Centre of Mass Help: Finding x and y Coordinates of a Parallelogram Lamina

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Discussion Overview

The discussion revolves around finding the x and y coordinates of the center of mass of a uniform lamina shaped like a parallelogram, as well as the center of mass of a modified body formed by folding triangles over specific lines. Participants also explore the implications of pivoting the body at a certain point and the calculations related to angles and torque in equilibrium conditions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant claims the center of mass of the parallelogram ABCD is at (0, b/2) and suggests the same for the newly formed body after folding.
  • Another participant questions the labeling of points A, B, C, and D, suggesting a possible mislabeling that could affect calculations.
  • A later reply corrects the coordinates of points A and D, indicating they are (-2a, 0) and (-a, b), respectively.
  • Participants discuss the implications of pivoting the body at point P (3a/4, b/3) and how it affects the center of mass when the body is in equilibrium.
  • One participant proposes a method involving drawing a diagram and hanging the body to visualize the center of mass directly beneath the pivot point.
  • Another participant raises questions about the coordinates of points B and T, expressing uncertainty about B's position.
  • One participant suggests calculating the angle between line BT and the vertical using the coordinates of points P and B.
  • A different participant calculates the tangent of the angle based on the gradient between the center of mass and the pivot point, leading to a specific expression for torque related to the weight of the parallelogram.
  • Two posts introduce a new problem regarding the center of mass of a semicircular plate mounted over a rectangle, seeking guidance on how to approach the solution.

Areas of Agreement / Disagreement

Participants express differing views on the labeling of points and the implications for calculations. There is no consensus on the correct coordinates or the approach to solving the problem, and the discussion remains unresolved regarding the new problem introduced.

Contextual Notes

Some participants express uncertainty about the coordinates and relationships between points, which may affect the calculations. The discussion includes assumptions about the configuration of the body and the conditions for equilibrium that remain unverified.

JJB11
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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.

(Same answer as previous one).

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:
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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:
Sorry, that was silly of me: A and D are actually (-2a,0) and (-a,b).
 
JJB11 said:
This body is now pivoted freely at the point P (3a/4, b/3) and is in equilibrium with its plane vertical.
"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. :smile:

(a) Find the tangent of the angle between the line BT and the vertical.
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.
 
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...
 
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.
 
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.
 
That's about what I make it.
 
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
hero.rohan said:
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?
How will you start to solve this?
 

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