A particle moving along a circle has velocity v -2 -3 -5 k- m - s and angular velocity -2 -2 -2 k- at a particular instant- Themagnitude of centripetal acceleration of the particle is A- 0 mm - s 2B- 2 -114 m - s 2- 3 -114 m - s 2D- 4 -114 m - s 2

The correct option is B 2 √(114) m/s^2Centripetal 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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

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}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0 m m / s^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

Q. 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

Q. The velocity-displacement graph for a car on a straight road is shown in figure

Determine the acceleration of the car at

s = 150 m

Q. A particle is moving along a circular path. The angular velocity, linear velocity, angular acceleration, and centripetal acceleration of the particle at any instant respectively are

\to ω, \to v, \to α

and

_{c}

. Which of the following relations is not correct?

Q. A particle moving in a circle changes its angular velocity uniformly from

20 rad/s

40 rad/s

in first 5 seconds. Find the angular acceleration of the particle.

Q. For a particle moving on a circular path of radius

5 c m

, the velocity at an instant ‘

t

’ is

3^i + 4^j m / s

. Assuming the particle to be in uniform circular motion, the value of centripetal acceleration is

Join BYJU'S Learning Program

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