# Circular Motion/Gravity Problems

## 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

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

## 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

Last edited by a moderator:

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

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

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
Homework Helper
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

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

mjsd
Homework Helper
yes, that will give you the cut-off point

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

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

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

therefore the car will not leave the road

when the car is traveling at 7.67 meters per second:
$$\[ \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}$$

therefore the car will not leave the road

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

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

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

does that look correct? thanks.

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

you legend! thanks