Rotating surface/friction question

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The discussion revolves around understanding the transition from kinetic slipping to static sticking on a rotating surface. The key point is determining the direction of friction, with the user contemplating whether it is from outside to inside or vice versa. The boundary condition for slipping is established by the inequality μ*m*g > m*ω²*r, indicating that slipping occurs when the radius exceeds μ*g/ω². This suggests that if the radius is larger than this threshold, the object will slip outward. The conclusion emphasizes the importance of correctly identifying the frictional direction in relation to the radius and angular velocity.
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Homework Statement


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The Attempt at a Solution


Hey guys, I understand how to find the static friction point (equate mv2/r to max static friction). However, I'm confused as to which direction (in to out or out to in) is the correct one ie. from kinetic slipping to 'sticking'. At the moment I'm thinking it's outside in (Jo's answer) but I can't justify it properly. Thanks!
 
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You've determined the boundary where it starts to slip. And the boundary is dependent on

μ*m*g > m*ω²*r

or

r < μ*g/ω²

This means that outside this distance it slips doesn't it? If the aim is to have it slip until it grabs then ...
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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