A 70 ; kg man stands in contact against the i | vedclass

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(C) Mass of the man,$m = 70 \; kg$. Radius of the drum,$r = 3 \; m$. Coefficient of friction,$\mu = 0.15$. Acceleration due to gravity,$g = 10 \; m/s^

Vedclass · Accepted Answer

$4.71$

Vedclass · Answer

(C) Mass of the man,$m = 70 \; kg$. Radius of the drum,$r = 3 \; m$. Coefficient of friction,$\mu = 0.15$. Acceleration due to gravity,$g = 10 \; m/s^2$. For the man to remain stuck to the wall,the frictional force $f$ must balance the weight of the man,$mg$. $f = \mu F_N = mg$,where $F_N$ is the normal force provided by the wall. The normal force provides the necessary centripetal force: $F_N = mr\omega^2$. Substituting $F_N$ into the friction equation: $\mu(mr\omega^2) = mg$. Simplifying for angular velocity $\omega$: $\omega^2 = \frac{g}{\mu r}$. $\omega = \sqrt{\frac{g}{\mu r}} = \sqrt{\frac{10}{0.15 	imes 3}} = \sqrt{\frac{10}{0.45}} = \sqrt{22.22} \approx 4.71 \; rad/s$.

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