Reference Frame Change
Reference frame change in planar kinematics involves analyzing motion from different observational perspectives, accounting for how the position, velocity, and acceleration of a point or rigid body appear in one frame relative to another.
Let’s consider a generic situation involving the velocities and acceleration of a point relative to two distinct frames or bodies. Take a point P, observed in two frames: the motion of P can be described relative to the body relative to either frames.
We will referring to one frame as the Fixed Frame, and the other as the Moving Frame. In this case as well, the term “Fixed Frame” doesn’t necessarily mean it is fixed in an inertial reference frame. It would only need to be fixed in an inertial reference frame if we were analyzing kinetics and applying Newton’s or Euler’s laws. For now, we’re focused on kinematics, dealing solely with the geometry of motion. So these frames simply represent two distinct frames.
In cases where multiple bodies are connected, we can analyze their kinematics by linking two frames at a time, transferring vectors between them. This approach becomes useful when studying the motion of a single point in multiple frames or bodies.
Unit vectors
The unit vectors of Frame 2 can be expressed in terms of Frame 1’s unit vectors using trigonometric relationships:
\begin{align*} \mathbf{i}_2 &= \cos(\theta) \, \mathbf{i}_1 + \sin(\theta) \, \mathbf{j}_1 \\ \mathbf{j}_2 &= -\sin(\theta) \, \mathbf{i}_1 + \cos(\theta) \, \mathbf{j}_1 \end{align*}
Since frame 2 is rotating relative to frame 1, the unit vectors \mathbf{i}_2 and \mathbf{j}_2 are functions of time. We’ll differentiate these vectors with respect to time t in frame 1.
The derivative of \mathbf{i}_2 with respect to t is:
\begin{aligned} \frac{\mathrm d\mathbf{i}_2}{\mathrm dt} & = \frac{\mathrm d}{\mathrm dt} \left( \cos(\theta) \, \mathbf{i}_1 + \sin(\theta) \, \mathbf{j}_1 \right) \\ & = -\sin(\theta) \, \dot{\theta} \, \mathbf{i}_1 + \cos(\theta) \, \dot{\theta} \, \mathbf{j}_1 \\ & = \dot{\theta} \left( -\sin(\theta) \, \mathbf{i}_1 + \cos(\theta) \, \mathbf{j}_1 \right) \\ & = \dot{\theta} \, \mathbf{j}_2 \end{aligned}
The derivative of \mathbf{j}_2 with respect to t is:
\begin{aligned} \frac{\mathrm d\mathbf{j}_2}{\mathrm dt} & = \frac{\mathrm d}{\mathrm dt} \left( -\sin(\theta) \, \mathbf{i}_1 + \cos(\theta) \, \mathbf{j}_1 \right) \\ & = -\cos(\theta) \, \dot{\theta} \, \mathbf{i}_1 - \sin(\theta) \, \dot{\theta} \, \mathbf{j}_1 \\ & = \dot{\theta} \left( -\cos(\theta) \, \mathbf{i}_1 - \sin(\theta) \, \mathbf{j}_1 \right) \\ & = -\dot{\theta} \, \mathbf{i}_2 \end{aligned}
If multiple bodies are connected and each has its own reference frame, it is possible to analyze their kinematics by linking two frames at a time. By systematically transferring vectors between adjacent frames, it is possible to describe the motion of a single point across all frames involved.
Vector derivative formula
Let’s now consider a vector \mathbf{V} expressed in two different reference frames.
In the context of kinematics, \mathbf{V} can represent various vector quantities such as position, velocity, or acceleration. We’ll express \mathbf{V} in Frame 2, which is our moving frame.
Let’s denote the unit vectors of Frame 2 as \mathbf{i}_2 and \mathbf{j}_2, and the coordinates as x_2 and y_2. Then, the vector \mathbf{V} can be expressed in Frame 2 as:
\mathbf{V} = V_{x_2} \, \mathbf{i}_2 + V_{y_2} \, \mathbf{j}_2
The objective is is to differentiate \mathbf{V} with respect to time t as observed from Frame 1. Since Frame 2 is moving relative to Frame 1, both the components V_{x_2} and V_{y_2}, as well as the unit vectors \mathbf{i}_2 and \mathbf{j}_2, may change with time.
