Very difficult hw problem; pls help

Think in conversation of total energy, at the point where the car is supposed to have the lowest value of velocity in the circular track..

 

indeed not, did u ever threw a ball in the air and it kept going up?

 

what's the potential energy of gravity?
what do you think of total energy at each point of the track, even during the circular part?
what are the two special cases regarding kinetic energy and potential energy here?

 

at each point of the track there is a conservation of total energy
Et=Kinetic + Gravity
Et=1/2mv^2 + mgh
if r is max, at the highest point of the track v equal what?
what does h equal regarding r?(if we take the flat part of the track the level where potential energy equal 0)

 

I have to go now, I have an exam in less then 2 hours.
again I tell you think of conservation of total energy at 2 distinct point of the track..

 

as ziad1985 has mentioned before, this situation can be solved by using the formulas of P.E. (potential energy) and K.E. (kinetic energy)
[tex]
\begin{array}{l}
P.E. = mgh \\
K.E. = \frac{{mv^2 }}{2} \\
\end{array}
[/tex]

now you need to make these two equations equal to each other, as the kinetic energy of the car at the base of the circular path would be equal to the potential energy at the top (conservation of energy)

[tex]mgh = \frac{{mv^2 }}{2}[/tex]

now by manipulating that equation you can get it ion terms of r (radius of the circle)
assuming h to be the diameter of the circle (2r)
[tex]
\begin{array}{l}
mgh = \frac{{mv^2 }}{2} \\
2mgh = mv^2 \\
m\left( {2gh} \right) = mv^2 \\
2gh = v^2 \\
h = 2r \\
2g\left( {2r} \right) = v^2 \\
4gr = v^2 \\
\frac{{v^2 }}{{4g}} = r \\
g_{acceleration} = 9.8\,m\,s^{ - 2} \\
\frac{{4^2 }}{{4\left( {9.8} \right)}} = 0.4081 \\
\end{array}
[/tex]

so the maximum radius of the circle is 0.4081 meters, any larger the energies would not be equal and the car will leave the track

 

other problems?