Physics Problem

"Rusty Vandura - www.tinyurl.com/keepoppo" (rustyvandura)
03/14/2016 at 17:40 • Filed to: None

From the film “The Martian:”

If we take the woman in the picture to be 5 feet tall, then we can assume the radius of the wheel on whose edge she is standing to be 50 feet. Give or take.

Here’s the question: If a human is standing on the inside of a rotating cylinder 100 feed in diamater (radius = 50 ft), now many RPMs would that cylinder have to be turning on its axis in order to approximate 1 gravity at the outside of the cylinder?

DISCUSSION (30)

RamblinRover Luxury-Yacht > Rusty Vandura - www.tinyurl.com/keepoppo
03/14/2016 at 17:47

I’m coming up with ~12.6ft/s angular, or about 2.4 RPM.

TheOnelectronic > Rusty Vandura - www.tinyurl.com/keepoppo
03/14/2016 at 17:48

Using

Kinja'd!!!

Where T is the period (seconds per rotation) and a is 9.8m/s^2 which would feel like normal gravity, and a radius of 50 ft, it would have to rotate at about 7.7rpm

Phyrxes once again has a wagon! > Rusty Vandura - www.tinyurl.com/keepoppo
03/14/2016 at 17:48

I will napkin math this out after dinner and not on my phone.

Tl;Dr physics teacher time will be later

TheHondaBro > TheOnelectronic
03/14/2016 at 17:49

...What he said.

HammerheadFistpunch > Rusty Vandura - www.tinyurl.com/keepoppo
03/14/2016 at 17:50

This might help

TheOnelectronic > TheOnelectronic
03/14/2016 at 17:52

Plugged numbers straight into Wolfram Alpha and got the same answer.

BvdV - The Dutch Engineer > Rusty Vandura - www.tinyurl.com/keepoppo
03/14/2016 at 17:54

F=ma with a=r*omega^2 so omega=sqrt(a/r)=sqrt(9.81/15.24)=0.8 rad/s=7.63 RPM

If I’m not missing a factor since it is late here and physics is a while ago.

DrJohannVegas > Rusty Vandura - www.tinyurl.com/keepoppo
03/14/2016 at 17:56

Nice pi day post. I am gonna go metric on my answer, but it won’t matter in the end. I decided to show my work so y’all can laugh if I get it wrong.

If the target F(grav) is based on g = (9.8 m/(s^2)), then we want the “centrifugal force” to equal g*m. Since F(cent) = m*((v^2)/r), where v is the velocity of the object at distance r, we can simplify it a bit.

What we want is g = (v^2)/r. We know r is 15.24 m, and we know that g = 9.8 m/s^2, we get 9.8 m/s^2 = (v^2)/15.24m. Move that around to get 149.4 (ish) m^2/s^2 = v^2. Take the root of both sides to get v = 12.2 (ish) m/s. So close now!

Last part is the period of rotation. If we know that the cylinder has a radius of 15.24m, it has a circumference of ~ 95.76 m. So, at that speed, it would make a full rotation in around 7.85 seconds. Since you asked for RPM, that gives us 7.64-ish RPM.

DrJohannVegas > HammerheadFistpunch
03/14/2016 at 17:58

Shhhhh. Internet points are at stake!

Rusty Vandura - www.tinyurl.com/keepoppo > DrJohannVegas
03/14/2016 at 18:02

So 8 seconds per rotation. That is way slower than I would have expected and the movie was probably accurate in that regard. Is this a freshman Physics problem?

DrJohannVegas > Rusty Vandura - www.tinyurl.com/keepoppo
03/14/2016 at 18:04

Big radius. If I had to guess, I remembered that version of “centrifugal force” from first-year Physics in college, but I didn’t take it freshman year. I was too busy bombing Calc for that.

Rusty Vandura - www.tinyurl.com/keepoppo > RamblinRover Luxury-Yacht
03/14/2016 at 18:04

Your answer is different from everyone else’s.

TheOnelectronic > Rusty Vandura - www.tinyurl.com/keepoppo
03/14/2016 at 18:07

Basically. Rotational shit gets complicated since you have two different forms of everything (angular vs. tangential) but there’s always the old standby- googling equations and then just plugging shit into Wolfram anyway.

