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

f= 0.48 s⁻¹

Step-by-step 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=(mL^2)/(3)+ Mr^2`

Here
`r=(L)/(2)`

`I=(mL^2)/(3)+ M(L^2)/(4)`

`I=(6* 2^2)/(3)+ 42* (2^2)/(4)\ kg.m^2`

I=50 kg.m²

The angular frequency f given as

`f=(1)/(2\pi)*\sqrt{ (m_t\ g D)/(I)}`

Here
`m_t= m +M`

`D=(L)/(2)\ m`

`D=(2)/(2)\ m`

D=1 m

`f=(1)/(2\pi)*\sqrt{ ((42+6)* 9.81* 1)/(50)}\ s^(-1)`

f= 0.48 s⁻¹