Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 7

A uniform film of oil \(0.13 \mathrm{~mm}\) thick separates two discs, each of \(200 \mathrm{~mm}\) diameter, mounted co-axially. Ignoring edge effects, calculate the torque necessary to rotate one disc relative to the other at a speed of \(44 \mathrm{rad} \cdot \mathrm{s}^{-1}\) (7 rev/s) if the oil has a dynamic viscosity of \(0.14 \mathrm{~Pa} \cdot \mathrm{s}\).

Short Answer

Expert verified
The required torque is approximately 297.37 N·m.

Step by step solution

01

Understand the Problem

A thin layer of oil separates two discs. To determine the torque required to rotate one disc relative to another, the dynamic viscosity of the oil and the angular velocity are given. Formula for viscous torque in a cylindrical film needs to be applied here.
02

Define Constants and Variables

Given: \(oL = 0.13 \text{ mm} = 0.13 \times 10^{-3} \text{ m}\), \(d = 200 \text{ mm} = 0.2 \text{ m}\), \(\omega = 44 \text{ rad/s}\), \(\mu = 0.14 \text{ Pa} \cdot \text{s}\). Radius \(R\) of the disc is half the diameter, so \(R = \frac{0.2}{2} = 0.1 \text{ m}\).
03

Formula for Viscous Torque

The torque \(T\) in a fluid film can be calculated using the formula: \[ T = 2 \pi \mu \frac{\omega R^3}{h} \] Where \(T\) is the torque, \(\mu\) is the dynamic viscosity, \(\omega\) is the angular velocity, \(R\) is the radius, and \(h\) is the thickness of the oil film.
04

Insert Known Values

Substitute the provided data into the formula: \( \mu = 0.14 \text{ Pa} \cdot \text{s}\), \(\omega = 44 \text{ rad} \cdot \text{s}^{-1}\), \(R = 0.1 \text{ m}\), \(h = 0.13 \times 10^{-3} \text{ m}\)\[ T = 2 \pi (0.14) \frac{(44) (0.1)^3}{0.13 \times 10^{-3}} \]
05

Calculate Torque

Evaluate the expression step-by-step: \[ T = 2 \pi (0.14) \frac{(44) (0.001)}{0.00013} \] \[ T = 2 \pi (0.14) \times 338.46 \] \[ T = 2 \pi (47.38) \] \[ T = (6.28) (47.38) \] \[ T \approx 297.37 \text{ N} \cdot \text{m} \]

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.

Start your free trial

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.

Fluid Dynamics
Fluid dynamics is the study of fluids (liquids and gases) in motion. Its principles explain various phenomena in engineering, meteorology, and natural sciences.
In this context, we are focusing on viscous flow, which involves the internal friction within a fluid when it moves. This friction is caused by the fluid's viscosity. Fluids resist motion more or less depending on their viscosity.
This exercise involves understanding how the thin film of oil between two discs creates this internal friction and how it affects the torque needed to rotate one disc relative to the other.
Dynamic Viscosity
Dynamic viscosity is a measure of a fluid's internal resistance to flow. It's a critical factor in computing the torque required in our exercise.
The dynamic viscosity, denoted by \( \mu \), has units of \( \mathrm{Pa} \cdot \mathrm{s} \) (Pascal-seconds). It essentially tells us how 'thick' or 'thin' a fluid is. Oil typically has a higher viscosity compared to water, meaning it provides more resistance to shear forces.
The given exercise presents a dynamic viscosity of \( 0.14 \mathrm{Pa} \cdot \mathrm{s} \). This value will be used in the torque formula to find the necessary force to rotate the disc.
Angular Velocity
Angular velocity, represented by \( \omega \), quantifies how fast an object rotates or revolves relative to another point, usually the center of a circle.
It is measured in \( \mathrm{rad} \cdot \mathrm{s}^{-1} \) (radians per second). One complete revolution corresponds to \( 2\pi \) radians. So, in our case, if the disc rotates at a speed of \( 44 \mathrm{rad} \cdot \mathrm{s}^{-1} \), it means it completes \( \frac{44}{2\pi} \) revolutions per second.
Angular velocity connects with torque when we consider rotational motion caused by a viscous fluid. The faster the disc rotates, the more torque is required to maintain that motion through the viscous medium.
Torque Formula
Torque is the rotational equivalent of force. In the context of this exercise, it quantifies how much force is necessary to make one disc rotate relative to another through the oil film.
The formula to calculate viscous torque \( T \) in a thin cylindrical film of fluid is given by:
\[ T = 2 \pi \mu \frac{\omega R^3}{h} \]
Here:
- \( T \): Torque
- \( \mu \): Dynamic viscosity
- \( \omega \): Angular velocity
- \( R \): Radius of the disc
- \( h \): Thickness of the oil film
The formula shows the dependencies of torque on each of these variables. By accurately plugging in the given values, we can calculate the required torque needed to rotate the disc through a viscous oil film.

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

Eight kilometres below the surface of the ocean the pressure is 81.7 MPa. Determine the density of sea-water at this depth if the density at the surface is \(1025 \mathrm{~kg} \cdot \mathrm{m}^{-3}\) and the average bulk modulus of elasticity is \(2.34 \mathrm{GPa}\).

Calculate the Reynolds number for a fluid of density \(900 \mathrm{~kg} \cdot \mathrm{m}^{-3}\) and dynamic viscosity \(0.038 \mathrm{~Pa} \cdot \mathrm{s}\) flowing in a \(50 \mathrm{~mm}\) diameter pipe at the rate of \(2.5 \mathrm{~L} \cdot \mathrm{s}^{-1}\). Estimate the mean velocity above which laminar flow would be unlikely.

A liquid of kinematic viscosity \(370 \mathrm{~mm}^{2} \cdot \mathrm{s}^{-1}\) flows through an \(80 \mathrm{~mm}\) diameter pipe at \(0.01 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\). What type of flow is to be expected?

Calculate the density of air when the absolute pressure and the temperature are respectively \(140 \mathrm{kPa}\) and \(50{ }^{\circ} \mathrm{C}\) and \(\mathrm{R}=\) \(287 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\)

At an absolute pressure of \(101.3 \mathrm{kPa}\) and temperature of \(20{ }^{\circ} \mathrm{C}\) the dynamic viscosity of a certain diatomic gas is \(2 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}\) and its kinematic viscosity is \(15 \mathrm{~mm}^{2} \cdot \mathrm{s}^{-1}\). Taking the universal gas constant as \(8310 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\) and assuming the gas to be perfect, calculate its approximate relative molecular mass.
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Translational Dynamics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Fields in Physics

Read Explanation

Famous Physicists

Read Explanation

Work Energy and Power

Read Explanation
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