Semicircle

Classical Mechanics Level 4

The moment of inertia of a uniform, semi-circular disc relative to the perpendicular axis through its center of mass is $I = \gamma m r^2$ , where $m$ is the mass and $r$ the radius of the semi-circular disc. Determine the constant $\gamma$ .

The answer is 0.319873.

1 solution

Arjen Vreugdenhil
Feb 7, 2016

According to the parallel axis theorem, $I' = I + m d^2,$ where $I$ is the moment of inertia about an axis through an object's center of mass, $I'$ the moment of inertia of an axis parallel to its, and $d$ the distance between the two axes. In this case, $I = I' - m d^2,$ where $I$ is the required moment of inertia, $I'$ is the moment of inertia relative to a perpendicular axis through the center of the circle , and $d$ the distance of the center of the circle to the semi-circle's center of mass.

I claim that

$I' = \tfrac12 m r^2$
$d = \tfrac 4{3\pi} r$

from which it follows that $\gamma m r^2 = I = I' - m d^2 = \frac12 m r^2 - m \left(\frac 4{3\pi}\right)^2 r^2$ so that $\gamma = \frac12 - \frac{16}{9\pi^2} \approx \boxed{0.319873}.$

Proof of first claim : $I' = \tfrac12 m r^2$ .

This is the same as the well-known formula for a circular disc. For a proof, let $\mu$ be the area density; then $I = \int_V \mu r^2 = \int_0^r dr \int_{-\pi/2}^{\pi/2} r d\phi (\mu r^2) = \mu \pi \int_0^r r^3 dr = \tfrac14\mu\pi r^4.$ The mass is, of course, $m = A \mu = \tfrac12 \pi r^2\cdot \mu,$ and we see immediately that $I = \tfrac12 m r^2$ as advertized.

Proof of second claim : $d = \tfrac 4{3\pi} r$ (distance of semi-circle's center of mass to circle center).

Naturally the center of mass lies on the line of symmetry, which we choose to be the $y$ -axis. The $y$ coordinate we find through integration: $d = \langle y\rangle_A = \frac1 A \int_0^r dr \int_{-\pi/2}^{\pi/2} r d\phi (r \cos\phi) \\ = \frac 1 A \left(\int_0^r r^2 dr\right)\left(\int_{-\pi/2}^{\pi/2} \cos\phi\:d\phi\right) \\ = \frac1 {\tfrac 12 \pi r^2} \cdot \tfrac13r^3\cdot 2 = \frac 4{3\pi} r,$ Q.E.D.

Thanks for beautiful problem.I get (C. O.M) Correct. But I want to know How I can get $I' = \tfrac12 m r^2$ ? And I try to find the moment of inertia by $I_{x}= \int_{Area} y^2 dA$ and I get the answer as $\frac{\pi R^{4}}{8} -\left( \frac{\pi R^{2}}{2} \right) \left( \frac{4R}{3\pi } \right) ^{2}\approx 0. 1098R^{4}$ which I couldn't separate the mass from this formula. How I can continue the solution? And when I start my solution by parallel axis theorem? .

Refaat M. Sayed - 5 years, 4 months ago

For the formula $I = \tfrac12mr^2$ , see "Proof of the first claim" in my answer.

Moment of inertia around the $z$ axis (!) is defined as $I_z = \int dm\:r^2 = \int dm\:(x^2 + y^2)$ . You'd get $I_z = \int_0^R dr \int_{-\pi/2}^{pi/2} r d\phi\: \mu (x^2 + y^2) \\ = \int_0^R dr \int_{-\pi/2}^{pi/2} r d\phi\: \mu \left((r\sin\phi)^2 + (r\cos\phi - \tfrac{4}{3\pi}r)^2\right) \\ =\int_0^R dr \int_{-\pi/2}^{pi/2} r d\phi\:\mu r^3 \left(1 - \tfrac{8}{3\pi}\cos\phi + (\tfrac{4}{3\pi})^2\right) \\ = \mu R^4 \left(\frac\pi 4 - \frac{4}{3\pi} + \tfrac{4}{9\pi}\right) = \mu R^4\left(\frac\pi 4 - \frac{8}{9\pi}\right).$ Divide by the mass, which is $m = A\mu = \tfrac12\pi R^2 \mu$ : $\frac{I_z}{\mu} = R^2 \left(\frac1 2 - \frac{16}{9\pi^2}\right).$

Arjen Vreugdenhil - 5 years, 4 months ago

Thanks sir for help. Just question. Are you Civil Engineer?

