A particle is moving in a circular path and its acceleration vector is making an angle of 30° with the velocity vector, then the ratio of centripetal acceleration to its tangential acceleration is

Question

A particle is moving in a circular path and its acceleration vector is making an angle of 30° with the velocity vector, then the ratio of centripetal acceleration to its tangential acceleration is (A) (1/2) (B) (√3/2) (C) (1/√3) (D) √3

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-kmarbe31ziwd.png" /> For a body undergoing circular motion, let net acceleration makes an angle θ with the tangent, then tan θ =(ac/at) ∴ (ac/at)=tan 30° =(1/√3)

Q. A particle is moving in a circular path and its acceleration vector is making an angle of $3 0^{\circ}$ with the velocity vector, then the ratio of centripetal acceleration to its tangential acceleration is

$\frac{1}{2}$

$\frac{3}{2}$

$\frac{1}{3}$

$3$

For a body undergoing circular motion, let net acceleration makes an angle $θ$ with the tangent, then
$t an θ = \frac{a _{c}}{a _{t}}$
$∴ \frac{a _{c}}{a _{t}} = t an 3 0^{\circ} = \frac{1}{3}$

Q. A particle is moving in a circular path and its acceleration vector is making an angle of 30∘ with the velocity vector, then the ratio of centripetal acceleration to its tangential acceleration is

21​

23​​

3​1​

3​

For a body undergoing circular motion, let net acceleration makes an angle θ with the tangent, then tanθ=at​ac​​ ∴at​ac​​=tan30∘=3​1​

Q. A particle is moving in a circular path and its acceleration vector is making an angle of $3 0^{\circ}$ with the velocity vector, then the ratio of centripetal acceleration to its tangential acceleration is

$\frac{1}{2}$

$\frac{3}{2}$

$\frac{1}{3}$

$3$

For a body undergoing circular motion, let net acceleration makes an angle $θ$ with the tangent, then
$t an θ = \frac{a _{c}}{a _{t}}$
$∴ \frac{a _{c}}{a _{t}} = t an 3 0^{\circ} = \frac{1}{3}$