# Homework Help: Centre of Mass Help

1. Sep 10, 2011

### JJB11

A uniform lamina ABCD is in the shape of a parallelogram and its mass per unit area is u. Axes Ox, Oy are chosen, with O the point on AB such that AO = 2/3AB. Points A, B, C and D
have coordinates (-2a, O), (a, O), (2a, b) and (-a, b) respectively. Find the x and y coordinates of the centre of mass of ABCD.

(I have done this, it is (0,b/2)).

Triangles ASD and BCT (right angled triangles at each end of the parallelogram) are folded over SD (stright line from D to x-axis) and BT (straight line from B to intersection with CD) respectively so that they lie flat on the rectangular region SBTD. Find the x and y coordinates of the centre of mass of the newly formed body.

This body is now pivoted freely at the point P (3a/4, b/3) and is in equilibrium with its plane vertical.

(a) Find the tangent of the angle between the line BT and the vertical.

(b) Find also the magnitude of the couple that must be applied to the body for it to rest in equilibrium with the edge DT vertical.

Help! :(

Last edited: Sep 10, 2011
2. Sep 10, 2011

### Staff: Mentor

Are you sure you have the coordinates of the points labelled correctly? It looks like being BADC not ABCD. And placing the Origin between 2 points yet without causing one of those points to have a negative ordinate has me vexed. :(

I'll be interested to see how this is solved.

Last edited: Sep 10, 2011
3. Sep 10, 2011

### JJB11

Sorry, that was silly of me: A and D are actually (-2a,0) and (-a,b).

4. Sep 11, 2011

### Staff: Mentor

"Its plane vertical" means that it is hanging straight up and down. Like, it is on the wall.

Draw a large diagram of the body on cardboard, and cut it out. Now, poke a hole in it at point P and hang the shape on the wall by pushing a pin through this hole. The body hangs with its C of M directly beneath this pivot point P, as do all hanging bodies.

You have enough information on the co-ordinates of P and B to allow you to calculate the angle that line BT makes with a vertical line, when the body is hanging thus.

5. Sep 12, 2011

### JJB11

What do the points of B and T end up being? I've got T as the old centre of gravity (0,b/2) but I'm not sure about B at all...

6. Sep 12, 2011

### Staff: Mentor

I didn't actually rotate the body. I drew a line from P through the C of M and said that's going to be the "vertical" so what is the angle between that line and BT? I think that's a valid approach.

7. Sep 12, 2011

### JJB11

Ah, that's a good idea. You then get the gradient of the tangent between the centre of gravity and the pivot point and because m=tan theta then tan theta = (9a/2b).

I've also got the answer to part (b). If DT is vertical then it is parallel to the weight which lies along line y=b/2. Therefore the torque that provides this shift is (weight of parallelogram)*(distance from pivot) = 3ab*(mass per unit area)*g*b/6 which gives a couple (same magnitude but acting in opposite turning direction) of ab^2(mass per unit area)g/2.

8. Sep 12, 2011

### Staff: Mentor

That's about what I make it.

9. Sep 13, 2011

### hero.rohan

i need answer for this problem

a semicircular plate is mounted over a rectangle of side 2a cm and other side of a cm. Taking
origin at one corner of rectangle calculate the centre of mass of the system?

10. Sep 14, 2011

### Staff: Mentor

How will you start to solve this?