Refaat M. Sayed - 5 years, 4 months ago

@Refaat M. Sayed – No, mathematician and physics professor.

Arjen Vreugdenhil - 5 years, 4 months ago

@Arjen Vreugdenhil – Hi. Isn't the centre of mass for a semicircular disk at a distance of $2R/\pi$ from the geometrical centre of the full circle? I've noticed that you've taken that distance to be $4R/3\pi$ , if I'm not wrong.

A Former Brilliant Member - 5 years, 4 months ago

@A Former Brilliant Member – I believe my calculation is correct. See also the Wikipedia list of centroids (under semicircular area).

Your equation $2R/\pi$ applies to the centroid of a semicircular arc .

Arjen Vreugdenhil - 5 years, 4 months ago

@Arjen Vreugdenhil – Well, i must have confused have the two! Thanks! Nice question, btw.

A Former Brilliant Member - 5 years, 4 months ago

nice one !! i made a blunder by applying $I = I' + md^2$ and was getting 0.68. Elegantly written solution by the ways :)

Rohith M.Athreya - 5 years, 4 months ago

Sir I tried this for a quarter of a circle and got the moment of inertia as negative. Is it possible to get negative moment of inertia?

Aditya Kumar - 5 years, 4 months ago

Consider a circle with radius $r$ centered at (0, 0), and cut out the quarter of that circle that lies in the first quadrant.

The $y$ coordinate of the center of mass is still $4r/3\pi$ , as before; by symmetry, the $x$ coordinate of the center of mass is also $4r/3\pi$ . Thus the distance between the axes satisfies $d^2 = 2\left(\frac{4r}{3\pi}\right)^2,$ so that $I = \tfrac12 mr^2 - m\cdot 2\left(\frac{4}{3\pi}\right)^2r^2$ and $\gamma = \frac12 - \frac{32}{9\pi^2} \approx 0.140.$

Arjen Vreugdenhil - 5 years, 4 months ago

I think you should recheck your calculations, cuz i tried for a quarter circle, and its positive, alright. Plus i dont think MOI can be negative, given its the product of mass and square of radius of gyration.

A Former Brilliant Member - 5 years, 4 months ago

sir ,can't we use the perpendicular axis theorem in the first claim?

SB Antu - 5 years, 1 month ago

What do you mean? How would you apply that here?

Arjen Vreugdenhil - 5 years, 1 month ago

I used that to find the moment of inertia with respect to an axis through the centre of the semi circle ... Ix+Iy=Iz ; Here Ix and Iy are both 1/4 MR^2 ...So Iz=1/2 MR^2 (which was actually I' in your solution) ...I solved it using that method & it worked!

SB Antu - 5 years, 1 month ago

Very easy question

vidhan singh - 5 years, 3 months ago

If you want a more challenging question, try a circle "wedge" with angle $\varphi$ instead of a simple semicircle with $\varphi = 180^\circ$ .

Arjen Vreugdenhil - 5 years, 3 months ago

Not that challenging. Centre of mass will be locted at (4/3)sin(x/2)(r/x) on axis of symmetry .Moment of inertia will remain same (mr^2)/2 about centre of disc wedge was part of. Then we will use parallel axis theorem.

vidhan singh - 5 years, 3 months ago

@Vidhan Singh – I said "more challenging", not "much more challenging"-- if you think this too easy, you master this topic better than many.

If you ask for more challenging yet, how about these:

Find the moment of inertia of a section of the circular disc: take a circle, intersect it with an arbitrary line, and rotate one of its pieces around the perpendicular axis through its centroid.
Find the moment of inertia of a half-sphere when rotating about an axis through its centroid, perpendicular to the line of symmetry.
Find the moment of inertia of the semi-circle (a) about the axis through its centroid parallel to its straight side, and (b) about its axis of symmetry.

Arjen Vreugdenhil - 5 years, 3 months ago

@Arjen Vreugdenhil – Ya solved them , not that challenging. Sir i didn't mean to offend you.

vidhan singh - 5 years, 3 months ago

Semicircle

The answer is 0.319873.

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