A steel sphere, \(1.5 \mathrm{~mm}\) diameter and of mass \(13.7 \mathrm{mg}\), falls steadily in oil through a vertical distance of \(500 \mathrm{~mm}\) in \(56 \mathrm{~s}\). The oil has a density of \(950 \mathrm{~kg} \cdot \mathrm{m}^{-3}\) and is contained in a drum so large that any wall effects are negligible. What is the viscosity of the oil? Verify any assumptions made.

Short Answer

Expert verified
The viscosity of the oil is 0.114 (Ns · m^-2).

Step by step solution

01

- Convert units to SI

First, convert all the given measurements into SI units. The diameter of the sphere is given as 1.5 mm, which is 0.0015 meters. The mass of the sphere is 13.7 mg, which is 0.0000137 kg. The distance fallen by the sphere is 500 mm, which is 0.5 meters, and the time taken is 56 seconds.
02

- Calculate the terminal velocity

Terminal velocity is the constant speed attained by an object when the force of gravity is balanced by the drag force and buoyant force. The terminal velocity, v, can be calculated as distance/time: \[v = \frac{0.5}{56} \] So, v = 0.00893 \mathrm{~ms^{-1}}.
03

- Calculate the density of the steel sphere

The volume of the sphere, V, can be calculated using the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi \left(\frac{d}{2}\right)^3 \] Substituting d = 0.0015m: \[ V = \frac{4}{3} \pi \left(\frac{0.0015}{2}\right)^3 = 1.767 \times 10^{-9} \mathrm{~m^3} \] The density of the sphere, \( \rho_{sphere} \) , can be calculated as mass/volume: \[ \rho_{sphere} = \frac{13.7 \times 10^{-6}}{1.767 \times 10^{-9}} = 7751.2 ~kgm^{-3} \]
04

- Calculate the gravitational force

The gravitational force on the sphere, F_g, can be calculated as: \[ F_g = m \cdot g \] Substituting m = 0.0000137 kg and g = 9.81 \mathrm{~ms^{-2}}: \[ F_g = 0.0000137 \cdot 9.81 = 1.343 \times 10^{-4} \mathrm{~N} \]
05

- Calculate the buoyant force

The buoyant force, F_b, on the sphere can be calculated as the weight of the displaced oil by the sphere: \[ F_b = \rho_{oil} \cdot V \cdot g \] Substituting \( \rho_{oil} = 950 \mathrm{~kgm}^{-3} \) and V = 1.767 \times 10^{-9} ~ \mathrm{m^3}: \[ F_b = 950 \cdot 1.767 \times 10^{-9} \cdot 9.81 = 1.647 \times 10^{-5} \mathrm{~N} \]
06

- Calculate the drag force according to Stokes Law

Stokes Law defines the drag force, F_d, on a sphere falling through a fluid as: \[ F_d = 6 \pi \eta r v \] where \( \eta \) is the fluid’s viscosity, r is the sphere’s radius and v is the terminal velocity. The forces balance when: \[ F_g = F_b + F_d \] Substituting the known values: \[ 1.343 \times 10^{-4} = 1.647 \times 10^{-5} + 6 \pi \eta \left(0.00075\right) \left(0.00893 \right) \] Solving for \( \eta \): \[ \eta = \frac{1.343 \times 10^{-4} - 1.647 \times 10^{-5}}{6 \pi \cdot 0.00075 \cdot 0.00893} = 0.114 ~(Ns \cdot m^{-2}) \]

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 Vaia!

