Newtons Laws EX 24

The apparent weight of the astronaut is the normal force acting on her. It is the only force acting on her in this situation. Assuming that the spacestation rotates counterclockwise, her acceleration points toward the center and it is parallel to the normal force.

The normal force causes an acceleration towards the centre of the spacecraft. The normal force has a magnitude 20% of her weight on Earth. Thus, we know that

$F_{N}=0.2\times mg=0.2\times m(10)=2m$

Since it is the only force acting on the astronaut we can write

${\overrightarrow {F_{N}}}=m{\overrightarrow {a}}$

$F_{N}=m({v^{2} \over r})$

$2m=m({v^{2} \over r})$

${\sqrt {2r}}=v$

Since the spacesatation rotates at constant speed we can use the equation:

$v={2\pi r \over T}$ (Solve for T)

$T={2\pi r \over v}$ (Replace with $v={\sqrt {2r}}$)

$T={2\pi r \over {\sqrt {2r}}}={2\pi (1000) \over {\sqrt {2(1000)}}}=140.4s$