Q.Imagine that the gravitational force between the Earth and the Moon is provided by an invisible string that exists between them. What is the tension that exists in this invisible string, given that it must supply the Earth's centripetal pull on the Moon? (Mass of the Moon $=7.34\times10^{22}\text{ kg}$; distance between the Moon and the Earth $=3.84\times10^8\text{ m}$; time period of the Moon's orbit = 27.3 days)

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