Guys what's the formula for finding centre of gravity of cylinders.?
Answers
You don’t need to use calculus for this problem.
You can solve it by finding the new place where the cylinder could be suspended with all of the torques about that point summing to zero (static equilibrium).
You can probably guess that for a hollow cylinder with both end caps, the center of gravity will be the center of the cylinder (along the axis of symmetry, halfway between the two end caps).
If you were suspending the cylinder from an exterior point directly “above” this center, it would hang in static equilibrium.
If you were to remove, say the “right” end cap, the cylinder would now sag on the “left” side, because there is a torque acting in that direction due to the presence of the “left” end cap which is no longer offset by the “right” end cap.
You can probably guess that you need to move the point of suspension an amount x to the left so that the system is once again in static equilibrium.
I am actually finding the center of mass of the system, but it will also be the center of gravity of the system if it is located in a uniform gravitational field.
Again, call the shift in the location of the center of gravity toward the remaining end cap x.
Now consider all of the torques that will be acting on the system relative to this new center of gravity.
The mass of the left end cap is sigma*pi*R^2, where sigma is the area mass density and R is the radius of the end cap.
Its distance from the center of gravity is (L/2 - x), where L/2 is half the length of the cylinder and x is the amount by which the new center of gravity has shifted toward the remaining end cap relative to the original center of gravity.
The torque exerted by the remaining end cap relative to the new center of gravity is therefore sigma*pi*(R^2)*(L/2 - x).
Now you have to figure out where the center of gravity of the cylindrical shell (no end caps) is relative to the new center of gravity. It is halfway from the new center of gravity to the left end of the cylinder, (L/2 - x)/2.
The mass of this portion of the cylindrical shell (no end caps) lying between the new center of mass and the remaining end cap is sigma*2*pi*R*(L/2 - x) (surface area of this surface * area mass density).
Therefore, the torque that the cylindrical shell between the new center of gravity and the remaining end cap is sigma*2*pi*R*(L/2 - x)*(L/2 - x)/2.
Similarly, the torque exerted by the cylindrical shell (no end caps) between the new center of gravity and the open end of the cylinder is sigma*2*pi*R*(L/2 + x)(L/2 + x)/2.
The equation that now represents the balancing of torques about the new center of gravity is
sigma*pi*R^2*(L/2 - x) + sigma*2*pi*R*[(L/2 - x)^2]/2 = sigma*2*pi*R*[(L/2 + x)^2]/2
You can cancel out one power of R, sigma, and pi to simplify this to
R(L/2 - x) + (L/2 - x)^2 = (L/2 + x)^2
and solve for x.
RL/2 - Rx + L^2/4 - Lx + x^2 = L^2/4 + Lx + x^2
You can cancel the x^2 terms, the L^2/4 terms and start grouping terms by power of x
(IF I haven’t made any algebra errors)
-(R + 2L)x = -RL/2
x = RL/[2(R + 2L)]
This has the correct dimension of [length] ([length]^2/[length]). It makes sense that it depends upon both the length of the cylinder the radius of the cylinder, as both of these figure into the torque that the remaining end cap causes about the new center of gravity (the radius because the mass of the remaining end cap and the mass of the cylindrical shell (no end caps) between the new center of mass and the remaining end cap depend on this, and the length of the cylinder because the “lever” arm of the remaining end cap and the center of mass of the cylindrical shell (no end caps) depends on this.
This is the location of the new center of gravity. It is a distance x from the center of the cylinder, toward the remaining end cap, still on the axis of symmetry of the cylinder.
If R >> L, then x reduces to L/2, which makes sense, as the cylinder is basically just two end caps barely separated, so removing one shifts the center of gravity all the way to the remaining end cap.
If L >> R, then x reduces to a very small amount, which also makes sense, as the cylindrical surface (no end caps) contains most of the mass of the system, and removing an end cap does little to shift the center of gravity.
That is, if I didn’t make any silly errors.
Hope this helps and does not mislead or confuse you.
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