Questions: The next two questions refer to the following situation. 19. A bucket of water is swung around in a vertical circle with enough speed so that the water does not leave the bucket. The total mass of the bucket and water is 5 kg. What is the minimum speed of the bucket so that the water does not leave the bucket? (A) 1.6 m / s B) 2.7 m / s C) 3.2 m / s D) 4.6 m / s E) 5.0 m / s

Solution Steps

The problem involves a bucket of water being swung in a vertical circle. We need to find the minimum speed at the top of the circle so that the water does not fall out. This requires that the gravitational force is equal to or less than the centripetal force needed to keep the water in the bucket.

At the top of the circle, the gravitational force must be equal to the centripetal force for the water to just stay in the bucket. The centripetal force \( F_c \) is given by:

\[ F_c = \frac{m v^2}{r} \]

where \( m \) is the mass, \( v \) is the velocity, and \( r \) is the radius of the circle.

At the minimum speed, the gravitational force \( F_g = mg \) must equal the centripetal force:

\[ mg = \frac{m v^2}{r} \]

Canceling \( m \) from both sides, we get:

\[ g = \frac{v^2}{r} \]

Solving for \( v \), we have:

\[ v = \sqrt{gr} \]

Assuming the radius \( r \) is such that the gravitational acceleration \( g = 9.81 \, \text{m/s}^2 \), we need to find the minimum speed \( v \). However, the problem does not provide \( r \), so we assume a typical value for a small circle, say \( r = 1 \, \text{m} \).

\[ v = \sqrt{9.81 \times 1} = \sqrt{9.81} \approx 3.13 \, \text{m/s} \]

The closest answer to our calculated minimum speed of approximately \( 3.13 \, \text{m/s} \) is option C, \( 3.2 \, \text{m/s} \).

Final Answer