To find the centre of gravity of a lamina

AI Thread Summary
The discussion focuses on determining the center of gravity of a trapezium-shaped lamina and the conditions for it to remain stable when placed vertically. The center of gravity is calculated to be at (7a/9, 4h/9), but the challenge lies in finding the minimum height (h) to prevent toppling. Initial reasoning incorrectly assumed that the trapezium's quadrilateral part was a square, leading to confusion about the tipping point. The correct approach emphasizes that for stability, the center of gravity must remain above the base BC, ensuring that any slight displacement does not cause the lamina to topple. Understanding the balance of moments around point C is crucial for determining the stability conditions of the lamina.
gnits
Messages
137
Reaction score
46
Homework Statement
To find the centre of gravity of a lamina
Relevant Equations
Balancing of moments
Could I please ask for help with the following:

A lamina ABCD is in the form of a trapezium in which DC is parallel to AB, AB = 2a, CD = a and AD = h and the angle BAD is 90 degrees. Find the position of the centre of gravity of the lamina from the edges AD and AB.

The lamina is placed vertically with edge BC on a horizontal plane. Find the minimum value of h for the lamina to remain in this position without toppling in its own vertical plane.

I've done the first part and get the book answers of:

Center of gravity = ( 7a/9, 4h/9)

It's the second part where I am stuck.

Here's a diagram:

trap.png


So I reasoned that the lamina would topple if h (= length AD) were reduced to a width where the centre of gravity were to fall on the line joining A to C, so that when resting on side BC the centre of gravity would then be above C.

Would that be correct?

But this leads to:

Equation of line joining A to C is y = hx/a

So if centre of gravity is on this line then the point (7a/9, 4h/9) should satisfy the equation. But this leads to:

4h/9 = 7a/9 * h/a

and all the values cancel.

Book's answer is a*sqrt(10)/5

Thanks for any help,
Mitch.

EDIT:

Alternative method. Here's a diagram of the rotated shape:
trap2.png


So condition for C.O.G. to be above C is:

(11a/9)*cos(t) + (4h/9)*sin(t) = sqrt(a^2+h^2)

and we know that cos(t) = a/(sqrt(a^2+h^2) and sin(t) = h/sqrt(a^2+h^2)

And this leads to the right answer.

So with my first attempt, what was my mistake of reasoning?

FURTHER EDIT: I think my first method implicitly assumes that the quadrilateral part of the trapezium is a square, which it need not be.
 
Last edited:
Physics news on Phys.org
gnits said:
So I reasoned that the lamina would topple if h (= length AD) were reduced to a width where the centre of gravity were to fall on the line joining A to C, so that when resting on side BC the centre of gravity would then be above C.

Would that be correct?
No. With reference to your 2nd diagram, you can't assume AC is vertical at the critical (tipping) point.

I'll call the centre of gravity 'P' for brevity.

For stability, P must be above the base BC.

If P is slightly to the left of C (i.e. above BC) the weight produces an anticlockwise moment (torque) about C, This will be balanced by a clockwise moment produced by the normal reaction from the ground on BC, giving stability.

But if P is slightly to the right of C (i.e. outside BC) the weight produces a clockwise moment about C - which can't be balanced. So the lamina topples (rotates clockwise about C).

If you consider the traingle PBC, you want P vertically above C so BPC is a right-triangle with ∠BCP = 90º.
 
Thanks very much for you help. Much appreciated.
 
  • Like
Likes jim mcnamara and Steve4Physics
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...

Similar threads

Back
Top