Calculating minimum velocity around a loop

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SUMMARY

The discussion focuses on calculating the minimum velocity required for an object to successfully navigate a loop while counteracting gravitational forces. The key concept involves centripetal acceleration, which is determined by the radius of the loop (R) and the angular velocity (ω). At the top of the loop, the acceleration must remain non-negative to ensure the object does not fall off. The position, velocity, and acceleration vectors are defined mathematically, providing a clear framework for solving the problem.

PREREQUISITES
  • Understanding of centripetal acceleration
  • Familiarity with angular velocity (ω)
  • Knowledge of vector mathematics
  • Basic principles of gravitational force
NEXT STEPS
  • Study the equations of motion for circular motion
  • Learn about the relationship between angular velocity and linear velocity
  • Explore the concept of forces acting on objects in circular motion
  • Investigate the effects of varying loop radius on minimum velocity
USEFUL FOR

Physics students, educators, and anyone interested in understanding dynamics of motion in circular paths, particularly in the context of gravitational effects on objects in loops.

Nouo
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So I had this question for physics about something going around a loop. The question asked what the minimum velocity to counteract gravity is if something is going around a loop. hint: One force will be zero.

I'm not sure how to figure this out, can someone help?
 
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Can you calculate the centripetal acceleration of a particle moving at some velocity around the loop? You can assume the loop is of radius r and treat it as a constant.
 
If an object is going around a loop of radius R at constant angular velocity, [itex]\omega[/itex] then its position vector (taking center of the loop to be (0, 0) and the objects position at t= 0 to be (1, 0)) is [itex]R cos(\omega t)\vec{i}+ R sin(\omega t)\vec{j}[/itex]. Its velocity vector will be [itex]-\omega R sin(\omega t)\vec{i}+ \omega R cos(\omega t)\vec{j}[/itex] and its acceleration vector will be [itex]-\omega^2 R cos(\omega t)\vec{i}- \omega^2 R sin(\omega t)\vec{j}[/itex]. In order to stay on the loop, the acceleration at the top ([itex]\theta= \pi/4[/itex]) must be non-negative.
 

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