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? )
http://www.jiskha.com/display.cgi?id=1256408250
so are you suggesting that there is no tension when the rock is at the top of the circle?
no tension at the absolute minimum velocity.
okay, thanks !
To solve for the minimum speed of the rock, you can set the centripetal force (Fc) equal to the gravitational force acting on the rock.
The centripetal force (Fc) is given by the equation Fc = mv²/r, where m is the mass of the rock, v is its velocity, and r is the radius of the circular path.
The gravitational force (Fg) acting on the rock is given by Fg = mg, where g is the acceleration due to gravity (approximately 9.8 m/s²).
Setting Fc equal to Fg, we have:
mv²/r = mg
Now, we can rearrange the equation to solve for v:
v² = rg
v = √(rg)
Substituting the given values, we have:
v = √(0.75m * 9.8m/s²)
v ≈ √7.35 ≈ 2.71 m/s
Therefore, the slowest speed that the rock can travel and still maintain a circular path is approximately 2.71 m/s.