Complete Question
What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm. The student has 70-cm-long arms
Answer:
The value is [tex]w__{rpm} } = 29.17 \ rpm[/tex]
Explanation:
From the question we are told
The distance from the handle to the bottom of the bucket is [tex]d = 35 \ cm = 0.35 \ m[/tex]
The length of the students arm is L = 70 cm = 0.70 m
Generally the acceleration due to gravity experienced by the bucket of water is mathematically represented as
[tex]g = w^2 * r[/tex]
Here is is the radius of the circle which swinging of the bucket makes and this is mathematically represented as
[tex]r = L + d[/tex]
So
[tex]g = w^2 * ( L + d )[/tex]
= > [tex]w = \sqrt{\frac{g }{ L + d } }[/tex]
= > [tex]w = \sqrt{\frac{ 9.8}{ 0.7 + 0.35} }[/tex]
= > [tex]w = 3.055 \ rad/s[/tex]
Generally the angular speed in revolution per minute is mathematically represented as
[tex]w__{rpm} } = \frac{w * 60 }{2 \pi }[/tex]
=> [tex]w__{rpm} } = \frac{3.055 * 60 }{2 * 3.142 }[/tex]
=> [tex]w__{rpm} } = 29.17 \ rpm[/tex]
