Centre Of Mass Of Hollow Cone MCQ - Practice Questions & Answers

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A hollow hemisphere and a hollow cone of same mass are arranged as shown in figure. find the position of center of mass from center of hemisphere

$\frac{11}{13} R$

$\frac{14}{9} R$

$\frac{11}{12} R$

$\frac{5}{7} R$

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A hollow body as shown in figure consist of right circular portion attached to a hemisphere portion of Radius R. Determine the height H of cone if the centre of mass of the composite body considers with the centre O of the circular base (Take $R = \sqrt{x} H_{)}$ )

$\frac{1 - \sqrt{37}}{18}$

$\frac{1 + \sqrt{37}}{18}$

$\frac{1 + 3 \sqrt{2}}{5}$

$\frac{3 \sqrt{2} - 1}{5}$

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Position of centre of mass for Hollow Cone

Have a look at the figure of Hollow Cone

Since it is symmetrical about the y-axis

So we can say that its $x_{c m} = 0$ and $z_{c m} = 0$
Now we will calculate its $y_{cm}$ which is given by

$y_{cm} = \frac{\int y \cdot d m}{\int d m}$

So Take a small elemental ring of mass dm of radius r at a vertical distance y from O as shown in figure.

And $r = x \sin θ$ , and $y = x \cos θ$
Since our element mass is ring so its C.O.M will lie on the $y$ -axis.
Now $d m = σ d A = σ (2 π x \sin θ) d x$

Where

$σ = \frac{M}{π R * \sqrt{R^{2} + H^{2}}}$

$d m = \frac{2 M x d x}{R^{2} + H^{2}}$

$y_{c m} = \frac{1}{M} \int y d m = \frac{1}{M} \int_{0}^{\sqrt{R^{2} + H^{2}}} x \cos θ * \frac{2 M x d x}{R^{2} + H^{2}} = \frac{2 H}{3}$

${So}_{cm} = \frac{2 H}{3} from o.$

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