Centripetal Motion Homework: Min Coeff of Friction

AI Thread Summary
To determine the minimum coefficient of static friction required for a rider to remain in place on a rotating cylindrical ride, the centripetal force must equal the frictional force. Given a radius of 3.30 m, a speed of 10.0 m/s, and a rider mass of 55 kg, the centripetal force is calculated as 1667 N. The normal force is equivalent to the weight of the rider, which is 539.5 N (mass times gravity). The coefficient of friction is then found by dividing the centripetal force by the normal force, leading to a value of approximately 3.09, indicating the need for a significant frictional force to prevent falling. Accurate calculations must consider both centripetal and gravitational forces to ensure the rider remains secured against the wall.
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Homework Statement



At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders "plastered against the wall. Suppose the radius of the room is 3.30 m and the speed of the wall is 10.0 m/s when the floor falls away. What is the minimum coefficient of static friction that must exist between a rider's back and the wall, if the rider is to remain in place when the floor drops away if the rider has a mass of 55 kg?


Homework Equations



Centripetal Force = mv^2/r = 1667

friction = u*Fn


The Attempt at a Solution



centripetal force = friction = coefficient of friction * Normal Force

Normal force = mg (weight)

coefficient of friction = centripetal force/Normal Force = 51.5?

Is this correct?
 
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You're missing a force. Friction will be resisting the motion of the person from falling, but Gravity is the one doing the pulling, you need to consider that in your force calculations
 
centripetal force = friction = coefficient of friction * Normal Force
That is incorrect. The centripetal force is equal to the normal force, not the frictional force.
Normal force = mg (weight)
This too is incorrect. The weight of the person should be equal to the frictional force.
Now try finding the coefficient of static friction.
 
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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