Conservation of Energy, Momentum

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SUMMARY

The discussion focuses on the conservation of energy and momentum in a physics problem involving a bullet and a pendulum bob. The bullet, with a mass of 0.010 kg, passes through a pendulum bob of mass 1.2 kg and emerges at half its initial speed. The minimum speed required for the pendulum bob to reach the top of its arc is derived from the principles of energy conservation. The force in the rod after the bullet emerges is calculated to be 58.8 N, which accounts for both the centripetal force and the gravitational force acting on the pendulum bob.

PREREQUISITES
  • Understanding of conservation of momentum
  • Knowledge of centripetal force equations (Fr = mv²/r)
  • Familiarity with gravitational force calculations (mg)
  • Basic principles of energy conservation in mechanical systems
NEXT STEPS
  • Study the conservation of momentum in elastic and inelastic collisions
  • Learn about energy conservation in pendulum motion
  • Explore the relationship between tension, centripetal force, and gravitational force in circular motion
  • Investigate the dynamics of forces acting on objects in motion through various mediums
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and tutors looking to enhance their understanding of energy and momentum conservation principles.

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Homework Statement


A bullet of mass m = .010 kg and speed v passes completely through a pendulum bob of mass M = 1.2 kg. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a rigid rod of length l = 0.50 m and negligible mass that can pivot about the center point.


What minimum speed vp must the pendulum bob have to just make it to the top of the arc at point A (top of the circle, pendulum starting at the bottom)?

What is the speed of the bullet just before hitting the pendulum bob for the situation described in part a?

What is the magnitude of the force in the rod just after the bullet emerges?



Homework Equations



Fr = mv2/r

The Attempt at a Solution


I solved for parts a and b already, and I also got C (answer is 58.8N), but I'm having trouble understanding why this number is the answer. At the bottom of the circle would the magnitude of the force in the rod be:

Fr + mg?

that gets the answer, but aren't these forces going in different directions at the bottom of the circle?
 
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Fr = mv2/r is the centripetal force required to keep the bob on the string. Since tension must provide this centripetal force and counteract gravity, T=mv^2/r + mg.
 
ideasrule said:
Fr = mv2/r is the centripetal force required to keep the bob on the string. Since tension must provide this centripetal force and counteract gravity, T=mv^2/r + mg.

if it was counteracting gravity, doesn't that mean it would be - mg? I don't think I'm quite understanding here.
 

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