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Stay Dry! You tie a cord to a pail of water, and you swing the pail in a vertical circle of radius 0.600 \(\mathrm{m}\) . What minimum speed must you give the pail at the highest point of the circle if no water is to spill from it?

Short Answer

Expert verified
The minimum speed is approximately 2.43 m/s.

Step by step solution

01

Understanding the Forces

At the highest point in the vertical circle, the only forces acting on the pail are the gravitational force and the tension in the cord. To ensure no water spills, the pail must maintain a minimum speed such that the gravitational force provides the necessary centripetal force, meaning the tension could be zero at this point.
02

Set Up the Equation

At the minimum speed, the centripetal force is given solely by the gravitational force. Use the equation for centripetal force \( F_c = \frac{mv^2}{r} \) where \( m \) is the mass of the pail, \( v \) is the velocity, and \( r \) is the radius. The gravitational force is \( F_g = mg \) where \( g \) is the acceleration due to gravity. Set these two equal: \( \frac{mv^2}{r} = mg \).
03

Simplifying the Equation

Divide both sides of the equation by \( m \) to simplify, giving \( \frac{v^2}{r} = g \). Solve for \( v^2 \) to find \( v^2 = rg \).
04

Calculate the Minimum Speed

Substitute the given radius \( r = 0.600 \) \( \, \mathrm{m} \) and \( g = 9.81 \) \( \, \mathrm{m/s^2} \) into the equation \( v^2 = rg \):\[ v^2 = 0.600 \times 9.81 \]. Calculate \( v \): \[ v = \sqrt{0.600 \times 9.81} \approx \sqrt{5.886} \approx 2.43 \] \( \, \mathrm{m/s} \).
05

Confirm the Solution

Ensure that the calculated velocity meets the physical requirements. At this minimum speed, the tension in the cord is zero, and the gravitational force alone allows the pail to continue in a vertical circular path, preventing water spilling.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertical Circular Motion
Vertical circular motion occurs when an object moves in a circle in a vertical plane. It can be observed in daily situations like swinging a bucket of water in a circle or amusement park rides such as a Ferris wheel.
When discussing vertical circular motion, it's crucial to understand the forces at play, particularly at different points of the circle.
The object must follow a path where the gravitational force and any tension or support provide enough centripetal force to maintain the circular trajectory.
  • At the top of the circle: Both gravity and any tension in the support work together to provide the necessary centripetal force.
  • At the bottom of the circle: Tension must counteract gravity and still provide the centripetal force. This typically requires more tension than at the top.
This interplay of forces makes vertical circular motion different from horizontal circular motion. The balance between gravity, tension, and centripetal force dictates the minimum speed needed for an object to stay in motion without falling out of its path.
Gravitational Force
Gravitational force is the force of attraction exerted by the Earth on an object. It acts downwards and is defined by the equation:\[ F_g = mg \]where:
  • \( F_g \) is the gravitational force,
  • \( m \) is the mass of the object, and
  • \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \mathrm{m/s^2} \) on Earth.
In vertical circular motion, gravitational force plays a dual role. It is necessary for providing the centripetal force required to keep the object in circular motion, especially at the highest point. At this point, gravitational force helps maintain the object's path without needing any additional support force.
Understanding gravitational force is key to solving problems involving vertical circular motion. For any object at the top of its path, ensuring gravitational force provides enough centripetal force is crucial, as tension can be zero under those conditions.
Minimum Velocity
Minimum velocity in the context of vertical circular motion is the lowest speed an object can have at the top of the circle and still maintain its circular path without falling. This concept is crucial in ensuring that the object, such as the pail of water, doesn't spill during its motion.The minimum velocity can be determined by setting the centripetal force equal to the gravitational force at the highest point, as seen in the equation:\[ \frac{mv^2}{r} = mg \]Simplifying gives us:\[ v^2 = rg \]Solving for \( v \), we get:\[ v = \sqrt{rg} \]where:
  • \( r \) is the radius of the circle.
  • \( g \) is the acceleration due to gravity.
This equation shows that the minimum velocity depends on both the radius of the circular path and gravity. Any speed below this calculated minimum would mean the object can't sustain the centripetal force just by gravity, leading to the object falling out of its circular path.

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Most popular questions from this chapter

You throw a baseball straight up. The drag force is proportional to \(v^{2} .\) In terms of \(g,\) what is the \(y\) -component of the ball's acceleration when its speed is half its terminal speed and (a) it is moving up? (b) It is moving back down?

A \(125-\mathrm{kg}\) (including all the contents) rocket has an engine that produces a constant vertical force (the thrust) of 1720 \(\mathrm{N}\) . Inside this rocket, a \(15.5-\mathrm{N}\) electrical power supply rests on the floor. (a) Find the acceleration of the rocket, (b) When it has reached an altitude of 120 \(\mathrm{m}\) , how hard does the floor push on the power supply? (Hint: Start with a free-body diagram for the power supply.)

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