Problem 62 - University Physics with Modern Physics

Question

An object of mass \(m\) is at rest in equilibrium at the origin. At \(t=0\) a new force \(\vec{F}(t)\) is applied that has components $$ F_{x}(t)=k_{1}+k_{2} y \quad F_{y}(t)=k_{3} t $$ where \(k_{1}, k_{2},\) and \(k_{3}\) are constants. Calculate the position \(\vec{r}(t)\) and velocity \(\vec{v}(t)\) vectors as functions of time.

Step-by-Step Solution

Answer

Position is \(\vec{r}(t)\) and velocity is \(\vec{v}(t)\) functions of time.

Step 1: Identify the Forces and Write Equations of Motion

The forces acting on the object are given as \(F_x(t) = k_1 + k_2 y\) and \(F_y(t) = k_3 t\). According to Newton's second law, \(m \frac{d^2 x}{dt^2} = F_x(t)\) and \(m \frac{d^2 y}{dt^2} = F_y(t)\). This provides a system of differential equations.For the x-direction: \(m \frac{d^2 x}{dt^2} = k_1 + k_2 y\).For the y-direction: \(m \frac{d^2 y}{dt^2} = k_3 t\).

Step 2: Solve the y-direction Differential Equation

We solve \(m \frac{d^2 y}{dt^2} = k_3 t\). Integrating with respect to \(t\) gives \(m \frac{dy}{dt} = \frac{k_3}{2} t^2 + C_1\), where \(C_1\) is an integration constant. Integrating again gives \(y(t) = \frac{k_3}{6m} t^3 + C_1 t + C_2\), where \(C_2\) is another integration constant. Since the object starts at rest, \(y(0) = 0\) and \(\frac{dy}{dt}(0) = 0\) both give \(C_1 = 0\) and \(C_2 = 0\), hence \(y(t) = \frac{k_3}{6m} t^3\).

Step 3: Solve the x-direction Differential Equation

Now, substitute \(y(t) = \frac{k_3}{6m} t^3\) into the x-direction equation: \(m \frac{d^2 x}{dt^2} = k_1 + k_2 \frac{k_3}{6m} t^3\). Simplifying gives \(\frac{d^2 x}{dt^2} = \frac{k_1}{m} + \frac{k_2 k_3}{6m^2} t^3\). Integrating with respect to \(t\), we find \(\frac{dx}{dt} = \frac{k_1}{m}t + \frac{k_2 k_3}{24m^2} t^4 + C_3\), where \(C_3\) is an integration constant. With initial condition \(\frac{dx}{dt}(0) = 0\), we find \(C_3 = 0\). Integrating again gives \(x(t) = \frac{k_1}{2m}t^2 + \frac{k_2 k_3}{120m^2}t^5 + C_4\). With \(x(0) = 0\), \(C_4 = 0\). Hence, \(x(t) = \frac{k_1}{2m}t^2 + \frac{k_2 k_3}{120m^2}t^5\).

Step 4: Write the Position Vector \(\vec{r}(t)\)

Based on the solutions for \(x(t)\) and \(y(t)\), the position vector is given by:\[\vec{r}(t) = \left( \frac{k_1}{2m} t^2 + \frac{k_2 k_3}{120m^2} t^5 \right) \mathbf{i} + \left( \frac{k_3}{6m} t^3 \right) \mathbf{j}\]

Step 5: Write the Velocity Vector \(\vec{v}(t)\)

The velocity vector is determined by taking the derivative of the position vector with respect to time:\[\vec{v}(t) = \left( \frac{k_1}{m} t + \frac{k_2 k_3}{24m^2} t^4 \right) \mathbf{i} + \left( \frac{k_3}{2m} t^2 \right) \mathbf{j}\]

Key Concepts

Differential Equations

In this exercise, the movement of an object is described using Newton's second law. This involves differential equations that connect forces to the motion. A differential equation involves derivatives of a function, showing how a change in one quantity affects another. The forces acting on the object are:

• For the x-direction: \(m \frac{d^2 x}{dt^2} = k_1 + k_2 y\).
• For the y-direction: \(m \frac{d^2 y}{dt^2} = k_3 t\).

These equations are classified as second-order because they involve second derivatives (\(\frac{d^2 x}{dt^2}\) and \(\frac{d^2 y}{dt^2}\)). Each direction's equation models how the respective component of force influences the acceleration of the object. Solving these differential equations helps us find the position and velocity of the object over time.

Integration

Integration is a crucial process in solving differential equations. It allows us to go backward from the motion's acceleration to velocity and then to position. In this task, integration is applied twice to find solutions.

• First, you integrate the acceleration to get the velocity. For example, for the y-direction: \( m \frac{d^2 y}{dt^2} = k_3 t \) becomes \( m \frac{dy}{dt} = \frac{k_3}{2} t^2 + C_1 \).
• Then, you integrate the velocity to find the position: \(y(t) = \frac{k_3}{6m} t^3 + C_1 t + C_2\).

Repeated integration helps solve the system by connecting derivatives of different orders, eventually expressing them in terms of the original variable \(t\). Integration constants \(C_1\) and \(C_2\) appear because integration introduces unknowns that depend on initial conditions.

Initial Conditions

Initial conditions describe the object's initial state and are vital for solving differential equations. They allow you to determine the integration constants introduced during integration. Without them, the solutions remain incomplete.In the exercise, the object starts at rest at the origin, stated as:

• \(y(0) = 0\) and \(\frac{dy}{dt}(0) = 0\) for the y-component.
• For the x-component, similar conditions apply: \(x(0) = 0\) and \(\frac{dx}{dt}(0) = 0\).

These initial conditions simplify the solutions for \(y(t)\) and \(x(t)\), helping to find the constants \(C_1\), \(C_2\), \(C_3\), and \(C_4\). Correctly applying initial conditions ensures the solutions align with the physical setup of the problem, representing real-world scenarios accurately.

Physics Problem Solving

Physics problem solving combines math and physics principles to find solutions. Newton's second law forms the core of this exercise, relating force, mass, and acceleration through differential equations. By understanding the forces and their components, we translate a complex physical situation into solvable equations. Here are steps to tackle physics problems effectively:

• Understand the forces and algebraically express them.
• Translate these into mathematical models (differential equations).
• Use integration to solve these equations for position and velocity.
• Apply initial conditions to refine and complete the solution.

By following these steps, we connect theoretical concepts to practical outcomes, gaining insight into the object's motion due to applied forces in a methodical way. This structured approach equips students with tools to solve similar problems in physics.