Using the product rule for differentiation, the derivative of \mathbf{V} with respect to Frame 1 is:
\begin{aligned} \frac{\mathrm d\mathbf{V}}{\mathrm dt} \bigg|_1 & = \frac{\mathrm dV_{x_2}}{\mathrm dt} \, \mathbf{i}_2 + V_{x_2} \, \frac{\mathrm d\mathbf{i}_2}{\mathrm dt} + \frac{\mathrm dV_{y_2}}{\mathrm dt} \, \mathbf{j}_2 + V_{y_2} \, \frac{\mathrm d\mathbf{j}_2}{\mathrm dt} \\ & = \left( \frac{\mathrm dV_{x_2}}{\mathrm dt} \, \mathbf{i}_2 + \frac{\mathrm dV_{y_2}}{\mathrm dt} \, \mathbf{j}_2 \right) + \left( V_{x_2} \, \frac{\mathrm d\mathbf{i}_2}{\mathrm dt} + V_{y_2} \, \frac{\mathrm d\mathbf{j}_2}{\mathrm dt} \right) \end{aligned}
The first term presents the derivative of \mathbf{V} in frame 2 and the second term accounts for the rotation of frame 2 relative to frame 1.
From the previous derivation, we have the time derivatives of the unit vectors in frame 2 with respect to frame 1 and substituting these into the second term:
\begin{aligned} V_{x_2} \, \frac{\mathrm d\mathbf{i}_2}{\mathrm dt} + V_{y_2} \, \frac{\mathrm d\mathbf{j}_2}{\mathrm dt} & = V_{x_2} \, \dot{\theta} \, \mathbf{j}_2 - V_{y_2} \, \dot{\theta} \, \mathbf{i}_2 \\ & = \dot{\theta} \left( V_{x_2} \, \mathbf{j}_2 - V_{y_2} \, \mathbf{i}_2 \right) \\ & = \dot{\theta} \, \mathbf{k} \times \mathbf{V} \end{aligned}
As the expression can be represented as the cross product with the unit vector \mathbf{k} (perpendicular to the plane of motion).
Combining the two terms, the derivative of \mathbf{V} with respect to Frame 1 is:
\frac{\mathrm d\mathbf{V}}{\mathrm dt} \bigg|_1 = \frac{\mathrm d\mathbf{V}}{\mathrm dt} \bigg|_2 + \dot{\theta} \, \mathbf{k} \times \mathbf{V}
This is the derivative formula, which is used in transforming the derivative of a vector from a moving frame to a fixed frame.
Velocity
Let’s begin by expressing the position vector. We’ll consider the position from point O (the origin of frame 1) to point P as follows:
\mathbf{r}_{O_1P} \big|_1 = \mathbf{r}_{O_1O_2} \big|_1 + \mathbf{r}_{O_2P} \big|_2
To determine the velocity, we need to differentiate the position vector with respect to time. Applying differentiation, we obtain:
\frac{\mathrm{d}\mathbf{r}_{O_1P}}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{r}_{O_1O_2}}{\mathrm{d}t} \bigg|_1 + \frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \bigg|_1
The left side represents the velocity of point \mathbf P with respect to frame 1; the first term on the right side represents the velocity of O_2 (the origin of frame 2) with respect to frame 1 and the second term on the right represents the derivative of \mathbf{r}_{O_2P}, which is a vector expressed in the moving frame 2. We need to differentiate this vector with respect to frame 1.
Differentiating \mathbf{r}_{O_2P} \big|_2 with respect to frame 1 requires attention because frame 2 is rotating relative to frame 1. To handle this, we use the derivative formula derived previously:
\frac{\mathrm{d}\mathbf{V}}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{V}}{\mathrm{d}t} \bigg|_2 + \boldsymbol{\omega} \times \mathbf{V}
Applying this formula to \mathbf{r}_{O_2P} \big|_2:
\frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \bigg|_2 + \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2
Substituting this back into our original velocity equation:
\mathbf{v}_P \big|_1 = \mathbf{v}_{O_2} \big|_1 + \mathbf{v}_P \big|_2 + \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2
where:
- \mathbf{v}_P \big|_1: velocity of point \mathbf P with respect to frame 1,
- \mathbf{v}_{O_2} \big|_1: velocity of the origin of frame 2 (O_2) with respect to frame 1,
- \mathbf{v}_P \big|_2: velocity of point \mathbf P with respect to frame 2,
- \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2: term accounting for the rotation of frame 2 relative to frame 1.
