Circular Motion/Gravity Problems

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In summary, a car traveling at 5.0 m/s or 7.67 m/s at the top of a circular bridge with a radius of 10m will not leave the road, while a car traveling at 20 m/s will leave the road. The car will land approximately 22 meters from the top of the bridge. The equations used were W = mg, a = v^2/r, F = ma, FG = Fcentripetal, and s = ut + 1/2at^2.
  • #1
unique_pavadrin
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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


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
 
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  • #2
Ad the second question - use the equation FG = Fcentripetal. Then rearrange it to get the formula for angular speed.
 
  • #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
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
yes, that will give you the cut-off point
 
  • #8
okay thanks for that, ill try it and see how i go
 
  • #9
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
haven't check your numerical values, but method looks good (which is the more important bit)
 
  • #12
you legend! thanks
 

1. What is circular motion?

Circular motion is the movement of an object along a circular path, where the object maintains a constant distance from a fixed point called the center of rotation.

2. What is centripetal force?

Centripetal force is the force that acts on an object moving in a circular path, directed towards the center of rotation. It is responsible for keeping the object moving in a circular motion.

3. How is circular motion related to gravity?

Circular motion is related to gravity because the centripetal force that keeps an object moving in a circular path is often provided by the force of gravity. This is seen in the orbit of planets around the sun or the moon around the Earth.

4. How can we calculate the speed of an object in circular motion?

The speed of an object in circular motion can be calculated using the formula v = √(rω), where v is the speed, r is the radius of the circular path, and ω is the angular velocity (measured in radians per second).

5. What is the difference between centripetal force and centrifugal force?

Centripetal force is the force that keeps an object moving in a circular path, while centrifugal force is the apparent outward force that is felt by an object in circular motion, but is not a real force. It is a result of the object's inertia trying to keep it moving in a straight line.

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