Question

What is the frequency of rotation of a body rotating under a force of F = 100N, tied to a sting of length L =
2m and mass of 8g. Your answer should be correct to one decimal.​

Answer 1

Given :

Centripetal force = 100 N

Length of string = 2 m

Mass of the body = 8 g = 0.008 kg

To Find :

Frequency of rotation of a body.

Solution :

Centripetal force acting on a body of mass m moving at a velocity of v in the circle of radius r is given by

$\dag\:\underline{\boxed{\bf{\gray{F_c=\dfrac{mv^2}{r}}}}}$

Relation between linear velocity and angular velocity is given by; v = r ω

$\sf:\implies\:F_c=\dfrac{m(r\omega)^2}{r}$

$\sf:\implies\:F_c=\dfrac{mr^2\omega^2}{r}$

$\sf:\implies\:F_c=mr\omega^2$

We know that :: ω = 2π f

where f denotes frequency of rotation

$\sf:\implies\:F_c=mr(2\pi f)^2$

$\sf:\implies\:F_c=4\pi^2mrf^2$

By substituting the given values;

$\sf:\implies\:100=4(3.14)^2(0.008)(2)f^2$

$\sf:\implies\:100=0.64\times f$

$\sf:\implies\:f=\sqrt{\dfrac{100}{0.64}}$

$\sf:\implies\:f=\dfrac{10}{0.8}$

$:\implies\:\underline{\boxed{\bf{\orange{f=12.5\:Hz}}}}$

Answer 2

Answer:

12.5 Hz

Explanation:

force = mv2/r

force = mr × 4π2 × f2

f2 = force / 4π2 × f2

f = √ .......

answer 12.5 hz