Problem 44
Question
(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
VerifiedKey Concepts
Differential Equations
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.
Frictional Force
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.
Air Resistance
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.
Velocity Modeling
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.