Rearranging the term and dropping some indexes this formula can be written as:
\mathbf{v}_P = \mathbf{v}_{O_2} + \mathbf{v}_{\text{rel}} + \boldsymbol{\omega} \times \mathbf{r}
where:
- \mathbf{v}_P: absolute velocity of point \mathbf P with respect to frame 1,
- \mathbf{v}_{O_2}: absolute velocity of the origin of frame 2 (O_2) with respect to frame 1,
- \mathbf{v}_{\text{rel}}: relative velocity of point \mathbf P with respect to frame 2,
- \boldsymbol{\omega} \times \mathbf{r}: angular velocity of frame 2 relative to frame 1 crossed with the position of point \mathbf P in the moving frame.
Assuming planar (two-dimensional) motion, the angular velocity vector \boldsymbol{\omega} can be expressed as \omega \, \mathbf{k} (or \dot \theta \, \mathbf k), where \mathbf{k} is the unit vector perpendicular to the plane of motion, the equation becomes:
\begin{aligned} \mathbf{v}_P & = \mathbf{v}_{O_2} + \mathbf{v}_{\text{rel}} + \omega \, \mathbf k \times \mathbf{r} \\ & = \mathbf{v}_{O_2} + \mathbf{v}_{\text{rel}} + \dot \theta \, \mathbf k \times \mathbf{r} \end{aligned}
Acceleration
To find the acceleration of point \mathbf P relative to the two frames, we’ll differentiate the velocity equation with respect to time in frame 1:
\begin{aligned} & \frac{\mathrm{d}\mathbf{v}_P}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{v}_{O_2}}{\mathrm{d}t} \bigg|_1 + \frac{\mathrm{d}\mathbf{v}_P}{\mathrm{d}t} \bigg|_1 + \frac{\mathrm{d}}{\mathrm{d}t} \left( \boldsymbol{\omega} \times \mathbf{r}_{O_2P} \right) \bigg|_1 \\ & \mathbf{a}_P \big|_1 = \mathbf{a}_{O_2} \big|_1 + \frac{\mathrm{d}\mathbf{v}_P}{\mathrm{d}t} \bigg|_1 + \frac{\mathrm{d}}{\mathrm{d}t} \left( \boldsymbol{\omega} \times \mathbf{r}_{O_2P} \right) \bigg|_1 \end{aligned}
In this formula:
- \frac{\mathrm{d}\mathbf{v}_P}{\mathrm{d}t} \big|_1 represents the acceleration of point P with respect to frame 1 (\mathbf{a}_P \big|_1),
- \frac{\mathrm{d}\mathbf{v}_{O_2}}{\mathrm{d}t} \big|_1 represents the acceleration of the origin of frame 2 (\mathbf{a}_{O_2} \big|_1),
- \frac{\mathrm{d}\mathbf{v}_P}{\mathrm{d}t} \big|_1 requires careful differentiation since \mathbf{V}_P \big|_2 is expressed in the moving frame 2,
- \frac{\mathrm{d}}{\mathrm{d}t} \left( \boldsymbol{\omega} \times \mathbf{r}_{O_2P} \right) \big|_1 is the differentiation of the rotational term involving angular velocity and position vector.
