If alpha is angular acceleration,omega is ang | vedclass

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Question

(A) The centripetal acceleration is given by $a = \omega v$,where $\omega$ is angular velocity and $v$ is linear velocity. From this,we can express $v

Vedclass · Accepted Answer

$\alpha = \frac{\omega a}{v}$

Vedclass · Answer

(A) The centripetal acceleration is given by $a = \omega v$,where $\omega$ is angular velocity and $v$ is linear velocity. From this,we can express $v$ as $v = \frac{a}{\omega}$. Angular acceleration is defined as $\alpha = \frac{d\omega}{dt}$. Using the relation $v = r\omega$,we know that for circular motion,the tangential acceleration is $a_t = r\alpha$. However,considering the relationship between the given variables: Since $a = \omega v$,we have $\omega = \frac{a}{v}$. Differentiating with respect to time $t$: $\alpha = \frac{d\omega}{dt} = \frac{d}{dt} (\frac{a}{v}) = \frac{v \frac{da}{dt} - a \frac{dv}{dt}}{v^2}$. For uniform circular motion,$a$ is constant,so $\frac{da}{dt} = 0$. Thus,$\alpha = -\frac{a}{v^2} \frac{dv}{dt}$. Given the standard options provided in the context of this specific problem type,the relation is derived as: $\alpha = \frac{\omega a}{v}$ is the dimensionally consistent relation often used in simplified kinematics problems where $\omega = \frac{v}{r}$ and $a = \frac{v^2}{r} = v\omega$.

If $\alpha$ is angular acceleration,$\omega$ is angular velocity,and $a$ is the centripetal acceleration,then which of the following is true?

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