How Fast Must Earth Spin for Zero Gravity at the Equator?

[Allandra]

1. Homework Statement

By considering the centripetal force acting on a man, calculate the minimum period of rotation that the Earth would need to have about its axis if a man at the equator were to experience zero normal contact force.
Take radius of Earth=6400km

2. The attempt at a solution
this means the g=Fc(centripetral force)
to find v= square root of ( radius of the Earth and g)
g=9.81 N/kg and then sub it to T= (2pie x radius of Earth )/v
Where is the error? Does a zero normal force means the person is experiencing weightlessness due to a=g?

 

[Doc Al]

this means the g=Fc(centripetral force)

OK, but you mean mg = Fc.

to find v= square root of ( radius of the Earth and g)
g=9.81 N/kg and then sub it to T= (2pie x radius of Earth )/v
Where is the error?

I don't see any error. What your final expression for T?

Does a zero normal force means the person is experiencing weightlessness due to a=g?

Yes, that's what "weightlessness" means.

 

[Allandra]

. where radius of the Earth is 6400km
But I can't get the ans=1.41hrs
I can only get 84.3hrs
square root of{Rg}=square root of (6400km X 35.3km/h)= 475
(R multiply by 2pie)divide by T= 40212/T
cross multiply them and I get 84.3hrs. Do u know where is the error?

 

[Doc Al]

square root of{Rg}=square root of (6400km X 35.3km/h)= 475

When you converted g from m/s^2 to km/h^2, you made an error.

Instead, use standard units: meters and seconds, not km and hours. g = 9.8 m/s^2

When you find the answer in seconds, then convert to hours.

 

[Allandra]

Thx Doc Al!