/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 (II) Is it possible to whirl a b... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(II) Is it possible to whirl a bucket of water fast enough in a vertical circle so that the water won't fall out? If so, what is the minimum speed? Define all quantities needed.

Short Answer

Expert verified
Yes, minimum speed is \( \sqrt{rg} \).

Step by step solution

01

Understanding the Problem

We need to find out if a bucket of water can be spun in such a way that the water won't fall out when spun vertically. This involves understanding the forces acting on the water in the bucket during circular motion.
02

Identifying the Forces

When the bucket is at the top of the circle, the water inside should experience a centripetal force that equals or exceeds the gravitational force pulling it downward. This ensures the water stays in the bucket.
03

Introducing Relevant Quantities

Let \( v \) be the speed of the bucket, \( r \) be the radius of the circular path, and \( g \) be the acceleration due to gravity (approximately \( 9.8 \ m/s^2 \)).

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 91Ó°ÊÓ!

Key Concepts

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

Circular Motion
Circular motion occurs when an object moves along a circular path. The object continuously changes direction, which means there is acceleration involved, even if the object's speed is constant. This type of motion is common in many everyday situations, such as a car turning around a curve or the Earth orbiting around the sun.
In circular motion, two essential components are necessary: the plane of the motion and the axis of rotation. In our case, as the bucket of water moves vertically, we must consider both gravity and centripetal force to ensure that the water remains inside the bucket.
  • The **velocity vector** is always tangent to the circle, meaning it points in the direction the object is moving at any moment.
  • The **centripetal force** is directed towards the center of the circle, keeping the object in its circular path. Without this inward force, the object would move in a straight line.
Remember, if an object's speed changes, it can result in changes to the radius or frequency of its circular path, but in this scenario, we're considering a steady speed to maintain circular motion.
Gravitational Force
Gravitational force is the force of attraction between two objects with mass. On Earth, gravity is a constant force acting downwards toward the planet's center. It has an acceleration of approximately 9.8 m/s², often denoted as **g**.
For the water in the bucket, gravitational force plays a crucial role in determining whether it stays or spills out as the bucket moves. At the top of the circular path, gravity acts downward on the water, trying to pull it out of the bucket. For the water to remain in the bucket, the centripetal force must counteract this pull.
  • When the bucket is at the circle's top, the gravitational force and centripetal force work together towards the center of the circle.
  • If the combined forces are enough to keep the water inside, the water won't spill out even as gravity tries to pull it down.
    This makes understanding the relationship between gravity and motion essential when examining if the water will stay inside the bucket.
In summary, gravitational force is a major player in this situation, influencing whether the centripetal force is strong enough to keep the water from falling out.
Minimum Speed Calculation
To keep the water from spilling out at the top of the circle, the speed of the bucket must be high enough to maintain the necessary centripetal force.
The centripetal force needed is provided entirely by gravitational force when at the circle's peak. Thus, the key to finding the minimum speed, **v_min**, is to set the gravitational force equal to the centripetal force.
  • The formula for centripetal force is given by \[ F_c = \frac{mv^2}{r} \]
  • The gravitational force formula is \[ F_g = mg \]
At the bucket's highest point in the circle, we need \[ F_c = F_g \], leading to \[ \frac{mv^2}{r} = mg \].
This simplifies to \[ v^2 = rg \] or, solving for speed, \[ v = \sqrt{rg} \].
Here, **r** represents the radius of the circle and **g** is the acceleration due to gravity. By plugging in these values, you can calculate the minimum speed required to keep the bucket's water in place.
Through this understanding, you see that maintaining enough speed ensures the centripetal force can overcome gravity’s pull, keeping the water safely inside the bucket even at challenging vertical positions.

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

(II) A child slides down a slide with a \(34^{\circ}\) incline, and at the bottom her speed is precisely half what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child.

A car starts rolling down a \(1-\mathrm{in}-4\) hill \((1-\mathrm{in}-4\) means that for each 4 \(\mathrm{m}\) traveled along the road, the elevation change is 1 \(\mathrm{m} ) .\) How fast is it going when it reaches the bottom after traveling 55 \(\mathrm{m}\) ? (a) Ignore friction. (b) Assume an effective coefficient of friction equal to \(0.10 .\)

(II) The terminal velocity of a \(3 \times 10^{-5} \mathrm{~kg}\) raindrop is about \(9 \mathrm{~m} / \mathrm{s}\). Assuming a drag force \(F_{\mathrm{D}}=-b v\), determine \((a)\) the value of the constant \(b\) and \((b)\) the time required for such a drop, starting from rest, to reach \(63 \%\) of terminal velocity.

A ball of mass \(m=1.0 \mathrm{~kg}\) at the end of a thin cord of length \(r=0.80 \mathrm{~m}\) revolves in a vertical circle about point \(\mathrm{O},\) as shown in Fig. \(5-56 .\) During the time we observe it, the only forces acting on the ball are gravity and the tension in the cord. The motion is circular but not uniform because of the force of gravity. The ball increases in speed as it descends and decelerates as it rises on the other side of the circle. At the moment the cord makes an angle \(\theta=30^{\circ}\) below the horizontal, the ball's speed is \(6.0 \mathrm{~m} / \mathrm{s}\). At this point, determine the tangential acceleration, the radial acceleration, and the tension in the cord, \(F_{\mathrm{T}}\). Take \(\theta\) increasing downward as shown.

(III) The force of air resistance (drag force) on a rapidly falling body such as a skydiver has the form \(F_{\mathrm{D}}=-k v^{2},\) so that Newton's second law applied to such an object is $$ m \frac{d v}{d t}=m g-k v^{2} $$ where the downward direction is taken to be positive. (a) Use numerical integration [Section \(2-9]\) to estimate (within \(2 \%\) ) the position, speed, and acceleraton, from \(t=0\) up to \(t=15.0 \mathrm{~s},\) for a \(75-\mathrm{kg}\) skydiver who starts from rest, assuming \(k=0.22 \mathrm{~kg} / \mathrm{m}\) (b) Show that the diver eventually reaches a steady speed, the terminal speed, and explain why this happens. (c) How long does it take for the skydiver to reach \(99.5 \%\) of the terminal speed?

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.