Circular Motion Notes for IOE Entrance — Derivations & Past Questions

1

Uniform Circular Motion

Circular path travelled with constant speed

A periodic motion along a circular path with constant speed. The speed never changes — only the direction of motion turns continuously around the centre. For one full revolution of period $T$ and frequency $f$:

$$v = \dfrac{2\pi r}{T} = 2\pi r f$$

Speed and kinetic energy remain constant.
Velocity, force and acceleration are constant in magnitude but change direction continuously.
Work done is zero, because the force is always $\perp$ to the displacement.
Force & acceleration always point towards the centre, $\perp$ to the velocity (angle $=90^\circ$).

$$a_c = \dfrac{v^2}{r} = 4\pi^2 f^2 r = \dfrac{4\pi^2}{T^2}\,r, \qquad F_c = \dfrac{mv^2}{r}$$

After turning through an angle $\theta$: change in velocity $=2v\sin\dfrac{\theta}{2}$ and change in displacement $=2r\sin\dfrac{\theta}{2}$.

● particle ↑ velocity $v$ (tangent, along the motion) → acceleration $a$ (to centre)

2

Non-Uniform Circular Motion

Circular path with variable speed

Velocity changes in both magnitude and direction, so the particle has both tangential and radial acceleration.

$$a_t = \dfrac{dv}{dt}, \qquad a_c = \dfrac{v^2}{r}, \qquad a = \sqrt{a_t^2 + a_c^2}$$

If $a_t=0,\ a_c\neq0$ → uniform circular motion; if $a_t\neq0,\ a_c\neq0$ → variable circular motion. When the speed is increasing, the angle between velocity and total acceleration is acute; when decreasing, it is obtuse.

→ radial $a_c=\dfrac{v^2}{r}$ ⇡ tangential $a_t=\dfrac{dv}{dt}$ (changing speed)

3

Motion of a Cyclist

Leaning into a curve

Turning without inclination: friction supplies the centripetal force. $$\dfrac{mv^2}{r}=\mu mg \;\Rightarrow\; v_{max}=\sqrt{\mu r g}$$
Leaning at angle $\theta$ to the vertical, the horizontal component of the ground reaction provides the centripetal force: $$\tan\theta = \dfrac{v^2}{rg}$$

Tendency to overturn $\propto v^2$ — double the speed, quadruple the tendency.

Condition for skidding

When the centripetal force required for the turn is greater than the frictional force the tyres can provide, skidding occurs. If $\mu$ is the coefficient of friction between the road and the tyre, the limiting friction (frictional force) is $f = \mu R$, where the normal reaction $R = mg$, so $\;\therefore f = \mu (mg)$.

Step 1 — Skidding requirement

For skidding, centripetal force $>$ frictional force:

$$\frac{mv^2}{r} > \mu (mg) \;\;\Rightarrow\;\; \frac{v^2}{rg} > \mu$$

Step 2 — Relate to the angle of banking

But $\dfrac{v^2}{rg} = \tan\theta$, therefore:

$\tan\theta > \mu$

i.e. when the tangent of the angle of banking is greater than the coefficient of friction, skidding occurs.

The cyclist leans by $\theta$ towards the centre O $R\sin\theta$ supplies the centripetal force → $\tan\theta=\dfrac{v^2}{rg}$

4

Body in a Vertical Circle

Tension & critical speeds

Lowest: $T_A=\dfrac{mv_A^2}{r}+mg$ Horizontal: $T=\dfrac{mv^2}{r}$ Top: $T_C=\dfrac{mv_C^2}{r}-mg$ $T_{max}-T_{min}=6mg$

Energy conservation: $v_A^2 - v_C^2 = 4gr$.
Critical speeds: lowest $v_A=\sqrt{5gr}$, top $v_C=\sqrt{gr}$.
If $v_A \le \sqrt{2gr}$ the body oscillates; if $\sqrt{2gr} < v_A < \sqrt{5gr}$ it leaves the circular path before reaching the top.