we use the derivative formula derived previously:
\frac{\mathrm{d}\mathbf{V}}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{V}}{\mathrm{d}t} \bigg|_2 + \boldsymbol{\omega} \times \mathbf{V}
Applying this formula to \mathbf{v}_P \big|_2:
\frac{\mathrm{d}\mathbf{v}_P}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{v}_P}{\mathrm{d}t} \bigg|_2 + \boldsymbol{\omega} \times \mathbf{v}_P\big|_2 = \mathbf{a}_P \big|_2 + \boldsymbol{\omega} \times \mathbf{v}_P\big|_2
Differentiating \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2 and using \boldsymbol{\alpha} for the angular acceleration of frame 2 relative to frame 1:
\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \left( \boldsymbol{\omega} \times \mathbf{r}_{O_2P} \right)\bigg|_1 & = \frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} \bigg|_1 \times \mathbf{r}_{O_2P}\big|_1 + \boldsymbol{\omega} \times \frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \bigg|_1 \\ & = \boldsymbol{\alpha} \times \mathbf{r}_{O_2P}\big|_1 + \boldsymbol{\omega} \times \frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \bigg|_1 \end{aligned}
For \frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \big|_1 we use again the derivative formula and a result previously derived:
\frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{r}_{O_2P}}{\mathrm{d}t} \bigg|_2 + \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2 = \mathbf{v}_P \big|_2 + \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2
Substituting:
\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \left( \boldsymbol{\omega} \times \mathbf{r}_{O_2P} \right) \bigg|_1 & = \boldsymbol{\alpha} \times \mathbf{r}_{O_2P}\big|_1 + \boldsymbol{\omega} \times \left( \mathbf{v}_P \big|_2 + \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2 \right) \\ & = \boldsymbol{\alpha} \times \mathbf{r}_{O_2P}\big|_1 + \boldsymbol{\omega} \times \mathbf{v}_P \big|_2 + \boldsymbol{\omega} \times \boldsymbol{\omega} \times \mathbf{r}_{O_2P}\big|_2 \\ & = \boldsymbol{\alpha} \times \mathbf{r}_{O_2P}\big|_1 + \boldsymbol{\omega} \times \mathbf{v}_P \big|_2 - \omega^2 \mathbf{r}_{O_2P}\big|_2 \end{aligned}
Putting all together:
\begin{aligned} \mathbf{a}_P \big|_1 & = \mathbf{a}_{O_2} \big|_1 + \mathbf{a}_P \big|_2 + \boldsymbol{\omega} \times \mathbf{v}_P\big|_2 + \boldsymbol{\alpha} \times \mathbf{r}_{O_2P}\big|_1 + \boldsymbol{\omega} \times \mathbf{v}_P \big|_2 - \omega^2 \mathbf{r}_{O_2P}\big|_2 \\ & = \mathbf{a}_{O_2} \big|_1 + \mathbf{a}_P \big|_2 + \boldsymbol{\alpha} \times \mathbf{r}_{O_2P}\big|_1 - \omega^2 \mathbf{r}_{O_2P}\big|_2 + 2\boldsymbol{\omega} \times \mathbf{v}_P \big|_2 \end{aligned}
where:
- \mathbf{a}_P \big|_1: acceleration of point \mathbf P with respect to frame 1,
- \mathbf{a}_{O_2} \big|_1: acceleration of the origin of frame 2 (O_2) with respect to frame 1,
- \mathbf{a}_P \big|_2: acceleration of point \mathbf P with respect to frame 2,
- \boldsymbol{\alpha} \times \mathbf{r}_{O_2P}\big|_1: tangential acceleration,
- - \omega^2 \mathbf{r}_{O_2P}\big|_2: normal or radial acceleration,
- 2\boldsymbol{\omega} \times \mathbf{v}_{O_2P}\big|_2: new term (the Coriolis acceleration) which is a new term that arise when there is angular velocity between frame 2 relative to frame 1.
Rearranging the term and dropping some indexes this formula can be written as:
\mathbf{a}_P = \mathbf{a}_{O_2} + \mathbf{a}_{\text{rel}} + \boldsymbol{\alpha} \times \mathbf{r} - \omega^2 \mathbf{r} + 2\boldsymbol{\omega} \times \mathbf{v}_{\text{rel}}
where:
- \mathbf{a}_P: absolute acceleration of point \mathbf P with respect to frame 1,
- \mathbf{a}_{O_2}: absolute acceleration of the origin of frame 2 (O_2) with respect to frame 1,
- \mathbf{a}_{\text{rel}}: relative acceleration of point \mathbf P with respect to frame 2,
- \boldsymbol{\alpha} \times \mathbf{r}: tangential acceleration of point \mathbf P with respect to frame 2,
- \omega^2 \mathbf{r}: normal acceleration of point \mathbf P with respect to frame 2,
- 2\boldsymbol{\omega} \times \mathbf{v}_{\text{rel}}: Coriolis acceleration of point \mathbf P with respect to frame 2.
