Question

You tie a cord to a pail of water and swing the pail in a vertical circle of radius 0.600 m. What minimum speed must you give the pail at the highest point of the circle to avoid spilling water?

Short answer

2.43 m/s

Step-by-step solution

Understanding the Problem

To avoid spilling, the centrifugal force must be equal to or greater than the gravitational force at the top of the circle. Thus, the minimum speed is such that the gravitational force is exactly balanced by the required centripetal force to maintain circular motion.

Applying Formulas

At the top of the circle, the required centripetal force is provided by the gravitational force. The gravitational force is calculated by \( F_g = mg \) and the centripetal force by \( F_c = \frac{mv^2}{r} \). To avoid spilling, set these equal: \( mg = \frac{mv^2}{r} \).

Simplifying the Equation

We can cancel out the mass \( m \) from both sides of the equation, leaving \( g = \frac{v^2}{r} \). This simplifies finding the minimum speed \( v \) the pail needs at the top to avoid spillage.

Solving for Minimum Speed

Rearrange the equation \( g = \frac{v^2}{r} \) to solve for \( v \): \( v^2 = gr \). Therefore, \( v = \sqrt{gr} \). Given \( r = 0.600 \) m and using \( g = 9.81 \) m/s², substitute these into the equation: \( v = \sqrt{9.81 \times 0.600} \).

Calculating the Result

Compute \( v = \sqrt{5.886} \) which gives \( v \approx 2.43 \) m/s. This is the minimum speed required at the top of the circle to avoid spilling water from the pail.

Key concepts

Gravitational Force

Gravitational force plays a crucial role in all kinds of motion, especially when objects move in a circle. This force is the invisible attraction between two objects with mass.
The more massive an object, the greater the gravitational pull it exerts on another object.

• The force of gravity acts downward towards the center of the Earth.
• It is calculated using the formula: \( F_g = mg \), where \( m \) is mass and \( g \) is the acceleration due to gravity.
• The standard value of \( g \) on Earth is approximately \( 9.81 \text{ m/s}^2 \).

For an object in vertical circular motion, like our pail of water, gravitational force is especially important at the top of the circle, providing the necessary force to keep it in motion.
In essence, at the highest point of the swing, gravity aids in providing the centripetal force required to maintain the pail's circular path.

Circular Motion

Circular motion is when an object travels along a curved path or a circle. In this kind of motion, a centripetal force is necessary to keep the object moving in a circle, constantly pulling it towards the center.
Examples include a satellite orbiting a planet or a car turning around a curved road.

• The force keeping the object in the circular path is called centripetal force.
• This force is directed towards the center of the circle and is responsible for changing the direction of the object's velocity without altering its speed.
• For a pail of water swung in a vertical circle, the centripetal force at the top of the circle is provided by gravity.
• Centripetal force can be calculated using the formula: \( F_c = \frac{mv^2}{r} \).

Circular motion isn't only confined to gravity. Friction or tension can also act as centripetal forces in other scenarios, like driving or swinging.

Minimum Speed Calculation

Determining the minimum speed in circular motion scenarios is necessary to keep an object moving along its path without faltering. In this exercise, our goal is to ensure the pail at the highest point does not spill the water.
This is achieved by having the gravitational force effectively turn into the centripetal force needed to maintain the motion.

• The key equation to determine minimum speed in this context is derived by setting gravitational force equal to centripetal force: \( mg = \frac{mv^2}{r} \).
• From this equation, you can simplify to find the speed: \( v = \sqrt{gr} \).
• This formula ensures that the speed is enough to counteract the gravitational pull that tries to spill the water.
• Substituting \( g = 9.81 \text{ m/s}^2 \) and \( r = 0.600 \text{ m} \) gives a minimum speed of approximately 2.43 m/s.

This calculation is important in various areas. For instance, roller coasters must reach minimum speeds at certain points to keep riders safely looped, and vehicles on curved roads require adequate speeds to prevent skidding.