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.

Answer 1

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  $w__{rpm} } = 29.17 \ rpm$

Explanation:

From the question we are told

    The distance from the handle to the bottom of the bucket is  $d = 35 \ cm = 0.35 \ m$

      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

       $g = w^2 * r$

Here is is the radius of the circle which swinging of the bucket makes and this is mathematically represented as

       $r = L + d$

So

         $g = w^2 * ( L + d )$

= >     $w = \sqrt{\frac{g }{ L + d } }$

= >     $w = \sqrt{\frac{ 9.8}{ 0.7 + 0.35} }$

= >     $w = 3.055 \ rad/s$

Generally the angular speed in revolution per minute is mathematically represented as

        $w__{rpm} } = \frac{w * 60 }{2 \pi }$

=>      $w__{rpm} } = \frac{3.055 * 60 }{2 * 3.142 }$

=>      $w__{rpm} } = 29.17 \ rpm$