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(A) The boy is in circular motion on the platform. The necessary centripetal force is provided by the static friction force between the boy and the pl

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(A) The boy is in circular motion on the platform. The necessary centripetal force is provided by the static friction force between the boy and the platform.For the boy to just slip,the maximum static friction must equal the required centripetal force:$f_{max} = F_c$$\mu N = m \omega^2 r$Since the platform is horizontal,the normal force $N = mg$.Substituting this into the equation:$\mu mg = m \omega^2 r$$\mu = \frac{\omega^2 r}{g}$Given values: $\omega = 1 \, rad/s$,$r = 5 \, m$,and $g = 10 \, m/s^2$.$\mu = \frac{(1)^2 	imes 5}{10} = \frac{5}{10} = 0.5$Therefore,the coefficient of friction is $0.5$.

$A$ boy is sitting on the horizontal platform of a joy wheel at a distance of $5 \, m$ from the center. The wheel begins to rotate and when the angular speed exceeds $1 \, rad/s$,the boy just slips. Find the coefficient of friction between the boy and the wheel. (Take $g = 10 \, m/s^2$)

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