# Calculating the center of mass as an astronaut moves on a shuttle

• ago01
In summary, the conversation discusses a problem involving the movement of an astronaut on a space shuttle and the resulting shift of the shuttle's center of mass. The participants consider different approaches, including tracking the center of mass of the shuttle and using conservation of momentum. The correct answer is determined to be the distance between the starting position of the astronaut and the center of mass of the shuttle. The conversation also addresses the significance of the center of mass shift in understanding the movement of the shuttle.
ago01
Homework Statement
An astronaut with mass M_a moves across a rectangular space shuttle of length d. If the space shuttle has mass M_s, how far does the shuttle move back?
Relevant Equations
Center of mass.
I had solved this question but it didn't seem to be appropriate to post in the classical physics problem as my question is still homework-based.

Originally I had thought this might be a conservation of momentum problem. But since we don't have any initial conditions it leaves too much to guess. A simpler approach would be to track the center of mass of the space shuttle, and it's shift would correspond to the movement of the shuttle backwards.

Admittedly the homework question appears poorly worded but from what I deduced (based on getting the correct) answer, this is occurring in space.

The center of mass of the shuttle itself should be at ##\frac{d}{2}## but I thought this wasn't necessary to use. I figured I could track the center of mass at the origin (where the astronaut starts on the shuttle), and I arrived at:

##\frac{0M_s + dM_a}{M_s + M_d}##

which gave me the correct answer. But after reviewing my thought process I confused myself...since shouldn't I have needed to track the center of mass of the shuttle to track it's movement?

What does this result actually mean in terms of a center of mass shift? I am having trouble understanding it without thinking about the midpoint center of mass of the shuttle.

It doesn't actually matter how you choose to distribute the mass inside the shuttle, because it moves as a rigid body. You can localise the mass ##M_{\mathrm{s}}## of the shuttle at a point initially located at starting position of the astronaut, if you like.

ago01 said:
Homework Statement:: An astronaut with mass M_a moves across a rectangular space shuttle of length d. If the space shuttle has mass M_s, how far does the shuttle move back?
Relevant Equations:: Center of mass.

I had solved this question but it didn't seem to be appropriate to post in the classical physics problem as my question is still homework-based.

Originally I had thought this might be a conservation of momentum problem. But since we don't have any initial conditions it leaves too much to guess. A simpler approach would be to track the center of mass of the space shuttle, and it's shift would correspond to the movement of the shuttle backwards.

Admittedly the homework question appears poorly worded but from what I deduced (based on getting the correct) answer, this is occurring in space.

The center of mass of the shuttle itself should be at ##\frac{d}{2}## but I thought this wasn't necessary to use. I figured I could track the center of mass at the origin (where the astronaut starts on the shuttle), and I arrived at:

##\frac{0M_s + dM_a}{M_s + M_d}##

which gave me the correct answer. But after reviewing my thought process I confused myself...since shouldn't I have needed to track the center of mass of the shuttle to track it's movement?

What does this result actually mean in terms of a center of mass shift? I am having trouble understanding it without thinking about the midpoint center of mass of the shuttle.
Conservation of momentum implies the common mass centre does not move.
If the astronaut's mass moves distance x in one direction then how far does the shuttle move in the opposite direction?

It's not clear what your thought process was in getting your answer (or why you think the center of mass of the shuttle wasn't used). Do it step by step. Imagine the astronaut moving left to right across the shuttle. With respect to the center of the shuttle, where is the center of mass when the astronaut (1) begins his walk and (2) completes his walk? Since the center of mass of the system is fixed, how much did the shuttle have to move?

ago01 said:
The center of mass of the shuttle itself should be at ##\frac{d}{2}## but I thought this wasn't necessary to use. I figured I could track the center of mass at the origin (where the astronaut starts on the shuttle), and I arrived at:

##\frac{0M_s + dM_a}{M_s + M_d}##

which gave me the correct answer.

Yep. That should work fine.

Your idea that fixing the center of mass of the shuttle to be at $\frac{d}{2}$ is unnecessary is correct. As @ergospherical points out, you can choose the center of mass of the shuttle to be at any point you like. You could choose it such that it is at $\frac{d}{2}$, but you don't have to. Your choice of placing it at the origin, along with the initial location of the astronaut, is just fine.

All that's necessary is that once you pick a mass distribution for the shuttle, stick with it through the end of the problem.

ago01 said:
But after reviewing my thought process I confused myself...since shouldn't I have needed to track the center of mass of the shuttle to track it's movement?

I assume you did. That's the answer to the problem. The movement of the center of mass of the shuttle is what you are being asked to find. That's the final answer.

ago01 said:
What does this result actually mean in terms of a center of mass shift? I am having trouble understanding it without thinking about the midpoint center of mass of the shuttle.

If it helps to wrap your brain around this problem, realize that there are three "movements" (measured in distance) that are of concern:
• The distance the astronaut moves (relative to the center of mass of the system)
• The distance the shuttle moves (relative to the center of mass of the system)
• The distance the center of mass of the system moves, which is zero. This is the only thing which does not move.

Of course, the distance that the shuttle moves (with respect to the center of mass of the system) is the final answer.

