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Magazine Chapter 4: Problem 62
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.
Short Answer
Expert verified
Position is \(\vec{r}(t)\) and velocity is \(\vec{v}(t)\) functions of time.
Step by step solution
01
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: \(m \frac{d^2 y}{dt^2} = k_3 t\).
02
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 \(y(t) = \frac{k_3}{6m} t^3\).
03
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, \(x(t) = \frac{k_1}{2m}t^2 + \frac{k_2 k_3}{120m^2}t^5\).
04
Write the Position Vector \(\vec{r}(t)\)
Based on the solutions for \(x(t)\) and \(y(t)\), the position vector is given by:\[\vec{r}(t) = \left( \frac{k_1}{2m} t^2 + \frac{k_2 k_3}{120m^2} t^5 \right) \mathbf{i} + \left( \frac{k_3}{6m} t^3 \right) \mathbf{j}\]
05
Write the Velocity Vector \(\vec{v}(t)\)
The velocity vector is determined by taking the derivative of the position vector with respect to time:\[\vec{v}(t) = \left( \frac{k_1}{m} t + \frac{k_2 k_3}{24m^2} t^4 \right) \mathbf{i} + \left( \frac{k_3}{2m} t^2 \right) \mathbf{j}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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: \(m \frac{d^2 y}{dt^2} = k_3 t\).
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: \(y(t) = \frac{k_3}{6m} t^3 + C_1 t + C_2\).
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 \(\frac{dx}{dt}(0) = 0\).
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.