Assuming planar (two-dimensional) motion, the angular acceleration vector \boldsymbol{\alpha} can be expressed as \alpha \, \mathbf{k} (or \ddot \theta \, \mathbf k) and the angular velocity \boldsymbol{\omega} can be expressed as \omega \, \mathbf{k} (or \dot \theta \, \mathbf k), where \mathbf{k} is the unit vector perpendicular to the plane of motion, the equation becomes:
\begin{aligned} \mathbf{a}_P & = \mathbf{a}_{O_2} + \mathbf{a}_{\text{rel}} + \alpha \, \mathbf k \times \mathbf{r} - \omega^2 \mathbf{r} + 2\omega \, \mathbf k \times \mathbf{v}_{\text{rel}} \\ & = \mathbf{v}_{O_2} + \mathbf{v}_{\text{rel}} + \ddot \theta \, \mathbf k \times \mathbf{r} - \omega^2 \mathbf{r} + 2\dot \theta \, \mathbf k \times \mathbf{v}_{\text{rel}} \end{aligned}
Example
Let’s make an example and solve an engineering problem related to determining the velocity of a point relative to two different reference frames in planar two-dimensional motion.
We will analyze the velocity relationships between two reference frames and a specific point within a mechanical system.
The problem involves a fixed reference frame 1, and a moving frame 2, centered in \text{O}_2, which is welded to a vertical bar (\text{bar 2}). The common point of interest, \mathbf P, moves relative to these two frames. In this scenario, there is a collar \text C pinned to \text{bar 1} and sliding along \text{bar 2}. The angular velocity of \text{bar 1} is given as \omega_1 = -0.2 \, \frac{\text{rad}}{\text{s}}, and the acceleration is \alpha_1 = -0.25 \, \frac{\text{rad}}{\text{s}^2}.
Velocity
The goal is to determine the angular velocity of \text{bar 2} at that instant and the relative velocity of \mathbf P with respect to the moving frame 2. To achieve this, we will use the velocity relationship:
\mathbf{v}_P = \mathbf{v}_{O_2} + \mathbf{v}_{\text{rel}} + \omega_1 \, \mathbf{k} \times \mathbf{r}
where:
- \mathbf{v}_P is the velocity of point \mathbf P,
- \mathbf{v}_{O_2} is the velocity of the origin O_2 of the moving frame,
- \mathbf{v}_{\text{rel}} is the relative velocity of \mathbf P in the moving frame,
- \omega_1 is the angular velocity of the moving frame relative to the fixed frame,
- \mathbf{r} is the position vector of \mathbf P relative to O_2.
Here, \mathbf{k} points out of the plane, and \mathbf{r} is expressed in terms of the moving frame coordinates x_2, y_2.
The fixed frame 1 has its origin at O_1 with basis vectors \mathbf{i}_1, \mathbf{j}_1. The moving frame is welded to \text{bar 2}, with its origin at O_2 and basis vectors \mathbf{i}_2, \mathbf{j}_2. The position vector of \mathbf P in the moving frame is:
\mathbf{r}_{O_2P} = r_x \mathbf{i}_2 + r_y \mathbf{j}_2
In the moving frame, the relative velocity \mathbf{v}_{\text{rel}} represents the motion of \mathbf P observed from O_2. Since \mathbf P slides along \text{bar 2}, its motion is constrained to the y_2-axis:
\mathbf{v}_{\text{rel}} = v_{\text{rel}} \, \mathbf{j}_2
where v_{\text{rel}} is the magnitude of the relative velocity along \mathbf{j}_2.
