Artificial Gravity: Benefits & Uses

[coltsamuel96]

 

[coltsamuel96]

Please help me with this problem, at a distance of 40m from the centre of a space station, is a round platform for jogging... compare the weights felt by two persons jogging at 12km/h, when one is jogging in the direction of the rotation of the space station and other in the opposite direction... attempt at the solution, i can get the centrifugal force from the r and v.. ´but after that?thanks in advance

 

[Bill Foster]

$$\vec{F}=m\vec{a}$$
$$\vec{a}=\frac{v^2}{r}\hat{r}$$

 

[Bill Foster]

You'll need to know how fast the station is rotating. If it's not given, then you'll have to write the answer in terms of $$\omega$$.

 

[coltsamuel96]

the velocity i think can be calculated with v^2/r=g... i think it has to do something with the motion of the person jogging with and against the rotation on the station...

 

[Bill Foster]

Combining the two equations I listed previously...

$$\vec{F}=m\vec{a}=m\frac{v^2}{r}\hat{r}$$

There is your basic equation for centrifugal force.

Now, assume the station is rotating at some angular velocity $$\omega$$ rad/sec.

You'll need to convert that into a velocity:

$$v_s=\omega r$$

Now, if the guy is running in the direction of rotation, then his total speed will be:

$$v_{tot}=v+v_s=v+\omega r$$

If he's running against the direction of rotation, then his total speed will be:

$$v_{tot}=v-v_s=v-\omega r$$

Now you can plug these two velocities into the first equation:

$$\vec{F_+}=m\vec{a_+}=m\frac{\left(v+\omega r\right)^2}{r}\hat{r}$$

$$\vec{F_-}=m\vec{a_-}=m\frac{\left(v-\omega r\right)^2}{r}\hat{r}$$

 

[coltsamuel96]

thanks for the help, now i get it...