Let's assume that a particular ring of radius $r$ is spinning at angular speed $\omega$, then a person of mass $m$ in this ring will experience a centrifugal force $F_c$ given in magnitude by $$ F_c = m r\omega^2 $$ and pointing radially outward.

Important Aside. The centrifugal force, which points radially outward, is what's termed a "fictitious force" in classical mechanics because it is not the result of interactions with another object. Instead, it is merely an apparent force that results from making observations from a non-intertial frame of reference. However, if a person were actually walking along the outside of one of the rings, then he would experience the apparent gravity by virtue of the contact force between his feet and the ring.

If this person were to apply Newton's second law in this rotating frame and treat this apparent outward force as he would a gravitational force on the surface of the Earth where the force is given in mangitude by $mg$ where $g$ is the acceleration due to gravity, then he would find that $$ mg = m r \omega^2 $$ and therefore, he would experience an effective gravitational acceleration which is a function of $r$ and is given by $$ \boxed{g_\mathrm{eff} = r\omega^2} $$ So we see that the effective acceleration due to gravity in a given ring increases linearly with the radius of the ring and quadratically with its angular speed.

I just saw that someone else also wrote and answer that takes it from here, so I hope this helps!

Cheers!

Let's assume that a particular ring of radius $r$ is spinning at angular speed $\omega$, then a person of mass $m$ in this ring will experience a centrifugal force $F_c$ given in magnitude by $$ F_c = m r\omega^2 $$ and pointing radially outward.

Important Aside. The centrifugal force, which points radially outward, is what's termed a "fictitious force" in classical mechanics because it is not the result of interactions with another object. Instead, it is merely an apparent force that results from making observations from a non-intertial frame of reference. However, if a person were actually walking along the outside of one of the rings, then he would experience the apparent gravity by virtue of the contact force between his feet and the ring.

If this person were to apply Newton's second law in this rotating frame and treat this apparent outward force as he would a gravitational force on the surface of the Earth where the force is given in mangitude by $mg$ where $g$ is the acceleration due to gravity, then he would find that $$ mg = m r \omega^2 $$ and therefore, he would experience an effective gravitational acceleration which is a function of $r$ and is given by $$ \boxed{g_\mathrm{eff} = r\omega^2} $$ So we see that the effective acceleration due to gravity in a given ring increases linearly with the radius of the ring and quadratically with its angular speed.

I just saw that someone else also wrote an answer that takes it from here, so I hope this helps!

Cheers!

Let's assume that a particular ring of radius $r$ is spinning at angular speed $\omega$, then a person of mass $m$ in this ring will experience a centrifugal force $F_c$ given in magnitude by $$ F_c = m r\omega^2 $$ and pointing radially outward. If this person were to apply Newton's second law in this rotating frame and treat this apparent outward force as he would a gravitational force on the surface of the Earth where the force is given in mangitude by $mg$ where $g$ is the acceleration due to gravity, then he would find that $$ mg = m r \omega^2 $$ and therefore, he would experience an effective gravitational acceleration which is a function of $r$ and is given by $$ \boxed{g_\mathrm{eff} = r\omega^2} $$ So we see that the effective acceleration due to gravity in a given ring increases linearly with the radius of the ring and quadratically with its angular speed.

I just saw that someone else also wrote and answer that takes it from here, so I hope this helps!

Cheers!

Let's assume that a particular ring of radius $r$ is spinning at angular speed $\omega$, then a person of mass $m$ in this ring will experience a centrifugal force $F_c$ given in magnitude by $$ F_c = m r\omega^2 $$ and pointing radially outward.

Important Aside. The centrifugal force, which points radially outward, is what's termed a "fictitious force" in classical mechanics because it is not the result of interactions with another object. Instead, it is merely an apparent force that results from making observations from a non-intertial frame of reference. However, if a person were actually walking along the outside of one of the rings, then he would experience the apparent gravity by virtue of the contact force between his feet and the ring.

If this person were to apply Newton's second law in this rotating frame and treat this apparent outward force as he would a gravitational force on the surface of the Earth where the force is given in mangitude by $mg$ where $g$ is the acceleration due to gravity, then he would find that $$ mg = m r \omega^2 $$ and therefore, he would experience an effective gravitational acceleration which is a function of $r$ and is given by $$ \boxed{g_\mathrm{eff} = r\omega^2} $$ So we see that the effective acceleration due to gravity in a given ring increases linearly with the radius of the ring and quadratically with its angular speed.

I just saw that someone else also wrote and answer that takes it from here, so I hope this helps!

Cheers!