From the velocity equation, we have:
\mathbf{v}_P = \mathbf{v}_{O_2} + v_{\text{rel}} \mathbf{j}_2 + \omega_2 \, \mathbf{k} \times \mathbf{r}_{O_2P}
The angular velocity \omega_2 of \text{bar 2} is unknown and must be determined. The position vector \mathbf{r}_{O_2P} is known from geometry:
\mathbf{r}_{O_2P} = -\sin\left(30 ^\circ\right) + \cos\left(30 ^\circ\right) = -0.5 \mathbf{i}_2 + 0.866 \mathbf{j}_2
The cross product yields:
\mathbf{k} \times \mathbf{r}_{O_2 P} = -0.866 \mathbf{i}_2 - 0.5 \mathbf{j}_2
Substituting into the velocity equation:
\mathbf{v}_P = \mathbf{v}_{O_2} + v_{\text{rel}} \mathbf{j}_2 - 0.866 \omega_2 \mathbf{i}_2 - 0.5 \omega_2 \mathbf{j}_2
Since O_2 is stationary in the fixed frame, \mathbf{v}_{O_2} = 0. Therefore:
\mathbf{v}_P = v_{\text{rel}} \mathbf{j}_2 - 0.866 \dot \theta_2 \mathbf{i}_2 - 0.5 \dot \theta_2 \mathbf{j}_2
From the perspective of \text{bar 1}, the velocity of point \mathbf P can be computed from the velocity of two points on the same body:
\mathbf{v}_P = \omega_1 \, \mathbf{k} \times \mathbf{r}_{O_1P} = -0.2 \, \mathbf{k} \times \mathbf{r}_{O_1P}
With \mathbf{r}_{O_1P} = -0.5 \mathbf{i}_1 + 0.866 \mathbf{j}_1, the cross product gives:
\mathbf{v}_P = 0.1732 \mathbf{i}_1 + 0.1 \mathbf{j}_1
Equating \mathbf{i}_1 and \mathbf{j}_1 components in the fixed frame gives two equations:
\begin{aligned} & -0.866 \omega_2 = 0.1732 \\ & v_{\text{rel}} - 0.5 \omega_2 = 0.1 \end{aligned}
Solving for \omega_2:
\omega_2 = -0.2 \, \frac{\text{rad}}{\text{s}}
Substitute into the second equation to find v_{\text{rel}}:
v_{\text{rel}} = 0.1 \, \frac{\text{m}}{\text{s}}
The angular velocity of \text{bar 2} is \omega_2 = -0.2 \,\frac{\text{rad}}{\text{s}}, and the relative velocity of \mathbf P in the moving frame is \mathbf{v}_{\text{rel}} = 0.1 \, \mathbf{j}_2 \, \frac{\text{m}}{\text{s}}.
Acceleration
To extend the previous problem to accelerations, we now incorporate the Coriolis term along with tangential and radial acceleration terms. The relationship for the acceleration of a point \mathbf P, relative to two different reference frames, is given as:
\mathbf{a}_P = \mathbf{a}_{O_2} + \mathbf{a}_{\text{rel}} + \alpha \, \mathbf{k} \times \mathbf{r} - \omega^2 \mathbf{r} + 2\omega \, \mathbf{k} \times \mathbf{v}_{\text{rel}}
where:
- \mathbf{a}_P is the acceleration of \mathbf P in the fixed frame,
- \mathbf{a}_{O_2} is the acceleration of the moving frame’s origin, O_2, relative to the fixed frame. In this case, \mathbf{a}_{O_2} = 0,
- \mathbf{a}_{\text{rel}} is the relative acceleration of \mathbf P in the moving frame,
- \alpha_2 is the angular acceleration of the moving frame (\text{bar 2}) relative to the fixed frame,
- \omega_2 is the angular velocity of the moving frame,
- \mathbf{r} is the position vector of \mathbf P relative to O_2,
- \mathbf{v}_{\text{rel}} is the relative velocity of \mathbf P in the moving frame.
The angular acceleration \alpha_2 and relative acceleration \mathbf{a}_{\text{rel}} of \mathbf P in the moving frame are unknown.
