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

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs 882 N at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

Block \(B\), with mass 5.00 kg, rests on block \(A\), with mass 8.00 kg, which in turn is on a horizontal tabletop \(\textbf{(Fig. P5.92).}\) There is no friction between block \(A\) and the tabletop, but the coefficient of static friction between blocks \(A\) and \(B\) is 0.750. A light string attached to block \(A\) passes over a frictionless, massless pulley, and block \(C\) is suspended from the other end of the string. What is the largest mass that block \(C\) can have so that blocks \(A\) and \(B\) still slide together when the system is released from rest?

You are standing on a bathroom scale in an elevator in a tall building. Your mass is 64 kg. The elevator starts from rest and travels upward with a speed that varies with time according to \(v(t) =\) (3.0 m/s\(^2\))\(t\) \(+\) (0.20 m/s\(^3\))\(t^2\). When \(t =\) 4.0 s, what is the reading on the bathroom scale?

A 3.00-kg box that is several hundred meters above the earth's surface is suspended from the end of a short vertical rope of negligible mass. A time- dependent upward force is applied to the upper end of the rope and results in a tension in the rope of \(T(t) =\) (36.0 N/s)\(t\). The box is at rest at \(t =\) 0. The only forces on the box are the tension in the rope and gravity. (a) What is the velocity of the box at (i) \(t =\) 1.00 s and (ii) \(t =\) 3.00 s? (b) What is the maximum distance that the box descends below its initial position? (c) At what value of \(t\) does the box return to its initial position?

A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hollow, transparent plastic sphere. After gaining sufficient speed, she travels in a vertical circle with radius 13.0 m. She has mass 70.0 kg, and her motorcycle has mass 40.0 kg. (a) What minimum speed must she have at the top of the circle for the motorcycle tires to remain in contact with the sphere? (b) At the bottom of the circle, her speed is twice the value calculated in part (a). What is the magnitude of the normal force exerted on the motorcycle by the sphere at this point?

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