Question

A 2.00 m-long 6.00 kg ladder pivoted at the top hangs down from a platform at the circus. A 42.0 kg trapeze artist climbs to a point where her center of mass is at the center of the ladder and swings at the system's natural frequency. The angular frequency (in s −1) of the system of ladder and woman is

Answer 1

Answer:

f= 0.48 s⁻¹

Explanation:

Given that

Mass of the ladder ,m= 6 kg

L= 2 m

Mass of artist ,M= 42 kg

The total moment of inertia of the both ladder and artist ,I

$I=\dfrac{mL^2}{3}+ Mr^2$

Here$r=\dfrac{L}{2}$

$I=\dfrac{mL^2}{3}+ M\dfrac{L^2}{4}$

$I=\dfrac{6\times 2^2}{3}+ 42\times \dfrac{2^2}{4}\ kg.m^2$

I=50 kg.m²

The angular frequency f given as

$f=\dfrac{1}{2\pi}\times\sqrt{ \dfrac{m_t\ g D}{I}}$

Here $m_t= m +M$

$D=\dfrac{L}{2}\ m$

$D=\dfrac{2}{2}\ m$

D=1 m

$f=\dfrac{1}{2\pi}\times\sqrt{ \dfrac{(42+6)\times 9.81\times 1}{50}}\ s^{-1}$

f= 0.48 s⁻¹