Centre of Mass; Physics.?

Question

A boat of 45kg and length 3m is floating in a still frictionless water. A dog of 15kg walks from the bow to the stern. Find the distance through which the boat gets displaced.



The answer would be 0.75m .....I just can't find out how.

Please dont give me pointless answers like "This is a question from statics" or "This is really a tricky question". I need the answer mathematically........plz help.

jamesrobinson here has given me a very precise answer...thanks buddy. i really dont think it has got anything to do with momentum. the velocity with which the dog moves is immaterial.

Answer 1

Easy! The trick is to realise that the centre of mass OF THE SYSTEM must remain in the same place.
Let's start by defining the origin of our coordinate system at the bow.
So the dog is located at the origin, and the centre of mass of the boat is located 1.5m away (half-the boat's length, assuming a uniform boat.)
So the centre of mass initially is simply the sum of the mass*distance values for each mass divided by the total mass:
c of m is at (45 * 1.5 + 15 * 0) / 60 = 1.125 m

So initially, the centre of mass of the system is 1.125m from our origin. THIS IS WHERE IT WILL STAY!

When the dog reaches the end, it is not 1.5m from the origin because the boat will also have moved.
Let x be the distance that the boat has moved.

Hence the position of the dog is (3 - x) metres.
Draw a diagram if you need to check.

And we know
1.125 = ( 45*x + 15(3 - x) ) / 60
Rearrange for x
1.125 = ( 45x + 45 - 15x) / 60
1.125 = (30x + 45) / 60
67.5 = 30x + 45
so 30x = 22.5
therefore x = 0.75 m.

I hope this helps you.

Answer 2

Because there are no forces applied from outside the closed system (boat + dog), the center of mass of the system must remain in the same place and the whole system (boat + dog) must move a distance that allows this to happen.

The total mass of the system is 60kg (45 + 15).
Starting with zero distance at the end where the dog is the center of mass is calculated by taking the total moments about the point of zero distance and dividing by the total mass
45*(3/2)/60 = 1.125mtrs

Now the center of mass will be the same distance from the other end of the boat after the dog has moved. This means that the boat will have to move a distance of
(3 - 1.125 - 1.125) = 0.75 mtrs.

Answer 3

Don't know if this will fly in class, but it will give the right answer every time. Its true that this has nothing much to do with centre of mass and all about momentum, which is product of mass times velocity. Centre of mass comes in with treating the dog as if it were a point of mass 15kg moving 3m in a system which has a mass of 60kg(dog +boat):

(15kg/60kg)*3m=0.75m

Answer 4

Where is the CM? Well, just say that it's at "0". Treat this "0" as the balance point on an ideal scale.

15*dog's position before (D_b)+45*CM of boat's position before (B_b)=0
eg the boat's initial position is at -1/3 the dog's initial position,
B_b=-1/3D_b

15*dog's position after+45*CM of boat's position after=0
The dog's position after=the dog's position before-3+however far the boat moves the opposite way
eg D_a=D_b-3+x

The boat's positiopn after=the boat's position before+however far the boat moves
eg B_a=B_b+x

Thus 15(D_b-3+x)+45(-1/3D_b+x)=0
(Notice how the D_b's cancel out)
-45+15x+45x=0
60x=45
x=3/4

Answer 5

i'm working on it, but i dont think it has to do with center of mass, I think it has more to do with conservation of momentum. I will update my answer in a few.

hmm, do you have any more information? displacement in what direction? was ther a picture that went along with the problem?

Answer 6

0.10m?