Planar rigid body kinematics focuses on analyzing the motion of rigid bodies confined to a two-dimensional plane without considering the forces that cause the motion. It encompasses both translational and rotational movements, examining parameters such as the position and orientation of the body, linear and angular velocity, and linear and angular acceleration.
Planar motion can be classified into three types. Translation occurs when all points of a rigid body move along parallel paths. If these paths are straight lines, the motion is rectilinear translation, while if the paths are curved, it is curvilinear translation. In both cases, there is no rotation, and the body maintains its orientation relative to its original position.
Wheel Rolling on a Fixed Horizontal Surface
Wheel Rolling on a Curved Surface
Rotation about a fixed axis describes a motion where every point in the rigid body moves in a circular path around a single, stationary axis. The axis remains fixed in space, and all points on the body have the same angular velocity and angular acceleration but different linear velocities depending on their distance from the axis.
General planar motion is a combination of translation and rotation. The body experiences both a change in position and a change in orientation simultaneously. This type of motion is common in many practical applications, where the motion cannot be classified purely as translation or rotation.
Lets consider a rigid body, and identify two points \mathbf P and \mathbf Q which have a position vector in relation to a fix reference frame; in kinematics this reference frame doesn’t necessarily need to be inertial, as there is no application of Newton or Euler equations, but we are considering only the geometry.
We can express the position of a point Q in terms of vector addition of position vectors. Let \mathbf{r}_{OQ} be the position vector from the origin O to point Q. This can be written as:
\mathbf{r}_{OQ} = \mathbf{r}_{OP} + \mathbf{r}_{PQ}
Here, \mathbf{r}_{OP} is the position vector from O to another point P, and \mathbf{r}_{PQ} is the position vector from P to Q.
To obtain the velocities, we take the time derivatives of these position vectors:
\mathbf{v}_Q = \frac{\mathrm d\mathbf{r}_{OQ}}{\mathrm dt} = \frac{\mathrm d\mathbf{r}_{OP}}{\mathrm dt} + \frac{\mathrm d\mathbf{r}_{PQ}}{\mathrm dt}
This simplifies to:
\mathbf{v}_Q = \mathbf{v}_P + \mathbf{v}_{QP}
where:
It’s important to note that \mathbf{v}_Q and \mathbf{v}_P are absolute velocities measured in the fixed reference frame. The relative velocity \mathbf{v}_{QP} accounts for the motion of Q relative to P, which both points can have with respect to the reference frame.
Next, let’s consider the relative velocity \mathbf{v}_{QP}. Consider the position vector \mathbf{r}_{PQ} expressed in terms of its magnitude and direction. Let \theta be the angle that \mathbf{r}_{PQ} makes with the i-direction (the horizontal axis). We can express \mathbf{r}_{PQ} as:
\mathbf{r}_{PQ} = r_{PQ} \cos\theta \, \mathbf{i} + r_{PQ} \sin\theta \, \mathbf{j} = r_{PQ} \mathbf{u}
where \mathbf{u} = \cos\theta \, \mathbf{i} + \sin\theta \, \mathbf{j} is the unit vector in the direction of \mathbf{r}_{PQ}.
Taking the derivative of \mathbf{r}_{PQ} with respect to time to find the relative velocity \mathbf{v}_{QP}:
\mathbf{v}_{QP} = \frac{\mathrm d\mathbf{r}_{PQ}}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt} \left( r_{PQ} \cos\theta \, \mathbf{i} + r_{PQ} \sin\theta \, \mathbf{j} \right)
Applying the product rule, we get:
\mathbf{v}_{QP} = \dot{r}_{PQ} \cos\theta \, \mathbf{i} + r_{PQ} (-\sin\theta \, \dot{\theta}) \, \mathbf{i} + \dot{r}_{PQ} \sin\theta \, \mathbf{j} + r_{PQ} \cos\theta \, \dot{\theta} \, \mathbf{j}
Simplifying, we can factor out \dot{\theta}:
\mathbf{v}_{QP} = \dot{r}_{PQ} \mathbf{u} + r_{PQ} \dot{\theta} (-\sin\theta \, \mathbf{i} + \cos\theta \, \mathbf{j})
In fixed body kinematics the magnitude of the distance between two points is fixed:
\dot{r}_{PQ} = 0 so:
\mathbf{v}_{QP} = r_{PQ} \dot{\theta} (-\sin\theta \, \mathbf{i} + \cos\theta \, \mathbf{j})
Notice that -\sin\theta \, \mathbf{i} + \cos\theta \, \mathbf{j} is equivalent to \mathbf{k} \times \mathbf{u}, where \mathbf{k} is the unit vector perpendicular to the plane (out of the plane):
\mathbf{v}_{QP} = r_{PQ} \dot{\theta} \, (\mathbf{k} \times \mathbf{u}) = \dot{\theta} \, (\mathbf{k} \times r_{PQ} \mathbf{u}) = \dot{\theta} \, (\mathbf{k} \times \mathbf{r}_{PQ})
Therefore we have:
\mathbf{v}_Q = \mathbf{v}_P + \dot{\theta} \, (\mathbf{k} \times \mathbf{r}_{PQ}) = \mathbf{v}_P + \omega \, (\mathbf{k} \times \mathbf{r}_{PQ})
with \omega = \dot{\theta} is the angular velocity.
