A man is standing on a rough (mu = 0.5) horiz | vedclass

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Question

(A) To prevent slipping,the required centripetal force must be provided by the static frictional force.The condition for no slipping is: $F_c \leq f_s

Vedclass · Accepted Answer

$R \leq 0.2 \ m$

Vedclass · Answer

(A) To prevent slipping,the required centripetal force must be provided by the static frictional force.The condition for no slipping is: $F_c \leq f_s$Where $F_c = m R \omega^2$ is the centripetal force and $f_s \leq \mu N$ is the maximum static friction.Since the disc is horizontal,the normal force $N = mg$.Therefore,$m R \omega^2 \leq \mu mg$.Canceling mass $m$ from both sides,we get $R \omega^2 \leq \mu g$.Solving for $R$,we get $R \leq \frac{\mu g}{\omega^2}$.Given $\mu = 0.5$,$\omega = 5 \ rad/s$,and taking $g = 10 \ m/s^2$:$R \leq \frac{0.5 	imes 10}{5^2} = \frac{5}{25} = 0.2 \ m$.Thus,the man must stand at a distance $R \leq 0.2 \ m$ from the center.

$A$ man is standing on a rough $(\mu = 0.5)$ horizontal disc rotating with a constant angular velocity of $5 \ rad/s$. At what distance from the center should he stand so that he does not slip on the disc?

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