How Far Is the Dog from the Shore After Walking on the Boat?

[anotherghost]

Homework Statement

All right, so: dog on a boat. It's the snoopy problem. There's a boat floating on some water, and there's a dog on the end - he starts at X away from the shore. Then, he walks a length L across the boat towards the shore. How far is he away from the shore at the end?

Homework Equations

m1r1 + m2r2 = (m1 + m2)rcm

The Attempt at a Solution

As far as I can get is to set the center of mass as X at the beginning.

Mboat * Xi + Mdog * Xi = (Mboat + Mdog)Xi

and then the dog walks L towards the shore and I am not sure how to set this equation up.

Mboat * (Xi + L) + Mdog * (Xi - L) = (Mboat + Mdog) Xi

See, I know the center of mass doesn't move, because the system is isolated and the only forces are internal. However, I don't know how to make the positions relative - this equation would work if the problem literally stated he walks a length L towards the shore (his new position obviously would be Xi - L) but it says he walks length L down the boat, which means the boat moves too and he doesn't quite make the distance L from an external observer's point of view.

 

[rl.bhat]

In the given problem, the center of mass of boat and dog are not at Xi. Only CM of dog is at Xi. Let d be the CM of boat.
Now CM of (boat + dog) will be -------(1)
As you have said CM of the system remains the same. So when dog walks a length L towards the shore, CM of boat must move away from the shore. Let this be x. Now new position of the dog is [(X - L) + x] and new CM of boat wiil be (d + x). Find the CM of ( dog + boat) in this position and equate it to eq(1) and find x.

 

[kerimek]

I understand your frustration with this problem. It can be challenging to visualize and set up equations for problems like these. However, it is important to remember that in physics, we often make simplifying assumptions in order to solve problems. In this case, we can assume that the boat is stationary and only the dog is moving. This means that the center of mass will remain at Xi at all times.

Using this assumption, we can set up the following equation:

Mdog * (Xi + L) = (Mdog) Xi

Solving for Xi, we get:

Xi = L/2

Therefore, at the end of the dog's walk, he will be L/2 away from the shore.

It's important to keep in mind that while this solution may not perfectly reflect the real world scenario, it allows us to solve the problem and gain a better understanding of the concepts involved. As scientists, we must be willing to make simplifying assumptions in order to make progress in our understanding of the natural world.