A particle at rest starts moving with a const | vedclass

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Question

(C) Given: Angular acceleration $\alpha = 4 \ rad/s^2$,initial angular velocity $\omega_0 = 0$.Centripetal acceleration is given by $a_c = r\omega^2$.

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$1/2 \ s$

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(C) Given: Angular acceleration $\alpha = 4 \ rad/s^2$,initial angular velocity $\omega_0 = 0$.Centripetal acceleration is given by $a_c = r\omega^2$.Tangential acceleration is given by $a_t = r\alpha$.We are given that $a_c = a_t$,so $r\omega^2 = r\alpha$.This simplifies to $\omega^2 = \alpha$,which means $\omega = \sqrt{\alpha} = \sqrt{4} = 2 \ rad/s$.Using the kinematic equation for rotational motion,$\omega = \omega_0 + \alpha t$.Substituting the known values: $2 = 0 + 4t$.Solving for $t$: $t = 2/4 = 1/2 \ s$.

$A$ particle at rest starts moving with a constant angular acceleration of $4 \ rad/s^2$ in a circular path. At what time will the magnitude of its centripetal acceleration and tangential acceleration be equal?

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