Circular Motion 2 (AQA A-Level Further Maths): Revision Notes

Worked Example: Finding Tension, Angular Speed, and Period

Problem: A conical pendulum has a bob P of mass 2 kg hanging at the lower end of a light, inextensible string of length 2 m. The upper end of the string is fixed to a point O. The bob moves in a horizontal circle with centre C and radius 1.2 m at constant speed. Find the tension in the string, the angular speed of the bob, and the time taken for one revolution.

Solution:

Step 1: Find the vertical height using geometry.

In triangle OPC, we can use Pythagoras' theorem: $OC = \sqrt{2^2 - 1.2^2} = \sqrt{4 - 1.44} = \sqrt{2.56} = 1.6 \text{ m}$

Step 2: Resolve forces vertically.

The bob is in vertical equilibrium, so: $T \sin \theta = 2g$

where $\sin \theta = \frac{1.6}{2} = 0.8$

Therefore:

$T \times 0.8 = 2 \times 9.8$

$T = \frac{19.6}{0.8} = 24.5 \text{ N}$

The tension in the string is 24.5 N.

Step 3: Apply the horizontal equation of motion.

The horizontal component provides centripetal force: $T \cos \theta = mr\omega^2$

where $\cos \theta = \frac{1.2}{2} = 0.6$

$24.5 \times 0.6 = 2 \times 1.2 \times \omega^2$

$14.7 = 2.4\omega^2$

$\omega^2 = 6.125$

$\omega = 2.47 \text{ rad s}^{-1}$

The angular speed is 2.47 rad s⁻¹.

Step 4: Calculate the period.

$\text{Time for one revolution} = \frac{2\pi}{\omega} = \frac{2\pi}{2.47} = 2.5 \text{ seconds}$

The time taken is 2.5 seconds.

Worked Example: Minimum Angular Velocity with String Constraint

Problem: A 0.5 kg mass P is tied to the midpoint of a 2 m string XY, where X is $\sqrt{3}$ m vertically above Y. The mass P rotates in a horizontal circle so that the string remains taut. Find the least possible angular velocity of P. Take $g = 9.81$ m s⁻².

Solution:

Step 1: Draw a labelled diagram.

The string forms an isosceles triangle XYP. Since X is $\sqrt{3}$ m above Y and the total string length XY is 2 m, with P at the midpoint, each segment XP and YP has length 1 m.

Using geometry, $\cos \theta = \frac{\sqrt{3}}{2}$, so $\theta = 30°$.

Step 2: Write equilibrium equation for vertical forces.

Let T be the tension in segment XP and U be the tension in segment YP.

For vertical equilibrium of P: $T \cos \theta = U \cos \theta + 0.5g$

Substituting $\cos \theta = \frac{\sqrt{3}}{2}$ and solving:

$T - U = \frac{0.5g}{\cos \theta} = \frac{0.5 \times 9.81}{\sqrt{3}/2} = 5.664 \text{ N}$ ... [1]

Step 3: Write horizontal equation of motion.

For P moving in a circle: $T \sin \theta + U \sin \theta = mr\omega^2$

where $r = 1 \times \sin \theta = 0.5$ m

$T \sin \theta + U \sin \theta = 0.5 \times 0.5 \times \omega^2$

$(T + U) \sin 30° = 0.25\omega^2$

$T + U = 0.5\omega^2$ ... [2]

Step 4: Apply constraint condition.

The string can only go slack between P and Y. For the string to remain taut, $U \geq 0$.

From equations [1] and [2]:

$2U = 0.5\omega^2 - 5.664$

For $U \geq 0$:

$0.5\omega^2 \geq 5.664$

$\omega^2 \geq 11.33$

$\omega \geq \sqrt{11.33} = 3.4 \text{ rad s}^{-1}$

The least possible angular velocity is 3.4 rad s⁻¹.

Key insight: When a particle is attached to a string at a point other than the end, the tensions on either side may be different. The string going slack provides a constraint that leads to an inequality for the angular speed.

Worked Example: Proving a Relationship in a Rod System

Problem: Particle Y is connected to two light rods XY and YZ, both of length l, such that end X pivots about a fixed point. End Z is attached to a ring which slides on a smooth vertical rod XZ. Y and Z have the same mass m. The system rotates so that Y performs horizontal circles with constant speed $\omega$. Prove that if $\angle ZXY = \theta$, then $\cos \theta = \frac{3g}{l\omega^2}$.

Solution:

Step 1: Identify the forces.

Triangle XYZ is isosceles (since XY = YZ = l). Let T be the force in rod XY and U be the force in rod YZ. The vertical rod XZ exerts a normal reaction R on the ring at Z.

Step 2: Write equation for vertical equilibrium of Z.

Z is in vertical equilibrium: $U \cos \theta = mg$ ... [1]

Step 3: Write equation for vertical equilibrium of Y.

Y is also in vertical equilibrium: $T \cos \theta = U \cos \theta + mg$

Substituting equation [1]: $T \cos \theta = mg + mg = 2mg$ ... [2]

Step 4: Write horizontal equation of motion for Y.

Y moves in a horizontal circle of radius $l \sin \theta$: $T \sin \theta + U \sin \theta = m \times l \sin \theta \times \omega^2$ ... [3]

Step 5: Substitute and simplify.

From equations [1] and [2]: $\frac{2mg}{\cos \theta} \times \sin \theta + \frac{mg}{\cos \theta} \times \sin \theta = ml \sin \theta \times \omega^2$

Factor out $\sin \theta$: $\frac{2mg \sin \theta}{\cos \theta} + \frac{mg \sin \theta}{\cos \theta} = ml \sin \theta \times \omega^2$

$\frac{3mg \sin \theta}{\cos \theta} = ml \sin \theta \times \omega^2$

Divide both sides by $m \sin \theta$: $\frac{3g}{\cos \theta} = l\omega^2$

Rearrange to give: $\cos \theta = \frac{3g}{l\omega^2}$

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