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(C) For the child not to slip, the required centripetal force must be provided by the static friction force.At the point of slipping, the centripetal

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) For the child not to slip, the required centripetal force must be provided by the static friction force.At the point of slipping, the centripetal force equals the maximum static friction force:$m \omega^2 r = f_{s, 	ext{max}}$Since $f_{s, 	ext{max}} = \mu N$ and the normal force $N = mg$, we have:$m \omega^2 r = \mu mg$$\omega^2 = \frac{\mu g}{r}$Given values: $\mu = 0.8$, $g = 10 \ m/s^2$, and $r = 2 \ m$.Substituting these values:$\omega^2 = \frac{0.8 	imes 10}{2} = \frac{8}{2} = 4$$\omega = \sqrt{4} = 2 \ rad/s$Thus, the maximum angular velocity is $2 \ rad/s$.

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