Conservation of Energy, Momentum

AI Thread Summary
The discussion centers on a physics problem involving the conservation of energy and momentum in a system where a bullet passes through a pendulum bob. The minimum speed required for the pendulum bob to reach the top of its arc is calculated, along with the bullet's speed before impact and the force in the rod after the bullet exits. The force in the rod is determined to be 58.8N, which combines the centripetal force and the gravitational force acting on the bob. There is confusion regarding the direction of forces at the bottom of the pendulum's swing, specifically whether gravitational force should be considered negative. Clarification is sought on how these forces interact to yield the correct tension in the rod.
abpandanguyen
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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.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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