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Question

(C) The child starts from rest with constant tangential acceleration $a$. After time $t$,the tangential velocity is $v = at$.The radial (centripetal)

Vedclass · Accepted Answer

$\frac{[a^{4} t^{4}+a^{2} r^{2}]^{1/2}}{g r}$

Vedclass · Answer

(C) The child starts from rest with constant tangential acceleration $a$. After time $t$,the tangential velocity is $v = at$.The radial (centripetal) acceleration is $a_r = \frac{v^2}{r} = \frac{(at)^2}{r} = \frac{a^2 t^2}{r}$.The total acceleration $a_{net}$ experienced by the child is the vector sum of tangential and radial accelerations: $a_{net} = \sqrt{a_t^2 + a_r^2} = \sqrt{a^2 + \left(\frac{a^2 t^2}{r}\right)^2} = \sqrt{a^2 + \frac{a^4 t^4}{r^2}}$.Slipping starts when the required frictional force equals the limiting friction,i.e.,$F_{net} = m a_{net} = \mu m g$.Thus,$\mu g = \sqrt{a^2 + \frac{a^4 t^4}{r^2}} = \sqrt{\frac{a^2 r^2 + a^4 t^4}{r^2}} = \frac{1}{r} \sqrt{a^2 r^2 + a^4 t^4}$.Therefore,$\mu = \frac{[a^4 t^4 + a^2 r^2]^{1/2}}{r g}$.

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