Motion under a viscous force (a) A particle of mass M acted...

Motion under a viscous force (a) A particle of mass $M$ acted on by only the viscous force of the medium $-b v$ is projected from a point with velocity $v_{0}$. Write down the velocity as a function of time. Remembering that $v=d x / d t$, find $x(t) .$ If $M=10 \mathrm{~g}$ $b=4.0 \mathrm{dyn}-\mathrm{s} / \mathrm{cm}$, and $v_{0}=100 \mathrm{~cm} / \mathrm{s}$, find the distance the mass will travel. (b) In the Millikan oil-drop experiment some drops had radius $2.0 \times 10^{-4} \mathrm{~cm} .$ The density of the oil was $0.92 \mathrm{~g} / \mathrm{cc}$, and the viscosity of air was $1.8 \times 10^{-2}$ centipoises. Find the relaxation time and the terminal velocity. Neglect the buoyant correction.

Motion under a viscous force
(a) A particle of mass $M$ acted on by only the viscous force of the medium $-b v$ is projected from a point with velocity $v_{0}$. Write down the velocity as a function of time. Remembering that $v=d x / d t$, find $x(t) .$ If $M=10 \mathrm{~g}$ $b=4.0 \mathrm{dyn}-\mathrm{s} / \mathrm{cm}$, and $v_{0}=100 \mathrm{~cm} / \mathrm{s}$, find the distance
the mass will travel.
(b) In the Millikan oil-drop experiment some drops had radius $2.0 \times 10^{-4} \mathrm{~cm} .$ The density of the oil was $0.92 \mathrm{~g} / \mathrm{cc}$, and the viscosity of air was $1.8 \times 10^{-2}$ centipoises. Find the relaxation time and the terminal velocity. Neglect the buoyant correction.

Key Concepts

Relaxation Time

The relaxation time is defined as the characteristic time over which the velocity of a system undergoing exponential decay decreases by a factor of 1/e. In viscous motion, it is given by the ratio of the mass to the drag coefficient, serving as a measure of how rapidly the particle’s motion is damped by the viscous force.

Terminal Velocity

Terminal velocity is the steady-state speed reached by an object falling through a fluid when the gravitational force is balanced by the resistive viscous drag. In scenarios such as the Millikan oil-drop experiment, this concept is used to describe the constant speed a small particle attains when the net force acting on it becomes zero.

Integration to Determine Displacement

Once the time-dependent velocity is determined, integrating this velocity function with respect to time provides the displacement of the particle. This integration step is essential to connect the kinematics of the particle (velocity) with its trajectory (position) in a viscously damped motion.

Equations of Motion under Linear Drag

When the only force acting on a particle is a linear drag that is proportional to its velocity, the resulting equation of motion is a first?order linear differential equation. Solving this yields an exponentially decaying velocity function, illustrating how the object's speed decreases over time due to the dissipative effects of the medium.

Viscous Drag Force

The viscous drag force is a resistive force experienced by an object moving through a fluid, which in the linear regime is proportional to its velocity and acts in the opposite direction. This concept is crucial in analyzing the motion of particles in viscous fluids and forms the basis for formulating the equations of motion in these scenarios.

Exponential Decay of Velocity

In the context of viscous damping, the differential equation governing the motion leads to a solution where the velocity decays exponentially with time. The decay constant in the exponential, typically expressed as the ratio of the drag coefficient to mass, characterizes how quickly the velocity decreases to a fraction of its initial value.

Transcript

Please give Ace some feedback