Since \mathbf{a}_{O_2} = 0, the equation simplifies to:
\mathbf{a}_P = \mathbf{a}_{\text{rel}} + \alpha_2 \, \mathbf{k} \times \mathbf{r} - \omega_2^2 \mathbf{r} + 2\omega_2 \, \mathbf{k} \times \mathbf{v}_{\text{rel}}
The position vector of \mathbf P relative to O_2 is:
\mathbf{r} = -0.5 \, \mathbf{i}_2 + 0.866 \, \mathbf{j}_2 It is possible to compute all the terms; the relative acceleration is to be determined:
\mathbf{a}_{\text{rel}} = a_{\text{rel}} \, \mathbf{j}_2 The tangential acceleration is:
\alpha_2 \, \mathbf{k} \times \mathbf{r} = \alpha_2 \, \mathbf{k} \times (-0.5 \, \mathbf{i}_2 + 0.866 \, \mathbf{j}_2) = 0.866 \alpha_2 \, \mathbf{i}_2 + 0.5 \alpha_2 \, \mathbf{j}_2
The centripetal acceleration is:
-\omega_2^2 \mathbf{r} = -(-0.2)^2 (-0.5 \, \mathbf{i}_2 + 0.866 \, \mathbf{j}_2) = -0.02 \, \mathbf{i}_2 + 0.03464 \, \mathbf{j}_2
Finally the Coriolis acceleration is:
2\omega_2 \, \mathbf{k} \times \mathbf{v}_{\text{rel}} = 2(-0.2) \, \mathbf{k} \times (0.1 \, \mathbf{j}_2) = -0.04 \, \mathbf{i}_2
Substituting these terms into the acceleration equation:
\begin{aligned} \mathbf{a}_P & = a_{\text{rel}} \, \mathbf{j}_2 + (0.866 \alpha_2 - 0.02 - 0.04) \, \mathbf{i}_2 + (0.5 \alpha_B + 0.03464) \, \mathbf{j}_2 \\ & = (0.866 \alpha_2 - 0.06) \, \mathbf{i}_2 + (a_{\text{rel}} + 0.5 \alpha_2 + 0.03464) \, \mathbf{j}_2 \end{aligned}
From the perspective of \text{bar 1} the acceleration of point \mathbf P can be computed from the acceleration of two points on the same body:
\mathbf{a}_P = \alpha_1 \, \mathbf{k} \times \mathbf{r} - \omega_1^2 \mathbf{r}
Substitute \alpha_1 = 0.25 \, \mathbf{k}, \omega_1 = -0.2 \, \text{rad/s}, and \mathbf{r} = -0.5 \, \mathbf{i}_1 + 0.866 \, \mathbf{j}_1:
\begin{aligned} \mathbf{a}_P & = (0.25 \, \mathbf{k}) \times (-0.5 \, \mathbf{i}_1 + 0.866 \, \mathbf{j}_1) - (-0.2)^2 (-0.5 \, \mathbf{i}_1 + 0.866 \, \mathbf{j}_1) \\ & = (0.2165 \, \mathbf{i}_1 + 0.125 \, \mathbf{j}_1) - (-0.02 \, \mathbf{i}_1 + 0.03464 \, \mathbf{j}_1)\\ & = 0.1965 \, \mathbf{i}_1 + 0.15964 \, \mathbf{j}_1 \end{aligned}
Equating \mathbf{i} and \mathbf{j} components, solve for \alpha_2 and a_{\text{rel}}:
\begin{aligned} & 0.866 \alpha_2 - 0.06 = 0.1965 \\ & a_{\text{rel}} + 0.5 \alpha_2 + 0.03464 = 0.15964 \end{aligned}
Solving:
\begin{aligned} & \alpha_2 = 2.73 \, \tfrac{\text{rad}}{\text{s}^2} \\ & a_{\text{rel}} = -0.125 \, \tfrac{\text{m}}{\text{s}^2} \end{aligned}
The angular acceleration of \text{bar 2} is \alpha_2 = 2.73 \, \frac{\text{rad}}{\text{s}^2} (counterclockwise). The relative acceleration of \mathbf P in the moving frame is \mathbf{a}_{\text{rel}} = -0.125 \, \mathbf{j}_2 \,\frac{\text{m}}{\text{s}^2}.