Conservation of Energy Problem - Very Tricky

AI Thread Summary
The discussion revolves around a conservation of energy problem involving a cart on a track. The key equation derived is mgR = mgh + 1/2mv^2, leading to the expression for velocity just before the cart leaves the track, v^2 = 2g(R-h). It is noted that at the moment the cart leaves the track, the normal force equals zero, and only gravitational force acts on it afterward. The centripetal force equation is also referenced, emphasizing the need to consider the direction of centripetal acceleration when determining the angle at which the cart departs. The conversation highlights the importance of analyzing forces and energy conservation principles to solve for the angle.
doug1
Messages
9
Reaction score
0

Homework Statement



http://imgur.com/ZGvC1

At what angle does the cart fly off of the track?

No other information is provided


Homework Equations



I know that energy is always conserved (total energy before = total energy afterwards).


The Attempt at a Solution



Equation for energy before = energy after:

mgR + Ek = mgh + 1/2mv^2
mgR = mgh + 1/2mv^2 (h is the height above the ground just before the cart leaves the track)

gR = gh + 1/2v^2

v^2 = 2g(R-h) --> This is the velocity just before the cart leaves the track



The dotted line that we see is equal to R.

Thus, Fc (centripetal force) = mv^2/R

If the cart were to leave the track, Fn (normal force) would equal 0

Where do I go from here? Am i on the right track?
 
Physics news on Phys.org
So far so good. At the point where the cart leaves the track, what are the forces acting on it? What can you say about its acceleration at that point?
 
After it leaves the track the only force acting on it is gravity. Would I do: mg = mv^2/R

g = v^2/R

R = v^2/g

Plug in v^2 = 2g(R-h)

R = 2g(R-h)/g

R = 2R - 2h

R = 2h

How do I find the angle then?
 
doug1 said:
After it leaves the track the only force acting on it is gravity.
Good.
Would I do: mg = mv^2/R
Almost. What direction is the centripetal acceleration at the point that it leaves the track? Consider forces along that axis.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top