mg=mv²/r
v=sqrt(rg)
A pail of water is rotated in a vertical circle of
radius r = 1.49 m.
The acceleration of gravity is 9.8 m/s
2. What is the minimum speed of the pail at
the top of the circle if no water is to spill out?
Answer in units of m/s
2 answers
F=ma
(In rotation problems, centripetal acceleration is equal to V^2/r, where v is the velocity and r is the radius)
F=mv^2/r
The only force acting on the water bucket is that of gravity (mg).
mg=mv^2/r
Now we solve for v.
g=v^2/r
gr=v^2
√gr=v
√(9.8)(1.49)=v
14.6 m/s = v
This is the minimum speed because if it goes any slower it can't overcome the force of gravity (mg) and it will fall out.
(In rotation problems, centripetal acceleration is equal to V^2/r, where v is the velocity and r is the radius)
F=mv^2/r
The only force acting on the water bucket is that of gravity (mg).
mg=mv^2/r
Now we solve for v.
g=v^2/r
gr=v^2
√gr=v
√(9.8)(1.49)=v
14.6 m/s = v
This is the minimum speed because if it goes any slower it can't overcome the force of gravity (mg) and it will fall out.