91Ó°ÊÓ

(a) \((i)\) Using Newton's second law of motion, \(F=m a=m d v / d t,\) the differential equation for the velocity \(v\) is $$m \frac{d v}{d t}=m g \sin \theta \quad \text { or } \quad \frac{d v}{d t}=g \sin \theta$$ where \(m g \sin \theta, 0< \theta< \pi / 2,\) is the component of the weight along the plane in the direction of motion. \((i i)\) The model now becomes $$m \frac{d v}{d t}=m g \sin \theta-\mu m g \cos \theta$$ where \(\mu m g \cos \theta\) is the component of the force of sliding friction (which acts perpendicular to the plane) along the plane. The negative sign indicates that this component of force is a retarding force which acts in the direction opposite to that of motion. \((i i i)\) If air resistance is taken to be proportional to the instantaneous velocity of the body, the model becomes $$m \frac{d v}{d t}=m g \sin \theta-\mu m g \cos \theta-k v$$ where \(k\) is a constant of proportionality. (b) \((i)\) With \(m=3\) slugs, the differential equation is $$3 \frac{d v}{d t}=(96) \cdot \frac{1}{2} \quad \text { or } \quad \frac{d v}{d t}=16$$ Integrating the last equation gives \(v(t)=16 t+c_{1} .\) since \(v(0)=0,\) we have \(c_{1}=0\) and so \(v(t)=16 t\). \((i i)\) With \(m=3\) slugs, the differential equation is $$3 \frac{d v}{d t}=(96) \cdot \frac{1}{2}-\frac{\sqrt{3}}{4} \cdot(96) \cdot \frac{\sqrt{3}}{2} \quad \text { or } \quad \frac{d v}{d t}=4$$ In this case \(v(t)=4 t\). (iii) When the retarding force due to air resistance is taken into account, the differential equation for velocity \(v\) becomes $$3 \frac{d v}{d t}=(96) \cdot \frac{1}{2}-\frac{\sqrt{3}}{4} \cdot(96) \cdot \frac{\sqrt{3}}{2}-\frac{1}{4} v \quad \text { or } \quad 3 \frac{d v}{d t}=12-\frac{1}{4} v$$ The last differential equation is linear and has solution \(v(t)=48+c_{1} e^{-t / 12} .\) since \(v(0)=0,\) we find \(c_{1}=-48,\) so \(v(t)=48-48 e^{-t / 12}\).

Step by step solution

01

Understand Newton's Second Law

Newton's second law states that force is the product of mass and acceleration: \( F = ma = m \frac{dv}{dt} \). For this problem, the force is given as \( mg \sin \theta \), which represents the component of gravitational force acting along the inclined plane.

02

Solve the Basic Motion Equation

From part (a)(i), the differential equation is given by \( \frac{dv}{dt} = g \sin \theta \). This simplifies further when parameters are known.

03

Include Frictional Force

In part (a)(ii), the frictional component \( \mu mg \cos \theta \) is included, leading to the equation \( \frac{dv}{dt} = g \sin \theta - \mu g \cos \theta \). Friction acts against motion along the plane.

04

Model with Air Resistance

In part (a)(iii), consider air resistance proportional to velocity, with equation \( m \frac{dv}{dt} = mg \sin \theta - \mu mg \cos \theta - kv \). Air resistance is a retarding force and thus reduces acceleration.

05

Solve for Specific Mass and Conditions (Without Friction or Air Resistance)

For part (b)(i), with \( m = 3 \) slugs, and frictionless, the equation \( 3 \frac{dv}{dt} = 96 \cdot \frac{1}{2} \) simplifies to \( \frac{dv}{dt} = 16 \). Integrating gives \( v(t) = 16t + c_1 \) and using initial condition \( v(0) = 0 \), \( c_1 = 0 \) gives \( v(t) = 16t \).

06

Add Friction Effects

For part (b)(ii), include friction: \( 3 \frac{dv}{dt} = 96 \cdot \frac{1}{2} - \frac{\sqrt{3}}{4} \cdot 96 \cdot \frac{\sqrt{3}}{2} \) which simplifies to \( \frac{dv}{dt} = 4 \). Integrating gives \( v(t) = 4t \).

07

Air Resistance Considered

For part (b)(iii), with air resistance: \( 3 \frac{dv}{dt} = 12 - \frac{1}{4} v \). Solving this linear differential equation, the solution is \( v(t) = 48 + c_1 e^{-t/12} \). Applying \( v(0) = 0 \), find \( c_1 = -48 \), resulting in \( v(t) = 48 - 48e^{-t/12} \).

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Key Concepts

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

Differential Equations

Differential equations are mathematical tools used to describe systems involving rates of change. In our exercise, we see Newton's second law coming into play as a differential equation. The rate of change of velocity, also known as acceleration, is derived from forces acting on the system. The basic form is represented as \( m \frac{dv}{dt} = F \), where different forces replace \( F \) depending on the situation, such as gravitational force, friction, and air resistance.
This is vital for understanding how an object in motion responds to these forces over time. Differential equations help us create models for the velocity and path of the object, making predictions about future states. Solving these equations involves finding a function, \( v(t) \), which describes the velocity at any time \( t \).

• Gravitational force: affects how \( v \) increases or decreases.
• Friction and air resistance: act to reduce \( v \) over time.

Understanding these key factors can aid in predicting the behavior of moving bodies, leading to insights into the systems they model.

Frictional Force

Frictional force is a crucial concept in physics, often acting as a retarding force against motion. When dealing with objects moving along a plane, such as in our exercise, frictional force is modeled as \( \mu mg \cos \theta \). Here, \( \mu \) is the coefficient of friction, a measure of how much friction is acting, while \( mg \cos \theta \) is the component of the gravitational force perpendicular to the plane.
Frictional force can dramatically alter the motion of objects, slowing them down or stopping them entirely. It's included in Newton's equation as a negative term because it works against the direction of motion. This reduction in velocity must be considered when predicting the behavior and trajectory of moving objects.

• Friction opposes motion, affecting energy and speed.
• It is crucial for stability and control of motion.

Understanding friction helps in designing systems where control over motion is necessary, such as in cars, machinery, and even sports equipment.

Air Resistance

Air resistance, often proportional to the velocity, acts as a force against motion, affecting the object’s speed and energy. In the problem, air resistance is included in the model as \( -kv \), where \( k \) is a constant that represents how much air resistance affects the object.
Air resistance is particularly important for objects moving at high speeds or over long distances. It continually works to slow the object down by converting kinetic energy into another form, often heat. This can greatly affect velocity, making it decrease more quickly than it would under friction alone.

• Resists motion by converting motion energy into other forms.
• Is especially relevant for objects moving through fluids or gases.

Understanding air resistance can help in various fields, from designing efficient vehicles and aircraft to studying the motion of natural phenomena like falling leaves.

Velocity Modeling

Velocity modeling is an application of differential equations that focuses on predicting the speed and direction of movement over time. Using Newton's second law, velocity models consider various forces like gravity, friction, and air resistance to provide a complete picture of how an object will move.
In our exercise, the model changes with each variable added. Initially, gravity alone determines the velocity; then friction and air resistance modify it, offering a more realistic view of the system. The exact form and solution of the differential equation depend on these factors, resulting in equations like \( v(t) = 16t \) and \( v(t) = 48 - 48e^{-t/12} \).

• Informs about changes in motion with time.
• Allows predictions about future behavior and position.

Velocity modeling is essential in engineering, physics, and any field where understanding the dynamics of motion is necessary, offering insights for design and analysis.