The minimum speed of the water must be 3.4 m/s
Explanation:
There are two forces acting on the water in the pail when it is at the top of its circular motion:
- The force of gravity, mg, acting downward (where m is the mass of the water and g the acceleration of gravity)
- The normal reaction, N also acting downward
Since the water is in circular motion, the net force must be equal to the centripetal force, so:
[tex]N+mg=m\frac{v^2}{r}[/tex]
Where:
[tex]g=9.8 m/s^2[/tex]
v is the speed of the pail
r = 1.2 m is the radius of the circle
The water starts to spill out when the normal reaction of the pail becomes zero:
N = 0
When this occurs, the equation becomes:
[tex]mg=m\frac{v^2}{r}\\v=\sqrt{gr}[/tex]
And substitutin the values of g and r, we find the minimum speed that the water must have in order not to spill out:
[tex]v=\sqrt{(9.8)(1.2)}=3.4 m/s[/tex]
Learn more about circular motion:
brainly.com/question/2562955
brainly.com/question/6372960
#LearnwithBrainly
