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Very difficult hw problem; pls help

  • Thread starter tahyus
  • Start date
  • #1
15
0
the drawing shows a version of the loop the loop trick for a small car. if the car is given an initial velocity of 4m/s what is the largest value that the radius r r can have if the car is to remain in contact with the circular track at all times? please refer to the attached file for a diagram.

thnx

ty
 

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Answers and Replies

  • #2
236
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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..
 
  • #3
15
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doesnt the velocity remain the same? or does it change due to gravity?

this is confusing

ty
 
  • #4
236
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indeed not, did u ever threw a ball in the air and it kept going up?
 
  • #5
236
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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?
 
  • #6
15
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isnt PE = mgh?

im not sure if i am making this more complicated

or is there anything to do with circular motion?

ty
 
  • #7
236
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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)
 
Last edited:
  • #8
236
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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..
 
  • #9
15
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so i would substitute 4m/s into the v part of the equation?

ty
 
  • #10
15
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ok

i will try

thnx
 
  • #11
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
 
  • #12
15
0
thnx alot

can u help me with my other problems?

ty
 
  • #13
other problems?
 

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