JEE Physics Circular Motion Notes Set 02

Centripetal Force

A body moving in a circle of radius r needs a net inward force \( F = \dfrac{mv^2}{r} = m\omega^2 r \). It is not a new force: it is the resultant of real forces such as tension, friction, gravity, normal reaction or electric and magnetic forces. Centrifugal force is only a pseudo force felt in the rotating frame; it should not be added in the ground frame.

Turning on a Level Road

Friction provides the centripetal force: \( \dfrac{mv^2}{r} \le \mu mg \), so the maximum safe speed is \( v_{max} = \sqrt{\mu rg} \). It does not depend on the mass of the vehicle.

Banked Roads

Without friction, the horizontal component of the normal force gives the centripetal force: \( N\sin\theta = \dfrac{mv^2}{r} \), \( N\cos\theta = mg \), hence \( \tan\theta = \dfrac{v^2}{rg} \), and the design speed is \( v = \sqrt{rg\tan\theta} \). With friction (\( \mu = \tan\phi \)), the allowed range is \( \sqrt{rg\tan(\theta - \phi)} \le v \le \sqrt{rg\tan(\theta + \phi)} \), i.e. \( v_{max} = \sqrt{rg\dfrac{\mu + \tan\theta}{1 - \mu\tan\theta}} \). A small banking can be described by \( \tan\theta = h/b \).

Conical Pendulum and Wall of Death

For a conical pendulum of length L, \( T\sin\theta = \dfrac{mv^2}{r} \) and \( T\cos\theta = mg \), so \( v = \sqrt{rg\tan\theta} \) and the time period is \( 2\pi\sqrt{\dfrac{L\cos\theta}{g}} \). In a death well or rotor, friction balances the weight (\( f = mg \)) and the normal reaction gives the centripetal force (\( N = m\omega^2 r \)). The minimum angular speed is therefore \( \omega = \sqrt{g/(\mu r)} \).

Motion in a Vertical Circle

For a particle on a string of length R with speed u at the lowest point: at angle \( \theta \) from the bottom, \( T = \dfrac{mv^2}{R} + mg\cos\theta \) and \( v^2 = u^2 - 2gR(1 - \cos\theta) \).
Complete loop: \( u \ge \sqrt{5gR} \); then the speed at the top is at least \( \sqrt{gR} \), and \( T_{bottom} - T_{top} = 6mg \).
Leaves the circle (slack string): \( \sqrt{2gR} < u < \sqrt{5gR} \).
Oscillates in the lower half: \( 0 < u \le \sqrt{2gR} \).
For a rod or a tube, the top speed can be zero, so \( u \ge \sqrt{4gR} \). For a body sliding on a smooth track, a loop of radius R needs a start height of \( \tfrac{5}{2}R \).

Worked Examples

Worked Example 1. A car travels round a curve of radius 50 m on a level road with \( \mu = 0.5 \). Find the maximum safe speed (g = 10 m/s²).
Answer:
\( v = \sqrt{\mu rg} = \sqrt{0.5\times 50\times 10} = \sqrt{250} \approx 15.8 \) m/s.

Worked Example 2. A 0.2 kg ball on a 0.5 m string is swung in a vertical circle and has speed 6 m/s at the bottom. Find the tension at the bottom and at the top (g = 10 m/s²).
Answer:
Bottom: \( T = mg + \dfrac{mu^2}{R} = 2 + \dfrac{0.2\times 36}{0.5} = 16.4 \) N.
Top: \( v^2 = 36 - 4gR = 36 - 20 = 16 \); \( T = \dfrac{mv^2}{R} - mg = 6.4 - 2 = 4.4 \) N.

Worked Example 3. A conical pendulum of string length 1 m makes an angle of 60° with the vertical. Find its time period (g = 10 m/s²).
Answer:
\( T = 2\pi\sqrt{\dfrac{L\cos\theta}{g}} = 2\pi\sqrt{\dfrac{1\times 0.5}{10}} \).
\( = 2\pi\times 0.2236 \approx 1.4 \) s.