During the course of a demonstration the professor is called away. When he returns he finds a beaker of water that was at room t
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The centripetal acceleration is given by
where v is the tangential speed and r the radius of the circular orbit.
For the car in this problem,
and r=40 m, so we can re-arrange the previous equation to find the velocity of the car:
where v is the tangential speed and r the radius of the circular orbit.
For the car in this problem,
The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude
while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude
where <em>x</em> is measured in m.
The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have
Solve for <em>x</em> :
The solution with the negative square root is negative, so we throw it out. The other is the one we want,