The instant center of rotation is a specific point on a body around which the entire body appears to be rotating at any given moment, much like a snapshot in time. This point itself has zero velocity, and there is only one such instant center for a body at each instant. The instant center can either lie on the body or somewhere outside it, in which case we refer to the “extended body”.
There are three primary methods to locate the instant center of rotation:
Consider the instant center of rotation \mathbf P and another point \mathbf Q on the body. The velocity of any point on the body can be described by the angular velocity of the body multiplied by the position vector from the instantaneous center to that point Q:
\mathbf{v}_Q = \mathbf{v}_P + \omega \, (\mathbf{k} \times \mathbf{r}_{PQ})
Since the instant center of rotation has zero velocity by definition, its contribution to the equation vanishes. Therefore, the velocity of point \mathbf Q is simply the angular velocity of the body crossed with the position vector from \mathbf P to \mathbf Q.
Let’s consider an example: a wheel that rolls without slipping. Imagine a wheel in contact with the ground. At the exact moment the wheel touches the ground, the contact point has zero velocity relative to the ground. This contact point is the instant center of rotation of the wheel at that instant.
A second method for determining the instant center of rotation involves examining the velocity vectors of various points on the moving body. To apply this technique, begin by identifying the velocity vectors of at least two distinct points on the body. For each of these velocity vectors, we can draw a line that is perpendicular to the direction of the velocity at that specific point.
Since the instant center of rotation is the point about which the entire body is momentarily rotating, it must lie along these perpendicular lines. The point where these perpendicular lines intersect is the instant center of rotation. This intersection occurs because, at the instant center, the velocity of that point is zero, and all other velocities in the body are purely rotational around it.
By using this geometric approach, you can accurately locate the instant center of rotation based on the directional properties of the velocity vectors of different points on the body.
The bar AB and the bar CD are rotating around a fix axes, and therefore their velocity can only go up and down or left or right.
It is possible to draw a perpendicular from the velocity vectors and, where that crosses, that defines the instant center of rotation.
A third method for determining the instant center of rotation involves applying geometric relationships and the principles of similar triangles. This approach begins by identifying the paths or lines along which different points on the body are moving.
By analyzing the motion, it is possible to construct triangles that represent the geometric relationships between these points and the desired instant center. These triangles are arranged in such a way that they are similar, meaning their corresponding angles are equal and their sides are in proportion.
By setting up these similar triangles, the proportions between corresponding sides can be used to solve for the unknown position of the instant center. The point where the corresponding sides intersect, based on the similarity criteria, indicates the location of the instant center of rotation. This geometric method leverages the inherent properties of similar triangles to accurately determine the position of the instant center, ensuring that the rotational relationships within the moving body are properly accounted for.
Consider a rigid body with two points, P and Q, where the position vector from P to Q is denoted by \mathbf{r}_{PQ}. Let \theta(t) represent the angular displacement of the body as a function of time. The angular velocity is given by \omega = \frac{\mathrm d\theta}{\mathrm dt}, and the angular acceleration is \alpha = \frac{\mathrm d\omega}{\mathrm dt} = \frac{\mathrm d^2\theta}{\mathrm dt^2}. The angular acceleration is symbolized by {\alpha}.
