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

Click For Summary
SUMMARY

The discussion focuses on the mechanics of a charged block sliding down a conducting hemisphere, specifically analyzing the conditions under which the block loses contact with the surface. The key conclusion is that when the normal force equals zero, the sum of the vertical forces (gravity and electrostatic force) must equal the centripetal force required for circular motion. The participants emphasize the importance of understanding the forces acting on the block, including gravitational force, normal force, and electrostatic force, to derive the correct equations of motion.

PREREQUISITES
  • Understanding of centripetal force and its role in circular motion.
  • Familiarity with free body diagrams (FBD) for analyzing forces.
  • Knowledge of Newton's laws of motion, particularly in non-uniform circular motion.
  • Basic principles of electrostatics, including the effects of electric forces on charged objects.
NEXT STEPS
  • Study the derivation of centripetal force equations in non-uniform circular motion.
  • Learn how to construct and analyze free body diagrams for complex systems.
  • Explore the work-energy theorem and its application in variable acceleration scenarios.
  • Investigate the effects of electrostatic forces on charged objects in motion.
USEFUL FOR

Physics students, educators, and anyone interested in classical mechanics, particularly those studying dynamics involving forces and motion in charged systems.

  • #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.
 
Physics news on Phys.org
  • #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 :)
 

Similar threads

Replies
7
Views
2K
Replies
8
Views
1K
Replies
6
Views
4K
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
17
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
4K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K