A particle moving along a circle has velocity v - 2 - - 3 - - 5 k- m-s and angular velocity - - 2 - - 2 - - 2k- at a particular instant- The magnitude of centripetal acceleration of the particle is

Centripetal acceleration nbsp; a_C = rw^2 = v w = √(2^2 + 3^2 + 5^2)×√(2^2 + 2^2 + ( 2)^2) = (√(4 + 9 + 25)) (√(4 + 4 +4)) = ( √(38))(√(12)) = √(456) = 2√(11 ...

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Standard XIII

Physics

Centripetal Acceleration

A particle mo...

Question

A particle moving along a circle has velocity $\to v = 2^i + 3^j + 5^k m / s$ and angular velocity $\to ω = 2^i + 2^j - 2^k$ , at a particular instant. The magnitude of centripetal acceleration of the particle is

2 \sqrt{114} m / s^{2}

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0 m m / s^{2}

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3 \sqrt{114} m / s^{2}

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4 \sqrt{114} m / s^{2}

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Solution

The correct option is A $2 \sqrt{114} m / s^{2}$
Centripetal acceleration
$a_{C} = r w^{2} = v w$
$= \sqrt{2^{2} + 3^{2} + 5^{2}} \times \sqrt{2^{2} + 2^{2} + (- 2)^{2}}$
$= (\sqrt{4 + 9 + 25}) (\sqrt{4 + 4 + 4}) = (\sqrt{38}) (\sqrt{12})$
$= \sqrt{456} = 2 \sqrt{114} m / s^{2}$

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A particle moving along a circle has velocity →v=2^i+3^j+5^k m/s and angular velocity →ω=2^i+2^j−2^k, at a particular instant. The magnitude of centripetal acceleration of the particle is

The correct option is A 2√114 m/s2Centripetal acceleration aC=rw2=vw =√22+32+52×√22+22+(−2)2 =(√4+9+25)(√4+4+4)=(√38)(√12) =√456=2√114 m/s2

A particle moving along a circle has velocity $\to v = 2^i + 3^j + 5^k m / s$ and angular velocity $\to ω = 2^i + 2^j - 2^k$ , at a particular instant. The magnitude of centripetal acceleration of the particle is

The correct option is A $2 \sqrt{114} m / s^{2}$
Centripetal acceleration
$a_{C} = r w^{2} = v w$
$= \sqrt{2^{2} + 3^{2} + 5^{2}} \times \sqrt{2^{2} + 2^{2} + (- 2)^{2}}$
$= (\sqrt{4 + 9 + 25}) (\sqrt{4 + 4 + 4}) = (\sqrt{38}) (\sqrt{12})$
$= \sqrt{456} = 2 \sqrt{114} m / s^{2}$