A circular room has a radius of 6 meters. A 60 kg person is pinned up against the wall of the room as it rotates. The room begins to decelerate, at the moment the person begins to slip the room is rotating at 50 rpm. What is the normal force exerted on the person by the wall?

Respuesta :

Answer:9.872 kN

Explanation:

Given

radius of room r=6 m

mass of person m=60 kg

N= 50 rpm

[tex]\omega =2\pi N=\frac{2\pi N}{60}[/tex]

[tex]\omega =\frac{2\pi \cdot 50}{60}=5.23 rad/s[/tex]

Centripetal Force will Provide the Normal Reaction thus

Normal Reaction[tex]=m\omega ^2\cdot r[/tex]

Normal Reaction[tex]=60\cdot (5.23)^2\cdot 6=9872.16 N[/tex]

Normal Reaction[tex]=9.872 kN[/tex]

Q&A Education