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The question is : "the flat surface of a hemisphere of radius 'r' is cemented to one flat surface of a cylinder of the same radius and of the same material. If the length of the cylinder be 'l' and the total mass be 'm', show that the moment of inertia of the combination about the axis of the cylinder is given by: $$ mr^2\Big\{(l/2)+(4r/15))(l+(2r/3)\Big\} $$ I have considered starting point as the vertex from the centre of the hemisphere. I have take an elementary volume strip 2πr×rcos¢×d¢×dr. Then I had done integration for R: varying from 0 to r and R to √(l^2 + r^2) and ¢ varying from -π to 0 and 0 to π. But I am not getting the answer!

bubba
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Manoj Dash
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4 Answers4

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Mass apportioning between hemisphere and cylinder should be first calculated. Using standard results for hemisphere and cylinder,

$$ m_1 = \rho \frac23 \pi r^3 ,\, m_2= \rho \pi r^2 l ,\, \frac{m_1}{m_1+m_2}= p=\frac{2r/3}{2r/3+l} , \frac{m_2}{m_1+m_2}=q=\frac{l}{2r/3+l} $$

$$ MI= m_1r^2/5 + m_2 r^2/2 = p r^2/5 + q r^2/2 $$

$$ = \frac{mr^2}{10} \cdot \frac{4r+15l}{2r+3l} $$

which is same answer as that of Rajput.

Narasimham
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\begin{align} &\;\text{ Moment of inertia of your object}\vphantom{\frac12} \\ = &\text{ MI of cylinder} + \text{MI of hemisphere} \\ = &\,\text{ MI of cylinder} + \frac12 \text{MI of sphere} \end{align} You can look up the moments of inertia of a cylinder and sphere in many places, like here, for example.

bubba
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Sounds like a homework problem, so my "answer"'s two hints:

(a) first, the question as you've worded it is unclear (at least to me) whether the "total mass" $m$ refers to the cylinder alone, or to the cylinder plus hemishpere. Ultimately, you need the density $\rho=m/V$ where either $V=\pi r^2l$ or that plus $\frac23\pi r^3$.

(b) why the $\theta$-dependence? Your cylinder-with-hemisphere-on-top is cylindrically symmetric around the axis along its center that you've described. At a distance $R,\ 0\leq R\leq r,$ from that axis, the "total height" is just $h(R)=l+\sqrt{r^2-R^2}$. So a volume element is just $dV(R)=2\pi R\ast h(R)\ast dR$. Multiply that by $\rho$ for mass, and by $R^2$ for moment of inertia, and integrate over $R$ from $0$ to $r$. That should give you the correct answer.

John Forkosh
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  • Sorry @john, I am not getting how u calculated the volume. Actually the mass m is the total mass of the system. If I take an elementary strip at a distance of R from the axis having a width of dR , then the volume of that elementary strip could be : ((π/2* √(r^2-R^2))+(2l√(r^2 -R^2))). Please clarify. PS: I have recently started using mathstack. So bear with my editing. – Manoj Dash Jan 28 '16 at 05:37
  • @ManojDash The cylinder-alone volume is the area of its base, a circle of radius $r$, times its height (or length) $l$, so that's $\pi r^2\times l$. Add on the hemisphere's volume, which is half a sphere's volume, or $\frac 23\pi r^3$. Without hemisphere, a volume element at distance $R$ from the axis would just be $2\pi R\times l\times dR$, right? But $l$'s not enough because the hemisphere's on top. And at distance $R$ from its center, the extra needed is just $\sqrt{r^2-R^2}$. – John Forkosh Jan 29 '16 at 07:05
  • @ManojDash MI required about axis of cylinder,it should be mentioned. Please check your work, physical dimension should tally as $ ML^2 $ When $l=0$ standard value expected for hemisphere part is $mr^2/5$ – Narasimham Feb 19 '17 at 05:03
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Here is a simple geometrical approach,

Let $\rho$ be the density of the material then the total mass of the composite solid $$m=\text{(total volume)}\times (\text{density})=\rho\left(\frac{2\pi}{3}r^3+\pi r^2 l\right)$$ $$\rho=\frac{3m}{\pi r^2(2r+3l)}$$

the mass of the sphere $$m_s=\rho \cdot \frac{2\pi}{3}r^3=\frac{3m}{\pi r^2(2r+3l)}\frac{2\pi}{3}r^3=\frac{2mr}{2r+3l}$$ The moment of inertia of hemisphere about the vertical axis passing through the center $$I_s=\frac12\left(\frac 25m_sr^2\right)=\frac 15\frac{2mr}{2r+3l}r^2=\frac{2mr^3}{5(2r+3l)}$$ the mass of the cylinder $$m_c=\rho \cdot \pi r^2l=\frac{3m}{\pi r^2(2r+3l)}\pi r^2 l=\frac{3ml}{2r+3l}$$ The moment of inertia of cylinder about the axis $$I_c=\frac 12m_cr^2=\frac 12\frac{3ml}{2r+3l}r^2=\frac{3mr^2l}{2(2r+3l)}$$ hence, the total moment of inertia of the composite solid about the axis of cylinder
$$I=I_s+I_c=\frac{2mr^3}{5(2r+3l)}+\frac{3mr^2l}{2(2r+3l)}$$ $$=\frac{mr^2}{2r+3l}\left(\frac {2r}5+\frac {3l}2\right)$$ I hope you can simplify the above expression.

Harish Chandra Rajpoot
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