/*! 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 30 A metal sphere of radius \(1 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

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A metal sphere of radius \(1 \mathrm{~mm}\) and mass \(50 \mathrm{mg}\) falls vertically in glycerine. Find (a) the viscous force exerted by the glycerine on the sphere when the speed of the sphere is \(1 \mathrm{~cm} \mathrm{~s}^{-1}\), (b) the hydrostatic force exerted by the glycerine on the sphere and (c) the terminal velocity with which the sphere will move down without acceleration. Density of glycerine \(=1260 \mathrm{~kg} \mathrm{~m}^{-1}\) and its coefficient of viscosity at room temperature \(=8 \cdot 0\) poise.

Short Answer

Expert verified
(a) Viscous force = \(1.51 \times 10^{-4} \text{ N}\), (b) Hydrostatic force = \(5.18 \times 10^{-5} \text{ N}\), (c) Terminal velocity = \(3.34 \times 10^{-2} \text{ m/s}\).

Step by step solution

01

Convert units for radius and viscosity

The radius of the sphere is given as \(1 \text{ mm}\). Convert it to meters: \(1 \text{ mm} = 0.001 \text{ m}\). The viscosity is given as \(8.0\) poise. To convert to SI units, \(1 \text{ poise} = 0.1 \text{ Ns/m}^2\), so \(8.0 \text{ poise} = 0.8 \text{ Ns/m}^2\).
02

Calculate the viscous force

Using Stokes' law, the viscous force \( F_v \) is given by \( F_v = 6 \pi \eta r v \), where \( \eta \) is the viscosity, \( r \) is the radius, and \( v \) is the velocity.Substituting the values: \( F_v = 6 \times \pi \times 0.8 \times 0.001 \times 0.01 = 1.51 \times 10^{-4} \text{ N}\).
03

Calculate the hydrostatic force

The hydrostatic force is equal to the buoyant force, which can be calculated using Archimedes' principle: \( F_b = \rho V g \), where \( \rho \) is the density of glycerine, \( V \) is the volume of the sphere, and \( g \) is the acceleration due to gravity.The volume of the sphere: \( V = \frac{4}{3} \pi r^3 \). Substituting, \( V = \frac{4}{3} \pi (0.001)^3 = 4.19 \times 10^{-9} \text{ m}^3\).Hence, \( F_b = 1260 \times 4.19 \times 10^{-9} \times 9.8 = 5.18 \times 10^{-5} \text{ N}\).
04

Calculate terminal velocity

At terminal velocity, the net force acting on the sphere is zero. This gives us: viscous force \( F_v \) plus buoyant force \( F_b \) equals weight \( mg \).Hence, \( mg = 6 \pi \eta r v_t + F_b \). Solving for \( v_t \), \( v_t = \frac{mg - F_b}{6 \pi \eta r}\).Mass \( m = 50 \text{ mg} = 50 \times 10^{-6} \text{ kg}\) and weight \( mg = 50 \times 10^{-6} \times 9.8 = 4.9 \times 10^{-4} \text{ N}\).Substitute the values: \( v_t = \frac{4.9 \times 10^{-4} - 5.18 \times 10^{-5}}{6 \pi \times 0.8 \times 0.001} = 3.34 \times 10^{-2} \text{ m/s}\).

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Key Concepts

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

Stokes' Law
Stokes' law is a principle in fluid mechanics that describes the force of viscosity on a spherical object moving through a fluid. It is a critical concept for understanding the behavior of objects in viscous fluids. According to Stokes' law, the viscous force \( F_v \) experienced by the sphere is given by:
  • \( F_v = 6 \pi \eta r v \)
where \( \eta \) represents the fluid's viscosity, \( r \) is the radius of the sphere, and \( v \) is its velocity.
For a small sphere falling through a viscous liquid, this formula allows us to calculate the drag force that opposes its motion. This drag force is directly proportional to the radius and velocity of the sphere, as well as the viscosity of the fluid, meaning that larger spheres or those moving faster through more viscous fluids experience greater drag. This law helps predict how small particles in a liquid will settle or how quickly they will stop moving.
Viscous Force
Viscous force is a type of frictional force that acts against the relative motion of fluid layers or the motion of an object through a fluid. It is a result of the fluid's internal resistance to flow.
When an object moves through a fluid, molecules in close contact with the object surface stick more strongly compared to those farther away. This creates a gradient of velocity within the fluid structure, leading to the viscous force. The Newtonian definition of viscosity captures this resistance as a linear relationship between shear stress and shear rate.
The greater the viscosity of the fluid, such as glycerine in our problem, the larger the force that resists the sphere's motion. This ensures a managed rate of movement, allowing scientists to accurately predict how substances interact within viscous environments. Understanding viscous force is essential in numerous industries, including food processing, pharmaceuticals, and automotive lubrication.
Archimedes' Principle
Archimedes' Principle is a fundamental concept in fluid dynamics, explaining how buoyancy works. It states that any object submerged in a fluid experiences an upward force equal to the weight of the fluid displaced by the object.
In our exercise, when we calculate the buoyant force acting on the sphere, we apply Archimedes' Principle. The buoyant force \( F_b \) is determined by:
  • \( F_b = \rho V g \)
where \( \rho \) is the fluid's density, \( V \) is the volume of the object, and \( g \) is the acceleration due to gravity.
This principle is crucial in knowing how objects float or sink. It is used to measure fluid displacement in ships, submarines, and various aquatic experiments. By understanding the interplay between the force of gravity pulling the object downwards and the buoyant force pushing it upwards, we can predict whether an object will remain afloat or submerge, as well as calculate the terminal velocity where these forces balance.

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Most popular questions from this chapter

Find the increase in pressure required to decrease the volume of a water sample by \(0-01 \%\). Bulk modulus of water \(=2 \cdot 1 \times 10^{3} \mathrm{~N} \mathrm{~m}^{-2}\).

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