Circular Motion/Gravity Problems

  • #1

Homework Statement


A car of mass 1000kg has just reached the top of a circular bridge of radius 6.0m (See the figure provided). Calculate where the car will reach the ground level if the car is traveling at
a) 5.0 m s-1
b) 7.67 m s-1
c) 20.0 m s-1
at the top of the curve
Image for first question:
http://img149.imageshack.us/img149/5581/1001rm5.jpg [Broken]


At what minimum speed should the Earth spin to enable you to float away from the Earth's surface at the equator?

Homework Equations


[tex]
\begin{array}{l}
W = mg \\
a = \frac{{v^2 }}{r} = \frac{{4\pi ^2 r}}{{T^2 }} = \frac{{2\pi v}}{T} \\
g = \frac{F}{m} = \frac{{GM}}{{r^2 }} \\
F = ma \\
F = \frac{{GMm}}{{r^2 }} \\
\end{array}
[/tex]


The Attempt at a Solution


I have no idea on where to start with these problems, so all help is greatly appreciated, many thanks in advance

unique_pavadrin
 
Last edited by a moderator:

Answers and Replies

  • #2
115
0
Ad the second question - use the equation FG = Fcentripetal. Then rearrange it to get the formula for angular speed.
 
  • #3
mjsd
Homework Helper
726
3
mmm... sounds like a projectile motion problem...?? do we assume that the car stick to the ground or what?
 
  • #4
mjsd, that is what ai am trying to find out, if the car will stick to the ground, past the peak of the bridge traveling at the different velocities
 
  • #5
mjsd
Homework Helper
726
3
ok, so in order for it to stick it mustn't move too fast... so that it will remain in uniform circular motion following the arc. obviously mv^2/r comes in here. if v is too high it will fly off
 
  • #6
so i have to find when the reaction force along the top of the bridge is zero?
 
  • #7
mjsd
Homework Helper
726
3
yes, that will give you the cut-off point
 
  • #8
okay thanks for that, ill try it and see how i go
 
  • #9
mjsd
Homework Helper
726
3
although in the question you said the circular bridge is of radius 6m, the diagram seems to indicate the radius is 10m instead??
 
  • #10
for the velocity of 5 meters per second:
[tex]
\begin{array}{l}
a = \frac{{v^2 }}{r} = \frac{{5^2 }}{{10}} = 2.5\,m\,s^{ - 2} \\
F = ma = 2.5\left( {1000} \right) = 2500 \\
W = mg = 9.8\left( {1000} \right) = 9800 \\
W > F \\
\end{array}
[/tex]

therefore the car will not leave the road

when the car is traveling at 7.67 meters per second:
[tex]\[
\begin{array}{l}
a = \frac{{v^2 }}{r} = \frac{{7.67^2 }}{{10}} = 5.88289\,m\,s^{ - 2} \\
F = ma = 5.88289\left( {1000} \right) = 5882.89 \\
W = mg = 9.8\left( {1000} \right) = 9800 \\
W > F \\
\end{array}
[/tex]

therefore the car will not leave the road

for the car traveling at 20 meters per second:
[tex]
\begin{array}{l}
a = \frac{{v^2 }}{r} = \frac{{20^2 }}{{10}} = 40\,m\,s^{ - 2} \\
F = ma = 40\left( {1000} \right) = 40000 \\
W = mg = 9.8\left( {1000} \right) = 9800 \\
W < F \\
\end{array}
[/tex]

therefore the car will leave the road.
where the car will land:
[tex]
\begin{array}{l}
s = ut + \frac{1}{2}at^2 \\
s_{vertical} = 6 \\
u_{vertical} = 0 \\
a_{vertical} = 9.8 \\
6 = \frac{1}{2}\left( {9.8} \right)t^2 \\
t = \sqrt {\frac{{2\left( 6 \right)}}{{9.8}}} = 1.1065 \\
s_{horizontal} = ut = 20\left( {1.1065} \right) = 22.1313\,m \\
\end{array}
[/tex]

therefore the car will land approximately 22 (horizontally) meters from the top of the bridge.

does that look correct? thanks.
 
  • #11
mjsd
Homework Helper
726
3
haven't check your numerical values, but method looks good (which is the more important bit)
 
  • #12
you legend! thanks
 

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