Question

current attempt in progress
the space station in the drawing is rotating to create artificial gravity. the speed of the inner ring is one half that of the outer ring. as an astronaut walks from the inner to the outer ring, what happens to her apparent weight?

her apparent weight does not change.
her apparent weight becomes four times as great.
her apparent weight becomes twice as great.
her apparent weight becomes one - fourth as great.
her apparent weight becomes half as great.

Explanation:

Step1: Recall centripetal - force formula

The apparent weight \(W\) of the astronaut is equal to the centripetal force \(F_c\) acting on her, and \(F_c=\frac{mv^{2}}{r}\), where \(m\) is the mass of the astronaut, \(v\) is her tangential speed, and \(r\) is the radius of the circular path. Let the speed of the inner - ring be \(v_i\) and the radius be \(r_i\), and the speed of the outer - ring be \(v_o\) and the radius be \(r_o\). Given that \(v_i = \frac{1}{2}v_o\).

Step2: Relate the centripetal forces at inner and outer rings

At the inner ring, \(F_{c,i}=\frac{mv_{i}^{2}}{r_{i}}\). At the outer ring, \(F_{c,o}=\frac{mv_{o}^{2}}{r_{o}}\). Since \(v_i=\frac{1}{2}v_o\), we substitute \(v_i\) into the inner - ring centripetal - force formula: \(F_{c,i}=\frac{m(\frac{1}{2}v_{o})^{2}}{r_{i}}=\frac{mv_{o}^{2}}{4r_{i}}\).
Assume the angular speed \(\omega\) is the same for both rings (since they are part of the same rotating system), and \(v = r\omega\). So \(v_i=r_i\omega\) and \(v_o=r_o\omega\), and since \(v_i=\frac{1}{2}v_o\), we have \(r_i=\frac{1}{2}r_o\).
Substitute \(r_i=\frac{1}{2}r_o\) into the formula for \(F_{c,i}\): \(F_{c,i}=\frac{mv_{o}^{2}}{4\times\frac{1}{2}r_{o}}=\frac{mv_{o}^{2}}{2r_{o}}\). And \(F_{c,o}=\frac{mv_{o}^{2}}{r_{o}}\).
We can see that \(F_{c,o} = 2F_{c,i}\).

Answer:

Her apparent weight becomes twice as great.