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`

Step-by-step 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{(g )/( L + d ) }`

= >
`w = \sqrt{( 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) } = (w * 60 )/(2 \pi )`

=>
`w__(rpm) } = (3.055 * 60 )/(2 * 3.142 )`

=>
`w__(rpm) } = 29.17 \ rpm`