Calculating the Height of a Frictionless Loop Launch

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SUMMARY

The discussion focuses on calculating the height from which a 75 kg ball bearing was released to successfully navigate a frictionless loop of 75 cm height while traveling at 1.9 m/s. The relevant energy equation is ΔE = ΔKv + ΔKr + ΔEtherm + ΔUpot, where ΔEtherm is negligible due to the absence of friction. By substituting the known values into the energy equations, participants can solve for the unknown height, leading to a clear understanding of the problem-solving approach in physics.

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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
 

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