The normal force exerted on a small sphere

AI Thread Summary
To find the normal force exerted on a small sphere rolling inside a hemispherical bowl, the approach involves using conservation of energy to determine the sphere's velocity at the bottom. The sphere starts from rest at the rim, and both translational and rotational energy must be considered. After calculating the velocity, the total force equation F_total = ma can be applied to find the normal force, where acceleration is determined at the point of contact. The discussion highlights the importance of recognizing the relationship between energy conservation and force calculations in this context. Understanding these principles is crucial for solving similar physics problems effectively.
thedaydreamer
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Homework Statement


A small sphere A of radius r and mass m rolls without slipping inside a large hemispherical bowl of radius R. The sphere A starts from rest at the top point on the rim of the hemisphere. Find the normal force exerted by small sphere when it touches the bottom of the hemisphere.




2. The attempt at a solution I'm unsure of how to approach this problem. I tried to use Conservation of Energy to find the velocity of the sphere but I don't know how that will help. I tried to use the fact that the relative acceleration to the inner surface of the bowl at the point of contact=0. This way I can find the frictional force but not the maximum normal force.
 
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hi thedaydreamer! :smile:
thedaydreamer said:
I tried to use Conservation of Energy to find the velocity of the sphere but I don't know how that will help. I tried to use the fact that the relative acceleration to the inner surface of the bowl at the point of contact=0. This way I can find the frictional force but not the maximum normal force.

yes, use conservation of energy (including rotational energy) to find the speed, then use Ftotal = ma to find N, where a is the https://www.physicsforums.com/library.php?do=view_item&itemid=27" :wink:
 
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Thank you tiny-tim. I should've noticed that myself. :)
 
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