JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Conservation of Momentum

The position of the centre of mass of a uniform semi - circular wire of radius ' R ' placed in

x - y

plane with its centre at the origin and the line joining its ends as

x

-axis is given by

(0, \frac{x R}{π})

. Then, the value of

∣ x ∣

is ______.

Answer: 2

Solution

For a uniform semi-circular wire, the center of mass lies on the axis of symmetry (y-axis in this case). By symmetry, the x-coordinate is 0. The y-coordinate is found by integration:

y_{c m} = \frac{\int y d m}{\int d m}

. For a semi-circle of radius R, using

y = R sin θ

and arc element

d s = R d θ

from

0

π

y_{c m} = \frac{\int _{0}^{π} R s i n θ \cdot R d θ}{\int _{0}^{π} R d θ} = \frac{R ^{2} [ - c o s θ ] _{0}^{π}}{R π} = \frac{R ^{2} \cdot 2}{R π} = \frac{2 R}{π}

. Therefore

x = 2

and

∣ x ∣ = 2

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Conservation of Momentum

Solution

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