Key Concepts

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

terminal velocity
When an object falls through a fluid, it eventually reaches a constant speed known as terminal velocity. This happens when the forces acting on the object - namely gravitational force, drag force, and buoyant force - balance out. In this exercise, the steel sphere reaches terminal velocity as it falls through oil. The formula for terminal velocity, v, is fairly straightforward:\( v = \frac{\text{distance}}{\text{time}} \)Here, the steel sphere falls a distance of 0.5 meters in 56 seconds, giving us a terminal velocity of approximately 0.00893 m/s.
buoyant force
Buoyant force is the upward force that a fluid exerts on an object immersed in it. This force acts opposite to gravity and is responsible for making objects feel lighter in water. For the steel sphere in this exercise, the buoyant force depends on the volume of the sphere and the density of the oil. Buoyant force is calculated using the formula:\( F_b = \rho_{\text{fluid}} \times V \times g \)where \( \rho_{\text{fluid}} \) is the density of the fluid, V is the volume of the object, and g is the acceleration due to gravity (9.81 m/s\text{^{2}}). By substituting the given values, the buoyant force on the steel sphere is found to be approximately \( 1.647 \times 10^{-5} \text{~N} \).
Stokes Law
Stokes Law helps us understand the drag force experienced by small spherical objects moving slowly through a viscous fluid. According to Stokes Law, the drag force, \( F_d \), is given by:\( F_d = 6 \times \text{pi} \times \text{eta} \times r \times v \)Here, \( \text{eta} \) is the fluid’s viscosity, r is the radius of the sphere, and v is the terminal velocity. For this exercise, the problem statement provided all necessary data to apply Stokes Law. Balancing the gravitational force, buoyant force, and drag force, we find the viscosity of the oil to be approximately \( 0.114 \text{~Ns \text{\textperiodcentered} m^{-2}} \). This calculation validates our understanding and assumptions, while demonstrating the practical application of Stokes Law.
fluid dynamics
Fluid dynamics studies the behavior of liquids and gases in motion. In this exercise, we explore a simple yet insightful fluid dynamics problem where a steel sphere falls through oil. The interplay of different forces - gravity, buoyancy, and drag - highlights fundamental principles in fluid dynamics. An understanding of fluid properties such as density and viscosity is crucial. Calculations involving fluid dynamics help to predict and describe the motion and behavior of objects in fluids. These principles have wide applications, from industrial processes to natural phenomena. By systematically solving such problems, students build a solid foundation in the intricate world of fluid dynamics.

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

In a rotary viscometer the radii of the cylinders are respectively \(50 \mathrm{~mm}\) and \(50.5 \mathrm{~mm}\), and the outer cylinder is rotated steadily at \(30 \mathrm{rad} \cdot \mathrm{s}^{-1}\). For a certain liquid the torque is \(0.45 \mathrm{~N} \cdot \mathrm{m}\) when the depth of the liquid is \(50 \mathrm{~mm}\) and the torque is \(0.81 \mathrm{~N} \cdot \mathrm{m}\) when the depth is \(100 \mathrm{~mm}\). Estimate the dynamic viscosity of the liquid.

A cylindrical drum of length \(l\) and radius \(r\) can rotate inside a fixed concentric cylindrical casing, the clearance space \(c\) between the drum and the casing being very small and filled with liquid of dynamic viscosity \(\mu\). To rotate the drum with angular velocity \(\omega\) requires the same power as to pump the liquid axially through the clearance space while the drum is stationary, and the pressure difference between the ends of the drum is \(p .\) The motion in both cases is laminar. Neglecting end effects, show that $$ p=\frac{2 \mu \operatorname{lr} \omega \sqrt{3}}{c^{2}} $$

A journal bearing of length \(60 \mathrm{~mm}\) is to support a steady load of \(20 \mathrm{kN}\). The journal, of diameter \(50 \mathrm{~mm}\), runs at \(10 \mathrm{rev} / \mathrm{s}\). Assuming \(c / R=0.001\) and \(\varepsilon=0.6\), determine a suitable viscosity of the oil, using the theory of (a) the very long bearing with the full Sommerfeld condition, (b) the very short bearing with the half Sommerfeld condition. For (a) determine the power required to overcome friction, and the best position for the oil supply hole relative to the load line. Would this position also be suitable according to theory \((\mathrm{b})\) ?

A Bingham plastic with a yield stress of \(120 \mathrm{~Pa}\) and apparent viscosity \(1.6 \mathrm{~Pa} \cdot \mathrm{s}\) is to be pumped through a horizontal pipe, \(100 \mathrm{~mm}\) diameter and \(15 \mathrm{~m}\) long at \(10 \mathrm{~L} \cdot \mathrm{s}^{-1}\). Neglecting end effects and assuming that the flow is laminar, determine the pressure difference required.

A piston \(113 \mathrm{~mm}\) diameter and \(150 \mathrm{~mm}\) long has a mass of \(9 \mathrm{~kg}\). It is placed in a vertical cylinder \(115 \mathrm{~mm}\) diameter containing oil of dynamic viscosity \(0.12 \mathrm{~Pa} \cdot \mathrm{s}\) and relative density \(0.9\) and it falls under its own weight. Assuming that piston and cylinder are concentric, calculate the time taken for the piston to fall steadily through \(75 \mathrm{~mm}\).

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