Q.A car takes a turn with velocity 50 ms^-1 on a circular road of radius of curvature 10 m. Calculate the centrifugal force experienced by a person of mass 60 kg inside the car.

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