Question

A particle is moving with the constant angular acceleration of 4 rad/s2 in a circular path. At time t = 0, the particle was at rest. Find the time at which the magnitudes of centripetal acceleration and tangential acceleration are equal.

Answer 1

Please find below your answer

Answer 2

Answer:

At 0.5 sec, the magnitudes of centripetal acceleration and tangential acceleration are equal.

Explanation:

Given that,

Angular acceleration $\alpha=4\ rad/s^2$

The centripetal acceleration is defined as,

$a = \dfrac{v^2}{r}$

Where, a = acceleration

v = angular velocity

r = radius

Using equation of motion

$\omega=\omega_{0}+\alpha t$....(I)

Put the value of acceleration in equation (I)

$\dfrac{v}{r} =4t$

$v=4rt$

Now,

Tangential acceleration = Centripetal acceleration

$\alpha r= a$.....(II)

Put the value of $\alpha$ in equation (II)

$4 r=\dfrac{v^2}{r}$....(III)

Now,put the value of v in equation (III)

$4r=\dfrac{16r^2t^2}{r}$

$4=16t^2$

$t^2=\dfrac{1}{4}$

$t=\dfrac{1}{2}$

$t =0.5\ sec$

Hence, At 0.5 sec, the magnitudes of centripetal acceleration and tangential acceleration are equal.