Full derivation — tension & critical speeds

A body of mass $m$ is whirled on a string of length $r$ in a vertical circle. Let A be the lowest point, C the highest point, and P a general point where the string makes an angle $\theta$ with the downward vertical.

Step 1 — Tension at a general point

At P, the net force towards the centre is the tension minus the radial component of the weight, and this must equal the centripetal force:

$$T - mg\cos\theta = \frac{mv^2}{r} \;\;\Rightarrow\;\; T = \frac{mv^2}{r} + mg\cos\theta$$

Step 2 — Tension at special points

Substituting $\theta = 0^\circ,\ 90^\circ,\ 180^\circ$:

$$T_A = \frac{mv_A^2}{r} + mg \;(\text{max}), \qquad T_{horiz} = \frac{mv^2}{r}, \qquad T_C = \frac{mv_C^2}{r} - mg \;(\text{min})$$

Step 3 — Energy conservation from A to C

Rising from bottom to top lifts the body a height $2r$. Mechanical energy is conserved:

$$\frac{1}{2}mv_A^2 = \frac{1}{2}mv_C^2 + mg(2r) \;\;\Rightarrow\;\; v_A^2 = v_C^2 + 4gr \tag{1}$$

Step 4 — Condition to complete the circle

The string must stay taut at the top, i.e. $T_C \ge 0$:

$$\frac{mv_C^2}{r} \ge mg \;\;\Rightarrow\;\; v_C \ge \sqrt{gr}$$

Step 5 — Minimum speed at the bottom

Putting the minimum top speed $v_C^2 = gr$ into (1):

$$v_A^2 = gr + 4gr = 5gr$$

$v_C = \sqrt{gr}$ (top) and $v_A = \sqrt{5gr}$ (bottom) are the critical speeds for completing the vertical circle.

Step 6 — Difference of extreme tensions

$$T_A - T_C = \frac{m(v_A^2 - v_C^2)}{r} + 2mg = \frac{m(4gr)}{r} + 2mg = 6mg$$

With the critical speeds, the bottom tension itself is $T_A = \dfrac{m(5gr)}{r} + mg = 6mg$, while the top tension just reaches zero.

—▶ tension $T$ (from the ball towards the centre) —▷ weight $mg$ (always straight down)

5

Conical Pendulum

Bob revolving in a horizontal circle

$$T\sin\theta = \dfrac{mv^2}{r}, \qquad T\cos\theta = mg, \qquad \tan\theta=\dfrac{v^2}{rg}$$ $$\text{Time period: } t = 2\pi\sqrt{\dfrac{l\cos\theta}{g}} = 2\pi\sqrt{\dfrac{h}{g}}$$

A perfectly horizontal string ($\theta=90^\circ$) would need infinite speed and zero time period — so a body can never be whirled in a truly horizontal circle.

Full derivation — speed, tension & time period

A bob of mass $m$ hangs from a string of length $l$ and is whirled so that it traces a horizontal circle of radius $r$ while the string sweeps a cone of semi-vertical angle $\theta$. The bob is a depth $h$ below the point of suspension, so $r = l\sin\theta$ and $h = l\cos\theta$.

Step 1 — Resolve the tension

Two forces act on the bob: the weight $mg$ (down) and the string tension $T$ (along the string). Resolving $T$ vertically and horizontally:

$$T\cos\theta = mg \tag{1}$$ $$T\sin\theta = \frac{mv^2}{r} \tag{2}$$

(1) holds because the bob stays at a constant height; (2) because the horizontal component must supply the centripetal force of the circular motion.

Step 2 — Divide (2) by (1)

$$\tan\theta = \frac{v^2}{rg} \;\;\Rightarrow\;\; v = \sqrt{rg\tan\theta}$$

A faster bob therefore swings out to a larger angle $\theta$.

