Conservation of energy of a car problem

AI Thread Summary
To determine the height \( h \) that a 200-kg roller coaster car reaches when launched at speed \( v_1 \), the conservation of energy principle is applied, equating kinetic and potential energy. The car must maintain enough speed at the top of the loop to ensure it remains on the track, which requires considering both gravitational and normal forces acting on it. At the top of the loop, the gravitational force must equal the centripetal force needed for circular motion, leading to the condition where the normal force is zero. This results in the equation \( mg = \frac{mv^2}{r} \) for the minimum speed required. The discussion emphasizes the importance of understanding forces in circular motion and the role of energy conservation in solving the problem.
myoplex11
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Homework Statement


Determine the height h to the top of the incline D to which the 200-kg roller coaster car will reach, if it is launched at B with a speed just sufficient for it to round the top of the loop at C without leaving the track. The radius of curvature at C is
ρC = 25 m. I have attached a picture of this problem below.



Homework Equations


KE1+PE1=KE2+PE2


The Attempt at a Solution


.5mv1^2+mgh=.5mv^2+mgh
.5mv1^2=mgh
0.5v1^2=gh
 

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… physics for "hug" is "force" …

myoplex11 said:

Homework Equations


KE1+PE1=KE2+PE2

The Attempt at a Solution


.5mv1^2+mgh=.5mv^2+mgh
.5mv1^2=mgh
0.5v1^2=gh

Your method would only give you the speed needed for the car to reach the top with zero speed.

The question requires the speed at the top to be so great that the car still "hugs the track".

Physics for "hug" is "force". :smile:

(Reminds me of: Principal "How is you mathematics?" Groucho Marx: "Sir, I speak it like a native!" :smile:)
​

So … what are the forces on the car at the top of the track? :smile:
 
centipetal force and the normal force i still have no idea how to do this
 
… the acceleration is always downward …

myoplex11 said:
centipetal force and the normal force i still have no idea how to do this

"Centipetal force" is not a force! :frown:

And you've missed out gravity …

The only two forces are gravity, and the normal force (the reaction force).

Those two forces determine the acceleration.

Paradoxically, they're both downward (at the top of the loop, C) … which means that the acceleration is always downward (centripetal)! :frown:

But that doesn't matter, because circular motion (at the top of the circle) requires downward acceleration! :smile:

This is geometry, not physics (and you can easily prove it using vectors) … an object with speed v in a circle or radius r has an acceleration of v^2/r towards the centre of the circle.

So, at C, you want gravity and the normal force together to equal m.v^2/r (Newtons' second law).

The question asks for the minimum case, which obviously must be when the normal force is zero: then the force from gravity alone equals mv^2/r.

And so … ? :smile:
 
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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