How to Analyze the FBD of a Toy Car on the Underside of a Dome?

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The discussion focuses on analyzing the free body diagram (FBD) of a toy car on the underside of a dome, emphasizing the forces acting on the car. Key forces include weight acting downward, the normal force directed toward the center of the dome, and the need for clarification on the frictional force's direction. Participants seek a labeled diagram to better visualize the scenario, questioning the clarity of the original problem statement regarding the car's motion and position. The conversation highlights the importance of precise definitions and diagrams in physics problems. Overall, a clear understanding of the forces involved is crucial for accurate analysis.
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Homework Statement
Draw the FBD including Kinetic Friction, Gravity, and Normal force
Relevant Equations
Fnet = mv^2/r
weight pointing down
Fn pointing down and to the left (to the center of the dome)
Fs pointing ..?? Should it not be off the page? Maybe up and off the page to oppose movement + gravity?
 
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bobbeh said:
Homework Statement: Draw the FBD including Kinetic Friction, Gravity, and Normal force
Relevant Equations: Fnet = mv^2/r

weight pointing down
Fn pointing down and to the left (to the center of the dome)
Fs pointing ..?? Should it not be off the page? Maybe up and off the page to oppose movement + gravity?
Can you post a labeled diagram of your "vertically horizontal circle on the underside of a dorm" please?
 
Hi @bobbeh and welcome to PF.

The title:
"FBD of a toy car moving in a vertically horizontal circle on the underside of a dome"
is not clear.

Is the plane in which the circle lies horizontal or vertical (or something else)?
Is the car'speed constant?
Are the car's wheel rolling without slipping?
At what position is the car on the dome?

You need to provide the complete and exact original question, word-for-word (and as already noted by @renormalize, a diagram).
 
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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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