Rusty Vandura - www.tinyurl.com/keepoppo > DrJohannVegas
03/14/2016 at 18:11

So the thing that I want to ponder is the effect of radius versus rotation speed. I thought that thing would’ve had to be spinning much faster. So even the ship in “2001: A Space Odessy” was realistic; the dude really could have been jogging on that toroidal segment.

ttyymmnn > Rusty Vandura - www.tinyurl.com/keepoppo
03/14/2016 at 18:14

More than one. What do I win?

ttyymmnn > Rusty Vandura - www.tinyurl.com/keepoppo
03/14/2016 at 18:15

I was told there wouldn’t be any math.

Phyrxes once again has a wagon! > DrJohannVegas
03/14/2016 at 18:24

-1 for using centrifugal force even in jest. I my answer agrees with yours and I did it both ways, posting my solution as a response to the OP.

Phyrxes once again has a wagon! > Rusty Vandura - www.tinyurl.com/keepoppo
03/14/2016 at 18:29

I assumed the object has constant angular velocity, that is it is revolving at the same constant rate. With this being true the relationship below can be used to find the desired velocity.

radial acceleration (centripetal acceleration) = linear velocity squared over radius.

a(r) = v^2/R

converting form linear velocity to angular velocity

v= r*omega means this becomes a(r) = (R*Omega)^2/R

A pair of R’s cancel and give you a(r) = (Omega^2)*R

Choose metric or English and plug in your gravity and radius to solve for the omega then convert it to revolutions per minute.

I did it in in both and arrived at 7.64 RPM just like DrJohannVegas.

DrJohannVegas > Phyrxes once again has a wagon!
03/14/2016 at 19:01

Admittedly, my physics prof was super old, and we were both very, very drunk at the time. But at least you got the scare quotes off my post. Sloppy work done hastily on the bus, just like in college...

Rusty Vandura - www.tinyurl.com/keepoppo > ttyymmnn
03/14/2016 at 19:07

Greater than zero would have sufficed.

Rusty Vandura - www.tinyurl.com/keepoppo > Phyrxes once again has a wagon!
03/14/2016 at 19:09

And, just like Dr. Vegas, you rock.

It’s all about the movie and I thought the rotating part just had to be rotating too slowly. As the direct result of distinguished Oppos, we see that they got the angular velocity right when they made the film.

Phyrxes once again has a wagon! > DrJohannVegas
03/14/2016 at 19:18

I had to take that shot at you as it is a standing rule in my class that if my students even put that word on anything I am taking points off.

DrJohannVegas > Phyrxes once again has a wagon!
03/14/2016 at 19:19

Yea, it seems like a generational thing. I know better, but I were lazy, I were.

Xyl0c41n3 > DrJohannVegas
03/14/2016 at 19:20

Except one of the assumptions is wrong. The average height for white American women is 5 feet, 5 inches tall. Assuming she’s five inches shorter than that average height means the guesstimates of the ring’s radius and diameter are incorrect, too.

Phyrxes once again has a wagon! > DrJohannVegas
03/14/2016 at 19:37

The funny thing was I just taught this stuff in class today. I will admit to using a calculator and not doing it in my head.

DrJohannVegas > Phyrxes once again has a wagon!
03/14/2016 at 22:41

I was on the tablet, and had to pull the phone out as a calculator to do the square root. I also didn’t teach shit today.

RamblinRover Luxury-Yacht > Rusty Vandura - www.tinyurl.com/keepoppo
03/15/2016 at 09:11

Yep, my ft/s number was all wrong (40.1 is correct), so my RPM was wrong also. I used the same process JohannVegas did, but slipped somewhere. The funny thing is, I’ve been watching a show (the anime Cowboy Bebop) which has an artificial gravity ring around the middle of the ship, and its angular speed appears to be much lower (people can step onto it like a moving sidewalk)... but with the ring being only about a 25ft radius. That would make it 28ft/s for full gravity, and ~14 ft/s for quarter gravity... which would be slow enough to step onto, under 10mph.

Rusty Vandura - www.tinyurl.com/keepoppo > RamblinRover Luxury-Yacht
03/15/2016 at 09:14

Don’t worry; we’re not going to sack you over this. Though the change to your avatar continues to shock me.

RamblinRover Luxury-Yacht > Rusty Vandura - www.tinyurl.com/keepoppo
03/15/2016 at 09:20

I’m like Nibby, I only change my avatar and title schtick every other year or so.

Rusty Vandura - www.tinyurl.com/keepoppo > RamblinRover Luxury-Yacht
03/15/2016 at 09:22

“Shock” was too strong a word. “Startle” would be better. Nothing Nibby does startles me, though he’s been rather sedate lately.