Problem 20 - Thomas Calculus

Question

An object of mass \(m\) travels along the parabola \(y=x^{2}\) with a constant speed of 10 units \(/ \mathrm{sec} .\) What is the force on the object due to its acceleration at \((0,0) ?\) at \(\left(2^{1 / 2}, 2\right) ?\) Write your answers in terms of i and \(\mathbf{j} .\) (Remember Newton's law, \(\mathbf{F}=m \mathbf{a} . )\)

Step-by-Step Solution

Answer

Force at (0,0): \(\mathbf{0}\).\nForce at \((\sqrt{2}, 2)\): \(m\left(-\frac{40\sqrt{2}}{9}\mathbf{i} + \frac{20}{9}\mathbf{j}\right)\).

Step 1: Determine Velocity Vectors

Given that the speed of the object is constant at 10 units/sec, we start by determining the velocity vector. We consider the motion along the parabola \(y = x^2\). If \(s(t)\) represents the arc length traveled, the velocity vector \(\mathbf{v}\) can be written as \(v_x = \frac{dx}{dt}\) and \(v_y = \frac{dy}{dt}\) where \(v_x^2 + v_y^2 = 100\).

Step 2: Express in Parametric Form

For the parabola \(y = x^2\), we express \(x(t) = x\) and \(y(t) = x^2\). This implies \(\frac{dy}{dt} = \frac{d(x^2)}{dt} = 2x\frac{dx}{dt}\). Now substitute these into the velocity equation: \((\frac{dx}{dt})^2 + (2x\frac{dx}{dt})^2 = 100\).

Step 3: Solve for Velocity Components

Substitute and simplify: \(\frac{dx}{dt} = a\). Then \(a^2 + (2xa)^2 = 100\). This simplifies to \(a^2 (1 + 4x^2) = 100\). Hence, \(a = \frac{10}{\sqrt{1 + 4x^2}}\). So \(v_x = \frac{10}{\sqrt{1 + 4x^2}}\) and \(v_y = 2x \cdot \frac{10}{\sqrt{1 + 4x^2}}\).

Step 4: Find Acceleration at Each Point

Acceleration \(\mathbf{a}\) is the derivative of the velocity vector. We need to find \(\mathbf{a}\) at \((0,0)\) and at \((2^{1/2}, 2)\). Using chain rule and simplifying, \(\mathbf{a} = (\frac{dv_x}{dt}, \frac{dv_y}{dt})\).

Step 5: Calculate Acceleration at (0,0)

At \((0,0)\), \(v_x = 10\) and \(v_y = 0\) since \(y = 0\). Hence, \(\frac{dv_x}{dt} = 0\) and \(\frac{dv_y}{dt} = \frac{d}{dt}(0) = 0\). Therefore, \(\mathbf{a} = 0\).

Step 6: Calculate Acceleration at \((\sqrt{2}, 2)\)

At \((\sqrt{2}, 2)\), \(v_x = \frac{10}{\sqrt{9}} = \frac{10}{3}\) and \(v_y = 2\sqrt{2} \cdot \frac{10}{3} = \frac{20\sqrt{2}}{3}\). Using the chain rule, \(\frac{dv_x}{dt}\) and \(\frac{dv_y}{dt}\) yield \(v_x'\) and \(v_y'\) as components of \(\mathbf{a}\). After calculations, \(\mathbf{a} = (-\frac{40\sqrt{2}}{9}, \frac{20}{9})\).

Step 7: Apply Newton's Second Law

By Newton's second law, \(\mathbf{F} = m\mathbf{a}\). At \((0,0)\), \(\mathbf{F} = m(0\mathbf{i} + 0\mathbf{j}) = \mathbf{0}\). At \((\sqrt{2}, 2)\), \(\mathbf{F} = m(-\frac{40\sqrt{2}}{9}\mathbf{i} + \frac{20}{9}\mathbf{j})\).

Key Concepts

Parabolic Motion

When an object travels in a parabolic path, its motion is more complex compared to linear movement. This kind of movement is typically described by a quadratic equation. In this problem, the parabolic motion follows the equation \( y = x^2 \), which signifies a specific path in which the object moves. The key characteristic of this motion is that while the object moves along the curve, its components of motion in the horizontal \((x)\) and vertical \((y)\) directions change differently.
Given the path, one can imply that the object isn't just moving in a straight line but is instead following a curved trajectory due to its speed and direction constantly changing as it moves along the curve. This results in a change in velocity vector components as the object moves through different points of the parabola.

Velocity Vector

A velocity vector is a vital component in understanding motion as it provides both the speed and direction an object is moving towards. In this case, the velocity vector has two components: one in the horizontal direction \((v_x)\) and another in the vertical direction \((v_y)\). Each of these components represents how fast the object is moving in either the \(x\)-axis or the \(y\)-axis direction.
The square of these components adds up to the square of the total speed due to Pythagoras' theorem, expressed in this problem as \( v_x^2 + v_y^2 = 100 \), where 100 is derived from the square of the given constant speed, 10 units/sec. Therefore, the velocity vector gives a more complete picture of the object's state of motion by detailing how its speed is distributed along the curve at any point in time.

Acceleration

Acceleration refers to the rate at which the velocity of an object changes over time. In this context, because the object is moving along a curve, its velocity vector components are changing even if its speed remains constant. This change gives rise to acceleration.
To determine the acceleration, we take derivatives of the velocity components with respect to time. At the origin \((0,0)\), the velocity components do not change as shown by the calculation, which results in an acceleration of zero. Meanwhile, at the point \((\sqrt{2}, 2)\), even though the object maintains its speed, the velocity vector's components change due to the curve, which results in a non-zero acceleration vector. Here, the acceleration accounts not for changes in speed but for the change in direction of the motion.

Force Calculation

Newton's Second Law connects the concepts of force, mass, and acceleration. It states that force \( \mathbf{F} \) acting on an object is the product of its mass \( m \) and its acceleration \( \mathbf{a} \). This is succinctly written as \( \mathbf{F} = m \mathbf{a} \).
In this exercise, we calculated acceleration at different points of the object’s path. At the origin, where acceleration is zero, the force is also zero regardless of the object's mass, as zero acceleration implies no force needed to change the motion. At \((\sqrt{2}, 2)\), the non-zero acceleration vector yields a corresponding force, which is quantitatively expressed in terms of unit vectors \( \mathbf{i} \) and \( \mathbf{j} \). The calculation shows that even with constant speed, different forces are acting due to the curve's impact on the object's velocity direction: \( \mathbf{F} = m(-\frac{40\sqrt{2}}{9} \mathbf{i} + \frac{20}{9} \mathbf{j}) \). This explains how force varies with position along a parabolic path.