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  • Moment of inertia (in MR2)of hollow cylinder mass M, Length R and Radius R about C.G and perpendicular to its own axis is (Give your answer till 2 places after the decimal point)</p

Moment of inertia (in MR2)of hollow cylinder mass M, Length R and Radius R about C.G and perpendicular to its own axis is (Give your answer till 2 places after the decimal point)

Option: 1

0.25


Option: 2

0.5


Option: 3

0.58


Option: 4

0.4


Answers (1)

best_answer

Consider a hollow cylinder of mass M, length 'l ' and radius 'r' capable of rotating about its geometrical axis. Let m be its mass per unit length.

\mathrm{m}=\mathrm{M} / l \quad \text { Hence } \mathrm{M}=\mathrm{m} \cdot l

A hollow cylinder can be regarded as a number of thin uniform rings of infinitesimal thickness piled on top of one another. Let us consider one such ring of thickness 'dx' at a distance of 'x' from the centre C of the cylinder.

\begin{aligned} &\text { Mass of such ring is given by }\\ &\text { Mass, } \mathrm{dm}=\mathrm{m} \cdot \mathrm{dx}=(\mathrm{M} / l) \mathrm{dx} \end{aligned}

\\ \text{The M.I. of such ring about a transverse axis (passing through C) i given by} \\ \mathrm{dI}=\mathrm{dm} \cdot \mathrm{R}^{2} \\ \text{The M.I. of a ring about diameter is given by} \\ \mathrm{dI}=\frac{\mathrm{dm} \cdot \mathrm{R}^{2}}{2}=\left(\frac{M}{l}\right) d x \cdot \frac{\mathrm{R}^{2}}{2} \\ \text{By parallel axes theorem} \\ \begin{array}{c} \mathrm{dI}_{\mathrm{c}}=\mathrm{dI}_{\mathrm{G}}+\mathrm{dm} \cdot \mathrm{x}^{2} \\ \mathrm{d} \mathrm{I}_{\mathrm{c}}=\left(\frac{M}{l}\right) d x \cdot \frac{\mathrm{R}^{2}}{2}+\left(\frac{M}{l}\right) d x \cdot \mathrm{x}^{2} \end{array}

\begin{aligned} &\text { Integrating the above expression in limits }\\ &\int \mathrm{d} \mathrm{I}_{\mathrm{c}}=\int_{-\frac{l}{2}}^{+\frac{l}{2}}\left(\frac{M}{l}\right) d x \cdot \frac{\mathrm{R}^{2}}{2}+\int_{-\frac{l}{2}}^{-\frac{l}{2}}\left(\frac{M}{l}\right) d x \cdot \mathrm{x}^{2}\\ \\ &\therefore \mathrm{I}_{\mathrm{C}}=\frac{M \mathrm{R}^{2}}{2 l} \int_{-\frac{l}{2}}^{+\frac{l}{2}} d x \cdot+\frac{M}{l} \int_{-\frac{l}{2}}^{+\frac{l}{2}} \mathrm{x}^{2} \cdot d x\\ \\ &\therefore \mathrm{I}_{\mathrm{c}}=\frac{M \mathrm{R}^{2}}{2 l}[x]_{-\frac{l}{2}}^{+\frac{l}{2}}+\frac{M}{l}\left[\frac{\mathrm{x}^{3}}{3}\right]_{-\frac{l}{2}}^{+\frac{l}{2}}\\ \\ &\therefore \mathrm{I}_{\mathrm{c}}=\frac{M \mathrm{R}^{2}}{2 l}\left[+\frac{l}{2}-\left(-\frac{l}{2}\right)\right]+\frac{M}{3 l}\left[\left(+\frac{l}{2}\right)^{3}-\left(-\frac{l}{2}\right)^{3}\right] \end{aligned}

\therefore \mathrm{I}_{\mathrm{c}}=\frac{M \mathrm{R}^{2}}{2 l}\left[\frac{l}{2}+\frac{l}{2}\right]+\frac{M}{3 l}\left[\frac{l^{3}}{8}+\frac{l^{3}}{8}\right]

\therefore I_{c}=\frac{M R^{2}}{2 l}\left[\frac{2 l}{2}\right]+\frac{M}{3 l}\left[\frac{2 l^{3}}{8}\right]

\therefore \mathrm{I}_{\mathrm{c}}=\frac{M \mathrm{R}^{2}}{2 l}[l]+\frac{M}{3 l}\left[\frac{l^{3}}{4}\right]

\therefore I_{c}=\frac{M R^{2}}{2}+\frac{M l^{2}}{12}

Given, l=R

So, \therefore I_{c}=\frac{7M R^{2}}{12}

So the answer is 0.58

 

Posted by

Anam Khan

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