Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Expert verified
2.43 m/s

Step by step solution

01

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.
02

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: \( mg = \frac{mv^2}{r} \).
03

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.
04

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: \( v = \sqrt{9.81 \times 0.600} \).
05

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 75.0-kg wrecking ball hangs from a uniform, heavy-duty chain of mass 26.0 kg. (a) Find the maximum and minimum tensions in the chain. (b) What is the tension at a point three-fourths of the way up from the bottom of the chain?

A 2.00-kg box is moving to the right with speed 9.00 m/s on a horizontal, frictionless surface. At \(t =\) 0 a horizontal force is applied to the box. The force is directed to the left and has magnitude \(F(t) = (6.00 N/s^2)t^2\). (a) What distance does the box move from its position at \(t =\) 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at \(t =\) 3.00 s?

On the ride "Spindletop" at the amusement park Six Flags Over Texas, people stood against the inner wall of a hollow vertical cylinder with radius 2.5 m. The cylinder started to rotate, and when it reached a constant rotation rate of 0.60 rev/s, the floor dropped about 0.5 m. The people remained pinned against the wall without touching the floor. (a) Draw a force diagram for a person on this ride after the floor has dropped. (b) What minimum coefficient of static friction was required for the person not to slide downward to the new position of the floor? (c) Does your answer in part (b) depend on the person's mass? (\(Note\): When such a ride is over, the cylinder is slowly brought to rest. As it slows down, people slide down the walls to the floor.)

A 50.0-kg stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane. (a) If the plane's speed at the lowest point of the circle is 95.0 m/s, what is the minimum radius of the circle so that the acceleration at this point will not exceed 4.00\(g\)? (b) What is the apparent weight of the pilot at the lowest point of the pullout?

(a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you can stop a car by locking the brakes when the car is traveling at 28.7 m/s (about 65 mi/h)? (b) On wet pavement the coefficient of kinetic friction may be only 0.25. How fast should you drive on wet pavement to be able to stop in the same distance as in part (a)? (\(Note\): Locking the brakes is \(not\) the safest way to stop.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free