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Laws of Motion - Dynamics of Uniform Circular Motion

Grade 11CBSEPhysics

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

🔑Concepts

Centripetal Force: For a body of mass mm moving in a circle of radius rr with constant speed vv, the net force acting towards the center is called centripetal force, given by Fc=mv2rF_c = \frac{mv^2}{r}.

Centripetal Acceleration: The acceleration directed towards the center of the circular path is ac=v2r=ω2ra_c = \frac{v^2}{r} = \omega^2 r, where ω\omega is the angular velocity.

Motion on a Level Road: When a car takes a turn on a horizontal road, the necessary centripetal force is provided by the static friction between the tires and the road (fsμsNf_s \le \mu_s N).

Maximum Velocity on a Level Road: To avoid skidding, the maximum speed is vmax=μsrgv_{max} = \sqrt{\mu_s r g}, where μs\mu_s is the coefficient of static friction.

Banking of Roads: To reduce dependence on friction and prevent skidding at high speeds, the outer edge of the road is raised above the inner edge. This is called banking, and the angle it makes with the horizontal is the banking angle θ\theta.

Optimum Speed on a Banked Road: The speed at which no frictional force is required is v0=rgtanθv_0 = \sqrt{rg \tan \theta}.

Maximum Safe Speed on a Banked Road: Considering friction μs\mu_s, the maximum speed is vmax=rg(μs+tanθ1μstanθ)v_{max} = \sqrt{rg \left( \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right)}.

Bending of a Cyclist: A cyclist leans inwards while taking a turn to provide the necessary centripetal force through the horizontal component of the normal reaction.

📐Formulae

Fc=mv2rF_c = \frac{mv^2}{r}

ac=v2r=ω2ra_c = \frac{v^2}{r} = \omega^2 r

vmax=μsrgv_{max} = \sqrt{\mu_s rg}

tanθ=v2rg\tan \theta = \frac{v^2}{rg}

vmax=rg(μs+tanθ1μstanθ)v_{max} = \sqrt{rg \left( \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right)}

vmin=rg(tanθμs1+μstanθ)v_{min} = \sqrt{rg \left( \frac{\tan \theta - \mu_s}{1 + \mu_s \tan \theta} \right)}

💡Examples

Problem 1:

A bend in a level road has a radius of 100m100\,m. If the coefficient of static friction between the tyres and the road is 0.40.4, what is the maximum speed with which a car can turn without skidding? (Take g=9.8m/s2g = 9.8\,m/s^2)

Solution:

Given: r=100mr = 100\,m, μs=0.4\mu_s = 0.4, g=9.8m/s2g = 9.8\,m/s^2. Using the formula for maximum speed on a level road: vmax=μsrg=0.4×100×9.8=39219.8m/sv_{max} = \sqrt{\mu_s rg} = \sqrt{0.4 \times 100 \times 9.8} = \sqrt{392} \approx 19.8\,m/s.

Explanation:

On a flat road, only friction provides the centripetal force. If the required force mv2r\frac{mv^2}{r} exceeds the maximum static friction μsmg\mu_s mg, the car will skid.

Problem 2:

A circular track of radius 300m300\,m is banked at an angle of 1515^\circ. What is the optimum speed of a race car to avoid wear and tear on its tyres? (Take g=9.8m/s2g = 9.8\,m/s^2, tan150.2679\tan 15^\circ \approx 0.2679)

Solution:

Given: r=300mr = 300\,m, θ=15\theta = 15^\circ, g=9.8m/s2g = 9.8\,m/s^2. The optimum speed is v0=rgtanθ=300×9.8×0.2679=787.62628.06m/sv_0 = \sqrt{rg \tan \theta} = \sqrt{300 \times 9.8 \times 0.2679} = \sqrt{787.626} \approx 28.06\,m/s.

Explanation:

The optimum speed is the speed at which the horizontal component of the normal reaction alone provides the necessary centripetal force, meaning friction is not needed.

Dynamics of Uniform Circular Motion Revision - Class 11 Physics CBSE