Problem 20 - Thomas Calculus

Question

An object of mass \(m\) travels along the parabola \(y=x^{2}\) with a constant speed of 10 units/ 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 j. (Remember Newton's law, \(\mathbf{F}=m \mathbf{a}\).)

Step-by-Step Solution

Answer

Force at (0,0) is \(200m\mathbf{j}\); Force at \((\sqrt{2}, 2)\) is calculated from \(\ddot{x}, \ddot{y}\).

Step 1: Determine velocity components

The object is moving along the parabola \(y = x^2\). Given that the speed \(v\) is constant at 10 units/sec, break down the velocity into components. At any point \((x, y)\), the velocity vector \(\mathbf{v}\) can be expressed in terms of its components \(\dot{x}\) and \(\dot{y}\) where \(\dot{y} = \frac{dy}{dt}\). Since \(y = x^2\), \(\dot{y} = 2x\dot{x}\). Therefore, the speed condition gives: \(v = \sqrt{\dot{x}^2 + (2x\dot{x})^2} = 10\).

Step 2: Solve for velocity component \(\dot{x}\)

We have the equation for speed: \[ 10 = \sqrt{\dot{x}^2 + 4x^2\dot{x}^2} = \dot{x}\sqrt{1 + 4x^2} \]Thus, \(\dot{x} = \frac{10}{\sqrt{1 + 4x^2}}\).

Step 3: Calculate acceleration components

Acceleration is the derivative of velocity. Since we have \(\dot{y} = 2x\dot{x}\), differentiate with respect to time:- \(\ddot{x} = \frac{d}{dt}\left( \frac{10}{\sqrt{1 + 4x^2}} \right)\)- \(\ddot{y} = \frac{d}{dt}(2x\dot{x}) = 2(\dot{x}^2 + x\ddot{x})\)You can use the chain rule and product rule in these calculations.

Step 4: Evaluate accelerations at \((0,0)\)

Substitute \(x = 0\) into the expressions for \(\ddot{x}\) and \(\ddot{y}\).- For \(x = 0\), \(\dot{x} = 10\), so \(\ddot{x} = 0\) (since \(\frac{d}{dt}(10) = 0\)).- \(\ddot{y} = 2(\dot{x}^2) = 2(10^2) = 200\).Thus, the acceleration vector \(\mathbf{a}\) at \((0,0)\) is \(0\mathbf{i} + 200\mathbf{j}\).

Step 5: Evaluate accelerations at \((\sqrt{2}, 2)\)

Substitute \(x = \sqrt{2}\) into the expressions for \(\ddot{x}\) and \(\ddot{y}\).- For \(x = \sqrt{2}\), the velocity component \(\dot{x} = \frac{10}{\sqrt{9}} = \frac{10}{3}\).- Distance derivatives yield \(\ddot{x} = -\frac{40x\dot{x}^2}{\sqrt{1+4x^2}^3}\).- Compute \(\ddot{y} = 2(\dot{x}^2 + x\ddot{x})\) using the above \(\dot{x}\) and calculated \(\ddot{x}\).Evaluate expressions to get the final \(\ddot{x}\) and \(\ddot{y}\) at the point.

Step 6: Calculate force vectors using Newton's Second Law

Newton's Second Law is \(\mathbf{F}=m \mathbf{a}\), where \(\mathbf{a}\) is the acceleration vector.At \((0,0)\), the force is \(\mathbf{F} = m(0\mathbf{i} + 200\mathbf{j}) = 200m\mathbf{j}\).At \((\sqrt{2}, 2)\), use the previously calculated \(\ddot{x}\) and \(\ddot{y}\) to determine \(\mathbf{F} = m(\ddot{x}\mathbf{i} + \ddot{y}\mathbf{j})\). Calculate values to get the force vector.

Key Concepts

Parabola Motion

When an object moves along a parabola, its path follows the equation of a quadratic curve. For instance, in the exercise where the object travels along the parabola \(y = x^2\), the path can be visually pictured as a U-shaped curve. Parabolic motion is a common occurrence in physics, often seen in projectile motion, where the trajectory of an object resembles a parabola.

An essential characteristic of parabola motion is that while the object moves, its speed can remain constant while its direction constantly changes. Constant speed means that the magnitude of velocity stays the same, but the vector direction alters, revealing a change in velocity components often influenced by gravity or other forces. This is why understanding parabolic motion is crucial; it allows us to explore how objects behave under specific motion constraints and the forces acting upon them.

Velocity Components

When analyzing motion, particularly on a curved path like a parabola, we must consider the object's velocity components. Velocity denotes speed and direction, and these components break down the velocity into two parts: horizontal and vertical directions.

In the context of a parabola motion, the horizontal component (\(\dot{x}\)) and vertical component (\(\dot{y}\)) of velocity can change even if the overall speed remains constant. For the parabola \(y = x^2\), \(\dot{y}\) depends on \(x\) because \(\dot{y} = 2x\dot{x}\). This relation shows that as \(x\) changes, so does \(\dot{y}\).

The solution to the exercise highlights the importance of determining these components. By simplifying the speed to remain constant at 10 units/sec, it is possible to calculate \(\dot{x}\) as \(\dot{x} = \frac{10}{\sqrt{1 + 4x^2}}\). With precise calculations, one can understand how the moving object's velocity changes as it traverses different points on the parabola.

Acceleration Components

Acceleration is the rate of change of velocity with time. Given a velocity function, finding the acceleration involves differentiating that function. In the case of an object moving along a parabola, both \(\ddot{x}\) (the horizontal acceleration) and \(\ddot{y}\) (the vertical acceleration) must be identified.

For the parabola \(y = x^2\), calculating acceleration components requires differentiating the velocity components. The rate of change of the horizontal velocity, \(\ddot{x}\), is derived from differentiating \(\dot{x}\). Similarly, the vertical acceleration \(\ddot{y} = 2(\dot{x}^2 + x\ddot{x})\), is noticed by applying the product and chain rules during differentiation.

Once we compute these derivatives, we gather information about the forces acting at specific points on the parabolic path. Using Newton's Second Law \(\mathbf{F} = m \mathbf{a}\), with \(\mathbf{a}\) being the vector inclusive of \(\ddot{x}\) and \(\ddot{y}\), provides insights into the object's dynamics. As illustrated in the solution, these acceleration components are key in understanding the force vectors exerted on the object at points like \((0,0)\) and \((\sqrt{2}, 2)\).