Q.Case study: Friction is a resistive force that arises when there is relative motion between two surfaces, provided the two surfaces are irregular relative to each other. Frictional force always opposes motion, i.e. it always acts in the direction opposite to the motion. Rubbing two objects against each other causes wear and tear due to friction, and heat energy is also generated. A cyclist travels on a circular track of radius 100 metres. If the coefficient of friction is .2, what is the maximum speed with which the cyclist can safely take the turn? (g = 9.8 m/s²)

$v_{\max}=\sqrt{\mu g r}=\sqrt{0.2\times9.8\times100}=\sqrt{196}=14$ m/s.

On a flat (unbanked) circular track, the centripetal force needed to keep the cyclist turning is provided entirely by friction between the tyres and the road. The maximum available friction force is $f_{\max}=\mu mg$ (limiting/static friction), so the maximum safe speed is when this friction exactly equals the required centripetal force: …

F = mv^2/r; doubling v gives 4x the force. Answer (B).

The centripetal force needed for uniform circular motion is F = mv^2/r.

With mass m and radius r unchanged, F is proportional to v^2. If v becomes 2v:

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Centrifugal force due to Earth's rotation is Fc = m ω² R cos λ, where λ is the latitude; it is zero at the poles and maximum at the equator.

A person standing on the rotating Earth moves in a circle of radius r = R cos λ about the Earth's rotation axis, where R is the Earth's radius and λ is the latitude of the place.

The centrifugal (pseudo) force experienced in the rotating frame is Fc = m ω² r = m ω² R cos λ.

At the poles, λ = 90°, so cos λ = 0, giving Fc = 0.

At the equator, λ = 0°, so cos λ = 1, giving the maximum value Fc = m ω² R.

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Centripetal force provides the centripetal acceleration a = ω^2 r (or v^2/r) required for circular motion; by Newton's second law, F = ma = m ω^2 r.

For a body of mass m moving in a circle of radius r with angular velocity ω, the linear speed is v = ωr.

The centripetal acceleration (directed toward the centre) is: …

The centripetal force needed to keep a mass m moving in a circle of radius r at speed v is $F=mv^2/r$.

Derivation. A body moving on a circular path of radius $r$ with uniform speed $v$ has centripetal (radial) acceleration directed toward the centre:

$a_c = \frac{v^2}{r}$ …

Centripetal force is the mass times the centripetal acceleration $v^2/r$.

A body moving in a circle of radius $r$ with uniform speed $v$ continuously changes direction, so it has a centripetal (radially inward) acceleration:

$a_c = \frac{v^2}{r}$

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Centripetal force is the net force directed toward the centre needed to keep a body moving in a circle; in terms of linear speed it is F = mv^2/r.

For a body of mass m moving in a circle of radius r with speed v, the centripetal acceleration is a = v^2/r, directed toward the centre.

By Newton's second law, the required centripetal force is F = m a = m v^2 / r.

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The tension supplies the centripetal force, T = mv^2/r = 4 N.

For a body moving in a circle, the net inward (centripetal) force required is F = mv^2/r. Here this force is supplied entirely by the string tension T.

Given: m = 0.5 kg, r = 2 m, v = 4 m/s

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