Calculating the Height of a Frictionless Loop Launch

AI Thread Summary
To determine the height from which a 75 kg ball bearing was released before entering a 75 cm high frictionless loop, energy conservation principles are applied. The initial potential energy is converted into kinetic energy as the ball descends, with the equation ΔE = ΔKv + ΔKr + ΔUpot being relevant. Since the loop is frictionless, the rotational energy can be neglected, simplifying the calculations. The ball's initial velocity of 1.9 m/s and the gravitational force are key factors in solving for the unknown height. This approach effectively leads to the solution for the release height.
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At the top of a "frictionless" loop that is 75 cm high, a 75 Kg ball bearing is traveling at 1.9m/s. From what height was the ball released?


I have absolutely no Idea where to start here.

Diameter= 0.075m
Velocity=1.9m/s
Force related to Gravity=0.74N
m=0.075Kg
 
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If you are using energy like your tag suggests then write out your energy equation:

ΔE = ΔKv + ΔKr + ΔEtherm + ΔUpot

Etherm can be neglected because the loop is frictionless. Substitute in your energy equations for your change in kinetic energy (HINT ball is assumed to start from rest unless the question otherwise states), the change in kinetic rotational energy (HINT that of a point mass) and your change in potential energy (which will have an unknown value). Solve for unknown and you should get your answer...
 


That makes much more sense, THANK YOU
 
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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