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Rotation Concept

If the circul...

Question

If the circular sector of radius $R$ (lamina) is uniform and subtends an angle $θ$ at the centre of curvature and it is placed symmetrical with x-axis, its CM (centre of mass) will be situated on x-axis from $O$ at a distance equals to :

\frac{2 R sin (\frac{θ}{2})}{3 θ}

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\frac{3 R sin θ}{(\frac{3 θ}{2})}

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\frac{4 R sin (\frac{θ}{2})}{3 θ}

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\frac{3 R sin (\frac{θ}{2})}{4 θ}

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Solution

The correct option is A $\frac{2 R sin (\frac{θ}{2})}{3 θ}$

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Similar questions

θ

\to a

\to b

sin (\frac{θ}{2})

\frac{x^{2}}{4} + \frac{y^{2}}{2} = 1

\frac{π}{4}

| O P | = r .

\frac{1 + {tan}^{2} θ}{1 + {cot}^{2} θ} = {(\frac{1 - tan θ}{1 - cot θ})}^{2}

θ

f (x)

y = f (x)

x = \frac{π}{4}

x = β > \frac{π}{4}

(β sin β + \frac{π}{4} cos β + \sqrt{2} β - \frac{π}{2})

f (\frac{π}{2})

{⎛ ⎜ ⎜ ⎜ ⎝ \frac{cos (\frac{π}{2} + x) + sin (\frac{π}{2} + x)}{cos (\frac{π}{2} - x) - sin (\frac{π}{2} - x)} ⎞ ⎟ ⎟ ⎟ ⎠}^{2}

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