3 \times 1000 \times 10 = 3000$ N. $F_c = mv^2/r$. Setting $F_c = f_{max}$: $3000 = 1000 v^2/100 \Rightarrow v^2 = 300 \Rightarrow v = \sqrt{300} \approx 17.3$ m/s (≈ 62 km/h). ### Hard — JEE Advanced Level **Q: A particle of mass m is attached to a string of length L. It is given a velocity $v_0 = \sqrt{4gL}$ at the bottom of a vertical circle. Does it complete the loop? Find the angle where the string becomes slack.** **Solution:** Minimum speed at top to complete loop: $v_{top,min} = \sqrt{gL}$. Using energy conservation from bottom to top: $\frac{1}{2}mv_0^2 = \frac{1}{2}mv_{top}^2 + mg(2L)$ $v_{top}^2 = v_0^2 - 4gL = 4gL - 4gL = 0$ So $v_{top} = 0 &lt; \sqrt{gL}$. The particle does NOT complete the loop. The string becomes slack when tension $T = 0$. At angle $\theta$ from the top (equivalently, at height $L + L\cos\theta$ from bottom): Energy: $v^2 = v_0^2 - 2g(L + L\cos\theta) = 4gL - 2gL(1+\cos\theta)$ At the slack point: $T = 0$, centripetal force = $mg\cos\theta$ component only (from gravity): $\frac{mv^2}{L} = mg\cos\theta \Rightarrow v^2 = gL\cos\theta$ Substituting: $gL\cos\theta = 4gL - 2gL - 2gL\cos\theta \Rightarrow 3\cos\theta = 2 \Rightarrow \cos\theta = 2/3$ The string becomes slack at $\theta = \cos^{-1}(2/3) \approx 48.2°$ from the top. <Callout type="exam"> The "string becomes slack" problem appeared in JEE Advanced 2015. The key is recognising that at the slack point, tension = 0 and gravity provides the entire centripetal force. These two conditions (energy conservation + force equation) give two equations to solve for angle and speed simultaneously. </Callout> ## Exam-Specific Tips **CBSE Class 11:** Focus on centripetal acceleration formula, deriving $a_c = v^2/r$, and banking of roads (optimum speed only). **JEE Main:** All of the above plus conical pendulum, vertical circular motion (minimum speed at top and bottom), banking with friction (maximum/minimum speed formulas). **NEET:** Centripetal force concept, satellite orbital speed ($v = \sqrt{GM/r}$), and the "which force provides centripetal force in this scenario" type questions. ## Common Mistakes to Avoid <Callout type="mistake"> **Mistake 1:** Drawing centrifugal force on a free body diagram in an inertial frame. There is no centrifugal force in an inertial frame. Free body diagrams in inertial frames should only show real forces (gravity, tension, normal force, friction). The net inward force IS the centripetal force — it's not an additional force. </Callout> <Callout type="mistake"> **Mistake 2:** For vertical circular motion, using $F_c = mv^2/r$ with the minimum speed condition at the top and forgetting to account for gravity. At the top, both tension AND gravity act centripetally. Setting only tension to zero (for minimum speed) is correct, but the force equation must include both. </Callout> <Callout type="mistake"> **Mistake 3:** In banking problems, confusing the angle with the horizontal vs angle with the vertical. The banking angle $\theta$ is measured from the horizontal. $\tan\theta = v^2/(rg)$ — if $\theta$ is from horizontal, this is correct. Check your geometry carefully. </Callout> <Callout type="mistake"> **Mistake 4:** Using $v = r\omega$ and then substituting wrong $r$ (using diameter instead of radius, or using the wrong radius for complex circular configurations). </Callout> <Callout type="mistake"> **Mistake 5:** Saying "centripetal force is in the outward direction." Centripetal force is ALWAYS directed inward (toward the centre). The word "centripetal" literally means "centre-seeking." </Callout> ## Practice Questions **1. A satellite moves in a circular orbit at radius R from Earth's centre. Show that its orbital speed is $v = \sqrt{GM/R}$.** <ExpandableAnswer> The centripetal force is provided by gravity: $F_c = F_g$. So $\frac{mv^2}{R} = \frac{GMm}{R^2}$. Dividing both sides by $m$ and multiplying by $R$: $v^2 = GM/R$, therefore $v = \sqrt{GM/R}$. Note: the mass of the satellite cancels out — orbital speed is independent of satellite mass. </ExpandableAnswer> **2. What is the difference between angular velocity and linear velocity?** <ExpandableAnswer> Angular velocity ($\omega$) measures how fast the angle changes: $\omega = d\theta/dt$, units = rad/s. Linear velocity ($v$) measures the actual speed of the object along the circular path: $v = r\omega$, units = m/s. All points on a rotating body have the same angular velocity, but points further from the centre have greater linear velocity ($v = r\omega$ — larger $r$ means larger $v$). </ExpandableAnswer> **3. A conical pendulum has length 1 m and the bob moves in a circle of radius 0.5 m. Find the time period.** <ExpandableAnswer> $\cos\theta = \sqrt{L^2 - r^2}/L = \sqrt{1 - 0.25}/1 = \sqrt{0.75} \approx 0.866$. $T = 2\pi\sqrt{L\cos\theta/g} = 2\pi\sqrt{1 \times 0.866/10} = 2\pi\sqrt{0.0866} = 2\pi \times 0.2943 \approx 1.85$ s. </ExpandableAnswer> **4. Why is the speed higher at the bottom of a vertical loop than at the top?** <ExpandableAnswer> Energy conservation. As the object descends from the top to the bottom of the loop (height difference = $2r$), it loses potential energy $mg(2r)$, which converts to kinetic energy. So $\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mv_{top}^2 + 2mgr$. Since kinetic energy at the bottom includes this extra potential energy, the speed is higher. The speed is NOT constant in vertical circular motion — only in horizontal circular motion can speed be constant. </ExpandableAnswer> ## FAQs **Q: Is centripetal force different from the "applied" force?** Centripetal force is not a new type of force — it's just the name we give to the net inward force in a specific context. When a stone is swung on a string, the tension IS the centripetal force. When a planet orbits the sun, gravity IS the centripetal force. The label "centripetal" describes the role, not the type of force. **Q: Why does water stay in a bucket swung in a vertical circle?** At the top of the circle (upside down position), if the speed is above the minimum $\sqrt{gr}$, the centripetal acceleration required ($v^2/r$) exceeds $g$. Gravity alone isn't enough to keep the water on its circular path — the bucket must push the water inward (toward the centre). So the bucket's bottom pushes the water toward the centre (normal force acts inward), keeping it in. At or below the minimum speed, the bucket can no longer push the water, and it falls out.