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# ​ Radius of Gyration - Solid of Uniform Density

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Apr 19, 2022

#4
+254
+5

Hi OldTimer here's one method.

Sketch the graph and take a strip parallel with the x - axis, distance y from the axis and width delta y, from the curve on the left to the vertical at x = a on the right.

Spin this around the x - axis, forming a cylinder.

Vol of cylinder =

$$\displaystyle \{ \pi(y+\delta y )^{2}-\pi y^{2}\}(a-x)=2\pi y\delta y(a-x),$$

(ignoring 2nd order terms, here and elsewhere).

All points on the cylinder are distance y from the x - axis so its moment of inertia (or second moment) about the x - axis

$$\displaystyle = 2\pi y \delta y(a-x).y^{2}=2\pi y^{3}(a-x)\delta y.$$

This now has to be summed for all cylinders between y = 0 and y = a and taken in the limit as delta y tends to zero.

That gets you the integral

$$\displaystyle \int^{a}_02\pi y^{3}(a-\frac{y^{2}}{a})dy=\frac{\pi a^{5}}{6}.$$

Now express this in the form Mk^2, where M is the mass of the solid and k is its radius of gyration.

If unit mass is equal to unit volume, then M is simply the volume of the solid. So,

$$\displaystyle M = \pi\int^{a}_{0}y^{2}dx = \pi \int^{a}_{0}ax dx = \frac{\pi a^{3}}{2}.$$

Finally then, the moment of inertia of the solid = $$\displaystyle \frac{\pi a^{5}}{6}=\frac{\pi a^{3}}{2}.\frac{a^{2}}{3}=M\left(\frac{a^{2}}{3}\right),$$

meaning that the square of the radius of gyration is (a^2)/3.

An alternative method would be to split the solid into circular discs perpendicular to the x - axis, and to sum these from x = 0 to x = a.

If you do that, you could make use of the standard result for the moment of inertia of a disc.

Apr 21, 2022

#4
+254
+5

Hi OldTimer here's one method.

Sketch the graph and take a strip parallel with the x - axis, distance y from the axis and width delta y, from the curve on the left to the vertical at x = a on the right.

Spin this around the x - axis, forming a cylinder.

Vol of cylinder =

$$\displaystyle \{ \pi(y+\delta y )^{2}-\pi y^{2}\}(a-x)=2\pi y\delta y(a-x),$$

(ignoring 2nd order terms, here and elsewhere).

All points on the cylinder are distance y from the x - axis so its moment of inertia (or second moment) about the x - axis

$$\displaystyle = 2\pi y \delta y(a-x).y^{2}=2\pi y^{3}(a-x)\delta y.$$

This now has to be summed for all cylinders between y = 0 and y = a and taken in the limit as delta y tends to zero.

That gets you the integral

$$\displaystyle \int^{a}_02\pi y^{3}(a-\frac{y^{2}}{a})dy=\frac{\pi a^{5}}{6}.$$

Now express this in the form Mk^2, where M is the mass of the solid and k is its radius of gyration.

If unit mass is equal to unit volume, then M is simply the volume of the solid. So,

$$\displaystyle M = \pi\int^{a}_{0}y^{2}dx = \pi \int^{a}_{0}ax dx = \frac{\pi a^{3}}{2}.$$

Finally then, the moment of inertia of the solid = $$\displaystyle \frac{\pi a^{5}}{6}=\frac{\pi a^{3}}{2}.\frac{a^{2}}{3}=M\left(\frac{a^{2}}{3}\right),$$

meaning that the square of the radius of gyration is (a^2)/3.

An alternative method would be to split the solid into circular discs perpendicular to the x - axis, and to sum these from x = 0 to x = a.

If you do that, you could make use of the standard result for the moment of inertia of a disc.

Tiggsy Apr 21, 2022
#5
+234
+2

Thanks for that Tiggsy its great...its the solution I was looking for!! . Unknowingly I went for your second alternative ...only cos I had found a formula.

Apr 22, 2022