A rock of a mass 0.2kg is attatched to a 0.75m long string and swung in a vertical plane. What is the slowest speed that the rock can travel and still maintain a circular path.
so, i started with the following equation: Fc= mv2/r
but where do i go from there to solve for v(what can i change Fc to? )
Oh, so the Centripetal force= mg
how though?
3 MI/HR
Oh, 'at the top'
great, thanks for your help :)
just to clairify,
you're suffesting that there is no tension when the rock is at the top of the swing?
To solve for the slowest speed that the rock can travel and still maintain a circular path, we can start with the equation Fc = mv^2 / r, where Fc represents the centripetal force required to keep the rock moving in a circular path.
The centripetal force can be expressed as Fc = mg, where m is the mass of the rock (0.2 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting this into the equation, we get mg = mv^2 / r. We can rearrange the equation to solve for v:
mg = (mv^2) / r
mg = mv^2r
Now, we can cancel out the mass of the rock, m, from both sides:
g = v^2 / r
Next, we can solve for v by multiplying both sides of the equation by r:
g * r = v^2
Finally, to solve for v, we take the square root of both sides of the equation:
v = sqrt(g * r)
Substituting the given values, where g is approximately 9.8 m/s^2 and r is 0.75 m:
v = sqrt(9.8 * 0.75)
v ≈ 3.05 m/s
Therefore, the slowest speed that the rock can travel and still maintain a circular path is approximately 3.05 m/s.
at the top, mv^2/r=mg
or v= well you solve it.