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Chapter 17: Problem 21

A bucket full of water is to be rotated in the vertical plane. What minimum angular velocity, in rad/s, is necessary to keep the water inside if the rotating arm is \(120 \mathrm{~cm}\) ? a) \(2.86\) c) \(3.86\) b) \(3.15\) d) \(4.26\)

Short Answer

Expert verified
Answer: 2.86 rad/s

Step by step solution

01

Identify the knowns and derive the formula for centripetal acceleration

The given length of the rotating arm (radius) is \(r = 120\,\text{cm} = 1.2\, \text{m}\). Since we aim to keep the water inside, the centripetal acceleration (towards the center) must at least equal the gravitational acceleration (\(g = 9.81\, m/s^2\)). The formula for centripetal acceleration, \(a_c\), is \(a_c = \omega^2r\), where \(\omega\) is the angular velocity.
02

Set the centripetal acceleration equal to the gravitational acceleration

We want the centripetal acceleration at the vertical point to be at least equal to the gravitational acceleration, so we need to find the minimum angular velocity \(\omega\) for which this condition holds. Setting \(a_c\) equal to \(g\), we have: \(\omega^2r = g\)
03

Solve for the angular velocity \(\omega\)

Now we need to solve the equation for \(\omega\). Dividing both sides by \(r\), we get: \(\omega^2 = \frac{g}{r}\) To find \(\omega\), we take the square root of both sides: \(\omega = \sqrt{\frac{g}{r}}\)
04

Plug in the values and calculate the angular velocity

Now we can plug in the values for \(r\) and \(g\): \(\omega = \sqrt{\frac{9.81\,m/s^2}{1.2\, \text{m}}}\) \(\omega \approx 2.86\, \text{rad/s}\)
05

Choose the correct answer

The minimum angular velocity to keep the water inside the bucket is approximately \(2.86\, \text{rad/s}\), which corresponds to option (a). Therefore, the correct answer is: a) \(2.86\)

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