Circular Motion/Gravity Problems

[unique_pavadrin]

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

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

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

 

[dobry_den]

Ad the second question - use the equation F_G = F_centripetal. Then rearrange it to get the formula for angular speed.

 

[mjsd]

mmm... sounds like a projectile motion problem...?? do we assume that the car stick to the ground or what?

 

[unique_pavadrin]

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

 

[mjsd]

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

 

[unique_pavadrin]

so i have to find when the reaction force along the top of the bridge is zero?

 

[mjsd]

yes, that will give you the cut-off point

 

[unique_pavadrin]

okay thanks for that, ill try it and see how i go

 

[mjsd]

although in the question you said the circular bridge is of radius 6m, the diagram seems to indicate the radius is 10m instead??

 

[unique_pavadrin]

for the velocity of 5 meters per second:
<br /> \begin{array}{l}<br /> a = \frac{{v^2 }}{r} = \frac{{5^2 }}{{10}} = 2.5\,m\,s^{ - 2} \\ <br /> F = ma = 2.5\left( {1000} \right) = 2500 \\ <br /> W = mg = 9.8\left( {1000} \right) = 9800 \\ <br /> W &gt; F \\ <br /> \end{array}<br />

therefore the car will not leave the road

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

therefore the car will not leave the road

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

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

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

does that look correct? thanks.

 

[mjsd]

haven't check your numerical values, but method looks good (which is the more important bit)

 

[unique_pavadrin]

you legend! thanks