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Laws of Motion - Dynamics of Uniform Circular Motion

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Uniform Circular Motion (UCM) is the motion of a body along a circular path with a constant speed, though its velocity changes continuously due to the change in direction.

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Centripetal Acceleration (aca_c): In UCM, a body experiences an acceleration directed towards the center of the circle, given by ac=v2r=ω2ra_c = \frac{v^2}{r} = \omega^2 r.

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Centripetal Force (FcF_c): It is the net force acting towards the center required to keep a body in circular motion. It is not a new kind of force but is provided by forces like gravity, friction, or tension.

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Centrifugal Force: This is a pseudo-force or fictitious force experienced by an observer in a rotating (non-inertial) frame of reference, acting radially outwards with magnitude mv2r\frac{mv^2}{r}.

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Motion on a Level Road: For a vehicle taking a turn on a flat road, the necessary centripetal force is provided by the static friction between the tyres and the road. The maximum safe speed is vmax=μsrgv_{max} = \sqrt{\mu_s rg}.

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Banking of Roads: To reduce wear and tear of tyres and prevent skidding at high speeds, the outer edge of the road is raised above the inner edge. The angle θ\theta is the angle of banking.

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Bending of a Cyclist: A cyclist leans inwards while taking a turn to generate a horizontal component of the normal reaction that provides the required centripetal force.

📐Formulae

ac=v2r=ω2r=vωa_c = \frac{v^2}{r} = \omega^2 r = v\omega

Fc=mv2r=mω2rF_c = \frac{mv^2}{r} = m \omega^2 r

vmax=μsgr (on a level road)v_{max} = \sqrt{\mu_s g r} \text{ (on a level road)}

tan⁡θ=v2rg (Angle of banking or bending)\tan \theta = \frac{v^2}{rg} \text{ (Angle of banking or bending)}

voptimum=rgtan⁡θ (Speed for zero friction on banked road)v_{optimum} = \sqrt{rg \tan \theta} \text{ (Speed for zero friction on banked road)}

vmax=rg(μs+tan⁡θ1−μstan⁡θ) (Max speed on banked road with friction)v_{max} = \sqrt{rg \left( \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right)} \text{ (Max speed on banked road with friction)}

💡Examples

Problem 1:

A car of mass 1500 kg1500 \text{ kg} is moving at a speed of 72 km/h72 \text{ km/h} on a circular track of radius 100 m100 \text{ m}. Calculate the centripetal force required to keep the car on the track.

Solution:

Given: m=1500 kgm = 1500 \text{ kg}, v=72 km/h=72×518=20 m/sv = 72 \text{ km/h} = 72 \times \frac{5}{18} = 20 \text{ m/s}, r=100 mr = 100 \text{ m}. Using the formula Fc=mv2rF_c = \frac{mv^2}{r}: Fc=1500×(20)2100F_c = \frac{1500 \times (20)^2}{100} Fc=1500×400100=1500×4=6000 NF_c = \frac{1500 \times 400}{100} = 1500 \times 4 = 6000 \text{ N}.

Explanation:

The centripetal force of 6000 N6000 \text{ N} must be provided by the friction between the car's tyres and the road to prevent it from sliding off the circular track.

Problem 2:

Calculate the angle of banking for a circular track of radius 60 m60 \text{ m} designed for cars moving at an average speed of 18 m/s18 \text{ m/s}. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

Given: v=18 m/sv = 18 \text{ m/s}, r=60 mr = 60 \text{ m}, g=10 m/s2g = 10 \text{ m/s}^2. Using the formula tan⁡θ=v2rg\tan \theta = \frac{v^2}{rg}: tan⁡θ=18260×10=324600\tan \theta = \frac{18^2}{60 \times 10} = \frac{324}{600} tan⁡θ=0.54\tan \theta = 0.54 θ=tan⁡−1(0.54)≈28.37∘\theta = \tan^{-1}(0.54) \approx 28.37^\circ.

Explanation:

To ensure that a car traveling at 18 m/s18 \text{ m/s} can negotiate the curve without relying on friction, the road should be banked at an angle of approximately 28.37∘28.37^\circ.

Dynamics of Uniform Circular Motion Revision - Class 11 Physics ICSE