What is the minimum frequency needed to spin a bucket without spilling water?

[devanlevin]

a bucket is filled with water and tied to a rope with a length of "L=1m", the bucket is then spun in a verticle circular motion ,
what is the minimum frequency the bucket can be spun at in order for the water to not spill?

i realize that i need to find the centripeutal force. is it mv^2/r??

from there i can find f=(v/L)*(1/2pi)

the correct answer is (1/2pi)*sqrt(g/L)--- where have i gone wrong

 

[Doc Al]

i realize that i need to find the centripeutal force. is it mv^2/r??

Yes.

from there i can find f=(v/L)*(1/2pi)

How did you get this result?

 

[devanlevin]

using the equation for frequency in circular motion, i realize its wrong,
do i need to compare mv^2/L to mg, saying that it will not fall out when mv^2/L=mg
ie when v^2/L=g
v=(sqrt(g*L)
still not right??

 

[Doc Al]

do i need to compare mv^2/L to mg, saying that it will not fall out when mv^2/L=mg
ie when v^2/L=g
v=(sqrt(g*L)

Yes, that's the minimum speed at the top to keep the water in the bucket. Use that speed to find the minimum frequency. Hint: It goes in a circle.

 

[devanlevin]

so, using equation for frequency (f)

$$\omega$$=2(pi)f

f=$$\frac{\omega}{2pi}$$

$$\omega$$=$$\frac{v}{L}$$

f=$$\frac{v/L}{2pi}$$


v=$$\sqrt{g*L}$$

f=$$\frac{\sqrt{g*L}/L}{2\pi}$$

 

[Doc Al]

f=$$\frac{\sqrt{g*L}/L}{2\pi}$$

Good. Now simplify that a bit so it looks like the given answer.