# Density and centre of mass problem!

1. Mar 3, 2014

### Muffintoast

1. The problem statement, all variables and given/known data

A machine part consists of a thin, uniform 4.00 kg (A) bar that is 1.50 m (X) long, hinged
perpendicular to a similar bar, placed vertically downward, of mass 3.00 kg (B) and length 1.80 m (Y).
The longer bar has a small but dense 2.00 kg (C) ball at one end (the ball can be considered as a
point mass). By what distance will the center of mass of this part move horizontally and vertically if
the vertical bar is pivoted 90° counterclockwise to make the entire part horizontal (8.51)?

2. Relevant equations
Centre of mass equation:
x(centre of mass) = Σm1+x1/Σm
y(centre of mass) = Σm2+y2/Σm

3. The attempt at a solution

So I tried finding the centre of mass before machine rotated, so the [mass of A x the (x) coordinate of A] + [mass of A x the (x) coordinate of B]/all masses together so [(4 x 1.4) + (4 x 0)]/(2 + 3 + 4) = 2/3

Then, [mass of B x the (y) coordinate of A] + [mass of B x the (y) coordinate of B]/all masses together so again, [(3 x 1/8) + (3 x 0)]/ ( 2 + 3 + 4) = 5.4/9

Then, I tried finding the centre of mass after the machine rotates so that the (y) coordinate of B becomes 0. We then know that y is 0, then to find x, it is [mass of A x (x) coordinate of A] + [mass of A x (x) coordinate of B]/ all the masses added together = [(4 x 1.5) + 94 x 1.800]/(2 + 3 + 0) = 1.4666

So to find the displacement, it is the centre of mass before the rotation - the centre of mass after the rotation = 2/3 - 1.46666 = 4/5 then, 5.4/9 - 0 = 5.4/9
So the answer is x(4/5), y(5.4/9)

This answer is completely wrong, I have the correct answer which is supposed to be Δx = 0.700m, Δy = 0.700m. Please tell me where I went wrong, this is my first week of uni and i'm just so confused. :(

2. Mar 3, 2014

### tiny-tim

Hi Muffintoast! Welcome to PF!
(so A is fixed, and B moves from vertical to horizontal)

I'm very confused …

I think you've use the full length instead of the half-length for the rods, and miseed out the ball.

Start again.

3. Mar 3, 2014

### Muffintoast

Okay, I'm going to start again but one question, why do you use half length instead of the full length? The formula that my lecturer gave us said to use the length of x (A) and y (B).

4. Mar 3, 2014

### SteamKing

Staff Emeritus
What formula?

5. Mar 4, 2014

### tiny-tim

Hi Muffintoast!

(just got up :zzz:)
For each large part of the total body, you must use the centre of mass of that part (times its mass) …

the centre of mass is in the middle, so you use half its length (instead of the whole distance as in eg a "point" part like the ball).

6. Mar 4, 2014

### Muffintoast

Ohh I see! To steamking, the formula that I posted on the original post. I asked my lecturer and she said that with this formula I use the half-length. I was not listening properly and I thought she said you use the full length. I'm doing all of this for the first time but I got the answer! Thanks everyone for the help!

So for the x coordinate I did, (4 x 0.75) + (3 x 1.5) + (2 x 1.5)/(4 + 3 + 2) = 10.5/9
For the y coordinate, (4 x 0) + (3 x -0.9) + (2 x -1.8)/(4 + 3 + 2) = -6.3/9

Then after the rotation, x = (4 x 0.75) + (3 x 2.40) + (2 x 3.30)/9 = 16.8/9
y = 0

The difference between the two is then Δx = 16.8 - 10.5/9 = 0.700m, Δy = 0 - - 6.3/9 = 0.700m! :D

7. May 3, 2015

### DaveTan

Hey! Sorry for reviving this old thread, but I am doing this same question and am confused on the x-coordinate part.

For before the rotation, you did

x-cord: (4 x 0.75) + (3 x 1.5) + (2 x 1.5)/(4 + 3 + 2) = 10.5/9

However, shouldn't it be

x-cord: (4 x 0.75) + (3 x 0) + (2 x 0)/(4 + 3 + 2) = 3/9

instead since the 3 kg bar and the 2kg ball lie on (0,0) assuming the hinge is the origin?

UPDATE:
Oh I found out that his origin would probably be on the left hand on the bar, so there is nothing wrong with the answer. Thanks!!

Last edited: May 3, 2015
8. May 3, 2015

### SteamKing

Staff Emeritus
DaveTan:
PF doesn't like it when users necropost to old threads. The original posters to the thread may be long gone by the time you make your post, and you may be waiting for an answer in vain.

It's better to start a new thread if you have a question, and you can quote or refer to the older thread if you need to.

9. May 3, 2015

### DaveTan

Oh no, I'm sorry! Won't do it again!