Question

A particle is moving with a constant angular acceleration of 4rad//s^(2) in a circular path. At time t=0 , particle was at rest. Find the time at which the magnitudes of centripetal acceleration and tangential acceleration are equal.

Answer 1

A particle is moving with a constant angular acceleration of 4rad//s^(2) in a circular path

The initial angular velocity ωi=0 at time t=0

Final angular velocity ωf at time t is given as

ωf= ωi+ αt , where α is the angular acceleration

ωf=0 +αt

Radial acceleration = ω^2/ r, where r is the radius of the circle

Radial acceleration= (αt)^2

Tangential acceleration = αr

when radial acceleration is equal to tangential acceleration

(αt)^2 = αr

t= $1/\sqrt{\alpha }$ =$1/\sqrt{4}$=0.5s

Answer 2

Given:

Angular acceleration $\alpha = 4$ $\frac{rad}{s^{2} }$

To Find:

The time at which the magnitudes of centripetal acceleration and tangential acceleration.

We know that,

Centripetal acceleration is given by,

 $a_{c} = r \omega ^{2}$ but $\omega = \alpha t$

 $a_{c} = r (\alpha t ) ^{2}$

Tangential acceleration is given by,

 $a_{t} = r\alpha$

At t = 0 tangential acceleration and centripetal acceleration are equal,

 $a_{t} = a_{c}$

$r\alpha = r\alpha ^{2} t^{2}$

  $t = \frac{1}{\sqrt{\alpha } }$

 $t = \frac{1}{\sqrt{4} }$

 $t = 0.5$ sec

Therefore, the time at which the magnitudes of centripetal acceleration and tangential acceleration are equal is 0.5 sec