/*! 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 16 A tube of iength \(L\) is filled... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A tube of iength \(L\) is filled completely with an incompressible liquid of mass \(M\) and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity \(\omega\). The force exerted by the liquid at the other end is:

Short Answer

Expert verified
The force exerted by the liquid is \(\frac{M \omega^2 L}{2}\).

Step by step solution

01

Understanding the Problem

We have a tube of length \(L\) filled with liquid of mass \(M\) and are rotating it about one end with angular velocity \(\omega\). We need to find the force exerted by the liquid at the opposite end of the tube. This problem involves concepts of rotational motion and dynamics.
02

Expressing Centrifugal Force per Element

When the tube is rotated, any given element of the liquid experiences a centrifugal force. Consider an infinitesimal mass element \(dm\) at a distance \(x\) from the axis of rotation. This element experiences a force \(dF = \rho A \cdot dx \cdot \omega^2 x\), where \(\rho\) is the density of the liquid and \(A\) is the cross-sectional area.
03

Integrating over the Length

Summing up the contributions of all infinitesimal elements across the length of the tube gives the total force. We can express the mass element as \(dm = \rho A dx\). The total force \(F\) exerted by the liquid is the integral \(\int_0^L \rho A \omega^2 x \, dx\).
04

Solving the Integral

Substituting and solving the integral, we perform the integration: \[ F = \rho A \omega^2 \int_0^L x \, dx = \rho A \omega^2 \left[ \frac{x^2}{2} \right]_0^L = \frac{\rho A \omega^2 L^2}{2}. \]
05

Using the Relationship Between Mass and Volume

Knowing that the total mass \(M = \rho A L\), we can substitute \(\rho A\) from this relationship into our force expression: \[ F = \frac{M \omega^2 L}{2}. \]
06

Conclusion

The force exerted by the liquid at the end of the tube is \(\frac{M \omega^2 L}{2}\), which arises from the integration of centrifugal forces along its length.

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.

Centrifugal Force
When objects are in rotational motion, a fascinating force called centrifugal force comes into play. This force is not a "real" force in the traditional sense, as it doesn't arise from a physical interaction. Instead, it is experienced due to the inertia of an object moving in a curved path.

As the object moves, it tends to travel a straight line, but the inward "centripetal" force required for circular motion creates a feeling of being pushed outward. When considering the liquid inside the rotating tube, centrifugal force acts on each small element, pushing it outward. The liquid doesn’t actually exert a force caused by its own motion, but its inertia when moving in a circle results in this effect.
  • Centrifugal force is proportional to the distance from the axis of rotation.
  • This force depends on the square of the angular velocity and the mass of the object.
  • It's given for each small element as \(dF = \rho A \cdot dx \cdot \omega^2 x\).
This outward force is crucial in understanding how liquids behave in rotating systems.
Dynamics of Rotation
The dynamics of rotation refers to how forces and torques interact to cause rotational motion. Just like linear dynamics focuses on forces causing linear motion, rotational dynamics deals with parts rotating about an axis.

Newton’s Second Law for Rotation states that torque is equal to the moment of inertia multiplied by angular acceleration, \(\tau = I\alpha\). This principle helps explain what happens within our rotating tube.
  • Torque produces angular acceleration, just as force does for linear acceleration.
  • In a rotating tube, elements of liquid exert forces due to inertia, which results in angular dynamics.
  • The interplay of these small forces across the length leads to a net force at its end.
Understanding rotational dynamics helps in evaluating the behavior of rotating fluids and structures.
Angular Velocity
Angular velocity is a central concept in rotational motion. It refers to the rate at which an object rotates around an axis. Essentially, it's how fast something spins and is measured in radians per second.

In our tube example, angular velocity \(\omega\) influences how much centrifugal force each element experiences.
  • Greater angular velocity means more force pushing elements outward.
  • It's an essential factor in computing centrifugal force, as seen in \(F = \frac{M \omega^2 L}{2}\).
  • Angular velocity defines how quickly the tube must complete rotations to maintain said force.
Comprehending how angular velocity impacts the components of rotational systems is vital in physics, guiding the calculations of forces like centrifugal force.
Integral Calculus in Physics
Integral calculus is a powerful mathematical tool, particularly useful in physics for calculating quantities that accumulate over a continuous range. In the context of rotational motion, part of our solution employed integrating small forces over a distance.

In the case of the rotating tube, integral calculus helps sum these infinitesimal centrifugal forces to find the total force at the tube’s end.
  • We set up an integral to account for all forces over the tube's length: \(\int_0^L \rho A \omega^2 x \, dx\).
  • This integral represents the additive effect of forces from countless tiny liquid elements.
  • After integration, we substitute known quantities to express this collectively as a single force: \( F = \frac{M \omega^2 L}{2}\).
The elegance of integral calculus allows us to handle complex problems by breaking them into simpler parts, which we can solve and combine for a complete picture.

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 person wants to drive on the vertical surface of a large cylindrical wooden well commonly known as death well in a circus. The radius of well is \(R\) and the coefficient of friction between the tyres of the motorcycle and the wall of the well is \(\mu_{s}\). The minimum speed, the motorcyclist must have in order to prevent slipping should be : (a) \(\sqrt{\frac{R_{g}}{\mu_{\mathrm{s}}}} \quad\) (b) \(\sqrt{\frac{\mu_{\mathrm{s}}}{\mathrm{Rg}}}\) (c) \(\sqrt{\frac{\mu_{i} g}{R}}\) (d) \(\sqrt{\frac{R}{\mu_{s}}}\)

The angular displacement of the rod is defined as \(\theta=\frac{3}{20} t^{2}\) where \(\theta\) is in radian and \(t\) is in second. The collar \(B\) slides along the rod in such a way that its distance from \(O\) is, \(r=0.9-0.12 t^{2}\) where \(r\) is in metre and \(t\) is in second. The velocity of collar at \(\theta=30^{\circ}\) is: (a) \(0.45 \mathrm{~m} / \mathrm{s}\) (b) \(0.48 \mathrm{~m} / \mathrm{s}\) (c) \(0.52 \mathrm{~m} / \mathrm{s}\) (d) \(0.27 \mathrm{~m} / \mathrm{s}\)

In circular motion: (a) radial acceleration is non-zero (b) radial velocity is zero (c) body is in equilibrium (d) all of the above

The position of a point \(P\) is \(\vec{r}=a \cos \theta \hat{1}+b \sin \theta \hat{j}\), where \(a\) and \(b\) are constants and \(\theta\) is angle between \(r\) and \(x\) -axis. If the rate of increasing of \(\theta\) is \(\omega\), the equation of path of particle is : (a) circle (b) parabola (c) ellipse (d) straight line

In the given figure, the square plate is at rest at position \(A\) at time \(t=0\) and moves as such \(\theta=1.5 t^{2}\), where angular displacement \(\theta\) is in radian and time \(t\) is in second. \(A\) small object \(P\) of mass \(0.4 \mathrm{~kg}\) is temporarily fixed to the plate with adhesive. The adhesive force \(F\) that the adhesive must support at time \(t=3 \mathrm{sec}\) is: (a) \(20 \mathrm{~N}\) (b) \(10 \mathrm{~N}\) (c) \(45.6 \mathrm{~N}\) (d) zero

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.