The tricky bit to this problem is realizing out how far the astronaut moves, relative to the center of mass of the system (hint: it's not $d$).

Doc Al
Doc Al said:
It's not clear what your thought process was in getting your answer (or why you think the center of mass of the shuttle wasn't used). Do it step by step. Imagine the astronaut moving left to right across the shuttle. With respect to the center of the shuttle, where is the center of mass when the astronaut (1) begins his walk and (2) completes his walk? Since the center of mass of the system is fixed, how much did the shuttle have to move?

collinsmark said:
I assume you did. That's the answer to the problem. The movement of the center of mass of the shuttle is what you are being asked to find. That's the final answer.
I'm sorry I wasn't clear. To be honest there wasn't much thought process to it... if I'm being honest it was a plug and chug fluke. I'm asking because I'm embarrassed about it being correct.

Walking through your example shows my problem clearly. I am having trouble separating not treating the two separately and instead treating them like a system.

If I just push through it mathematically:

Letting the left side of the shuttle be ##A## and right side be ##B##. Positive x to the right. Astronaut starts at ##A## and goes to ##B##.

If we let the CM of the shuttle be at the center initially as well as the origin, then the system's center of mass initially is:

##\frac{-\frac{d}{2} + 0m_s}{m_a+m_s}##

So the center of mass will sit somewhere between the CM of the shuttle and where the astronaut is standing.

If we look at the final position:

##\frac{\frac{d}{2} + 0m_s}{m_a+m_s}##

Which once again will have the CM sitting between the astronaut and the CM of the shuttle.

So the difference between final and initial gives us

##\frac{\frac{d}{2} + 0m_s}{m_a+m_s} - \frac{-\frac{d}{2} + 0m_s}{m_a+m_s}##

Which is the same answer as I had stumbled upon wrongly. This ends up being reconciled by realizing that without the presence of external force the center of mass doesn't move. Since all the forces of the astronaut pushing against the space shuttle and the shuttle against the astronaut are internal in the shuttle-astronaut system the only way to come to this conclusion is that this difference is exactly the shuttle moving. My problem really came from the OpenStax book I am using not really covering the center of mass not moving in the center of mass section (it was touched on in detail but this detail appears to be left out and instead implied during the discussion of momentum).

When viewed from the origin it appears that the shuttle does not move (indicated by the ##0m_s## in both places), but when viewed from the outside the shuttle moves in order to keep the center of mass from moving. The difficulty in accepting the result is probably because internal forces can't change momentum (you can't lift yourself up by pulling on your feet), but in this system two components are moving but when taken together the system has no momentum and the center of mass of the system doesn't move.

By lining up the two on graph paper I can see the astronaut doesn't actually cover ##d## if the center of mass is to stay equal. I haven't taken a stab at it, but I would suppose that even though he's at ##B## he's covered less than ##d## because for some portion of his movement the shuttle is moving against him. So possibly ##d - d_{cm}## as a first pass where ##d_{cm}## represents the shuttles negative movement.

I'm actually embarrassed at such a "simple" explanation (assuming I am correct) completely alluding me. Physics is a mind-bender. We essentially look at the center of mass of a system based on the centers of mass of it's parts and their distance from the origin (since we can treat the center of mass of each as a "representation" of the place the forces act on an object).

Last edited:
Lnewqban

## What is the center of mass and why is it important for an astronaut on a shuttle?

The center of mass is the point in an object or system where its mass is evenly distributed in all directions. For an astronaut on a shuttle, it is important because it helps determine the stability and balance of the shuttle as the astronaut moves around, and it also affects the trajectory and maneuverability of the shuttle.

## How is the center of mass calculated for an astronaut on a shuttle?

The center of mass can be calculated by finding the weighted average of all the individual masses in the system. For an astronaut on a shuttle, this involves taking into account the mass of the astronaut, the mass of the shuttle, and the distribution of these masses as the astronaut moves around.

## What factors affect the center of mass for an astronaut on a shuttle?

The center of mass for an astronaut on a shuttle can be affected by several factors, including the mass and distribution of the astronaut's body, the mass and distribution of objects the astronaut is carrying, and the movement and position of the astronaut within the shuttle.

## How does the center of mass change as an astronaut moves on a shuttle?

As an astronaut moves on a shuttle, the center of mass will also change. This is because the distribution of mass within the system is constantly shifting. For example, as the astronaut moves from one side of the shuttle to the other, the center of mass will shift towards the side the astronaut is on.

## Why is it important to constantly monitor the center of mass for an astronaut on a shuttle?

It is important to constantly monitor the center of mass for an astronaut on a shuttle because any significant changes in the center of mass can affect the stability and balance of the shuttle, which can impact the safety and success of the mission. By monitoring the center of mass, adjustments can be made to maintain the stability and balance of the shuttle.

• Introductory Physics Homework Help
Replies
3
Views
479
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
19
Views
3K
• Introductory Physics Homework Help
Replies
13
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
993
• Introductory Physics Homework Help
Replies
6
Views
6K
• Introductory Physics Homework Help
Replies
4
Views
810
• Introductory Physics Homework Help
Replies
19
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
3K