How is centripetal force involved here? (charged mass sliding down a conducting hemisphere)

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Homework Help Overview

The discussion revolves around a block sliding down a conducting hemisphere, with a focus on the role of centripetal force and the conditions under which the block loses contact with the surface. The problem involves forces acting on a charged mass and the dynamics of motion along a curved path.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the forces acting on the block, including gravitational and normal forces, and question how these relate to the block losing contact with the hemisphere. There are attempts to derive expressions for the normal force and to understand the conditions under which it becomes zero. Some participants express confusion about the physical interpretation of these forces and their contributions to centripetal motion.

Discussion Status

The discussion is active, with participants raising various questions about the relationships between forces and motion. Some guidance has been offered regarding the need to analyze free body diagrams and the conditions for the normal force to equal zero. However, there is no explicit consensus on the interpretations or the mathematical relationships being explored.

Contextual Notes

Participants note that the block is charged, which adds complexity to the problem. There is an emphasis on the need for a clear statement of the problem to facilitate better understanding and analysis. The discussion also highlights the importance of considering both vertical and horizontal components of forces in the context of circular motion.

  • #31
haruspex said:
It can be surprising, though. On the thread about a planet in an elliptical orbit it struck me that the centripetal force is only towards the star at the apogee and perigee. At other points the gravitational attraction has a tangential component.
Sure. This is similar to 2D projectile motion near the surface of the Earth. There is always a centripetal component that becomes equal to the weight at the top of the trajectory.
 
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  • #32
haruspex said:
It often happens that there is an analytic way to find velocity as a function of position (e.g. by energy conservation) but not as a function of time. In such cases, the substitution ##\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v## can be useful.

To be exact, the component of that vector sum normal to the velocity is the centripetal force.
I see... Thankyou :)
 

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