Step 3 — Magnitude of the tension

Squaring and adding (1) and (2), using $\sin^2\theta+\cos^2\theta=1$:

$$T^2 = (mg)^2 + \left(\frac{mv^2}{r}\right)^2 \;\;\Rightarrow\;\; T = m\sqrt{g^2 + \frac{v^4}{r^2}}$$

Step 4 — Time period

One revolution covers the circumference $2\pi r$ at speed $v$:

$$t = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{rg\tan\theta}} = 2\pi\sqrt{\frac{r}{g\tan\theta}}$$

Substituting $r = l\sin\theta$ and $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$:

$$t = 2\pi\sqrt{\frac{l\sin\theta\cos\theta}{g\sin\theta}}$$

$t = 2\pi\sqrt{\dfrac{l\cos\theta}{g}} = 2\pi\sqrt{\dfrac{h}{g}}$

As $\theta \to 90^\circ$, $\cos\theta \to 0$: the required speed $v \to \infty$ and $t \to 0$ — the string can never become horizontal.

String length $l$, semi-vertical angle $\theta$ — the bob traces a horizontal circle of radius $r=l\sin\theta$

6

Centripetal & Centrifugal Force

Real vs. pseudo force

Centripetal force is real and acts radially inward: $F_c=\dfrac{mv^2}{r}$.
Centrifugal force is a pseudo force felt in the rotating frame, directed radially outward — it is what presses a car passenger against the door on a turn.
Release the string and the stone flies off tangentially, along its instantaneous velocity.

Cut the string when the stone passes the top — it escapes along the tangent, in the direction it was already moving

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High-Yield Points

Quick revision

In uniform circular motion the work done is zero — the force is always $\perp$ to the displacement.

Centripetal acceleration $a=\dfrac{v^2}{R}=4\pi^2f^2R=\dfrac{4\pi^2}{T^2}R$.

Change in velocity $=2v\sin\frac{\theta}{2}$; change in displacement $=2R\sin\frac{\theta}{2}$.

Vertical circle — bottom $v=\sqrt{5gr}$, top $v=\sqrt{gr}$, and $T_{max}-T_{min}=6mg$.

Centrifugal force is a pseudo force directed radially outward.

A coin on a rotating table will not slip while its speed stays below $v=\sqrt{\mu g r}$ — independent of the coin's mass.

A particle slipping off a smooth sphere leaves the surface at height $h=\dfrac{2R}{3}$ above the centre.

Derivation — particle slipping off a smooth sphere

A particle starts from rest at the top of a smooth (frictionless) sphere of radius $R$ and slides down the outside. Let P be a point where the radius to the particle makes an angle $\theta$ with the upward vertical; the particle has then descended a height $R(1-\cos\theta)$.

Step 1 — Speed from energy conservation

The surface is smooth, so mechanical energy is conserved between the top and P:

$$\frac{1}{2}mv^2 = mgR(1-\cos\theta) \;\;\Rightarrow\;\; v^2 = 2gR(1-\cos\theta) \tag{1}$$

Step 2 — Radial (centripetal) equation at P

The component of gravity along the radius, less the normal reaction $N$, provides the centripetal force:

$$mg\cos\theta - N = \frac{mv^2}{R} \tag{2}$$

Step 3 — Condition for leaving the surface

The particle leaves the sphere where the contact is lost, i.e. $N = 0$:

$$mg\cos\theta = \frac{mv^2}{R} \;\;\Rightarrow\;\; v^2 = gR\cos\theta \tag{3}$$

Step 4 — Equate (1) and (3)

$$gR\cos\theta = 2gR(1-\cos\theta) \;\;\Rightarrow\;\; \cos\theta = 2 - 2\cos\theta \;\;\Rightarrow\;\; \cos\theta = \frac{2}{3}$$

Step 5 — Height above the centre

The height of P above the centre of the sphere is $h = R\cos\theta$, so:

$h = R\cos\theta = \dfrac{2R}{3}$ above the centre $\left(\cos\theta=\tfrac{2}{3}\right)$.

The particle therefore flies off after dropping a vertical height $\dfrac{R}{3}$ from the top.

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Challenge Questions

Slightly harder problems from the chapter

These higher-order questions go a step beyond the past-paper set — combining energy, friction and vector ideas from circular motion. Try each one before revealing the answer.