The absolute velocity of point Q, \mathbf{v}_Q, can be expressed in terms of the absolute velocity of point P, \mathbf{v}_P, and the angular velocity {\omega} as follows:
\mathbf{v}_Q = \mathbf{v}_P + \omega \mathbf{k} \times \mathbf{r}_{PQ}
Differentiating both sides with respect to time, we have:
\frac{\mathrm d\mathbf{v}_Q}{\mathrm dt} = \frac{\mathrm d\mathbf{v}_P}{\mathrm dt} + \frac{\mathrm d}{\mathrm dt} \left( \omega \mathbf{k} \times \mathbf{r}_{PQ} \right)
This gives:
\mathbf{a}_Q = \mathbf{a}_P + \frac{\mathrm d}{\mathrm dt} (\omega \mathbf{k}) \times \mathbf{r}_{PQ} + \omega \mathbf{k} \times \frac{\mathrm d\mathbf{r}_{PQ}}{\mathrm dt}
The term \frac{\mathrm d}{\mathrm dt} (\omega \mathbf{k}) results in the angular acceleration \alpha \mathbf{k}, so the equation becomes:
\mathbf{a}_Q = \mathbf{a}_P + (\alpha \mathbf{k}) \times \mathbf{r}_{PQ} + \omega \mathbf{k} \times \frac{\mathrm d\mathbf{r}_{PQ}}{\mathrm dt}
Next, we evaluate \frac{\mathrm d\mathbf{r}_{PQ}}{\mathrm dt}. From the relative velocity equation, we have:
\frac{\mathrm d\mathbf{r}_{PQ}}{\mathrm dt} = \omega \mathbf{k} \times \mathbf{r}_{PQ}
Substituting this into the acceleration equation, we get:
\mathbf{a}_Q = \mathbf{a}_P + (\alpha \mathbf{k}) \times \mathbf{r}_{PQ} + \omega \mathbf{k} \times (\omega \mathbf{k} \times \mathbf{r}_{PQ})
Now, simplify the double cross product using the vector identity:
\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B} (\mathbf{A} \cdot \mathbf{C}) - \mathbf{C} (\mathbf{A} \cdot \mathbf{B})
Here, \mathbf{A} = \omega \mathbf{k}, \mathbf{B} = \omega \mathbf{k}, and \mathbf{C} = \mathbf{r}_{PQ}. Applying the identity:
\omega \mathbf{k} \times (\omega \mathbf{k} \times \mathbf{r}_{PQ}) = (\omega \mathbf{k}) (\omega \mathbf{k} \cdot \mathbf{r}_{PQ}) - \mathbf{r}_{PQ} (\omega \mathbf{k} \cdot \omega \mathbf{k})
Since \mathbf{k} \cdot \mathbf{r}_{PQ} = 0 (as \mathbf{r}_{PQ} lies in the xy-plane and \mathbf{k} is along the z-axis), the first term vanishes. The second term simplifies to:
\omega \mathbf{k} \times (\omega \mathbf{k} \times \mathbf{r}_{PQ}) = -\omega^2 \mathbf{r}_{PQ}
Substituting this back into the equation for \mathbf{a}_Q, we obtain:
\mathbf{a}_Q = \mathbf{a}_P + (\alpha \mathbf{k}) \times \mathbf{r}_{PQ} - \omega^2 \mathbf{r}_{PQ}
Here, \mathbf{a}_P is the absolute acceleration of point P, \alpha \mathbf{k} \times \mathbf{r}_{PQ} is the tangential acceleration due to angular acceleration, and -\omega^2 \mathbf{r}_{PQ} is the normal (centripetal) acceleration due to angular velocity.
As an example of planar rigid body motion, we can analyze the dynamics of a piston mechanism, which is commonly used in engines. In this system, the piston moves linearly, while it is connected to a rotating crankshaft and connecting rod. Such systems exemplify how rotation and translation interact in mechanical devices. By studying the angular velocities, angular accelerations, and lengths of the links, we can determine the motion of all components, providing insights into the kinematics of rigid bodies in a plane.
Given that O is fixed and using the velocity formula for a rigid body:
\mathbf{v}_A = \mathbf{v}_O + \omega_{OA} \, \mathbf{k} \times \mathbf{r}_{OA}
where:
The velocity is:
\begin{aligned} \mathbf{v}_A & = (-10) \, \mathbf{k} \times (-0.072 \, \mathbf{i} - 0.096 \, \mathbf{j}) \, \tfrac{\text{m}}{\text{s}} \\ & = -10 \cdot (-0.096 \, \mathbf{i} - 0.072 \, \mathbf{j}) \, \tfrac{\text{m}}{\text{s}}\\ & = 0.96 \, \mathbf{i} + 0.72 \, \mathbf{j} \, \tfrac{\text{m}}{\text{s}} \end{aligned}
\mathbf{r}_{AB} can be computed using Pythagoras’s theorem. The length of AB (hypotenuse) is given as 0.33 \, \text{m}. The vertical component (height) is:
R_{ABv} = -0.12 \cdot \frac{4}{5} \mathbf j = -0.096 \, \mathbf j \, \text{m}
The horizontal component (base) is:
R_{ABh} = \sqrt{AB^2 - R_{ABv}^2} = \sqrt{0.33^2 - 0.072^2} = \sqrt{0.1089 - 0.009216} = \sqrt{0.099484} \approx 0.316 \, \mathbf i \, \text{m}
Therefore:
\mathbf{r}_{AB} = 0.316 \, \mathbf i - 0.096 \, \mathbf j \, \text{m}
The velocity of B is:
\mathbf{v}_B = \mathbf{v}_A + \omega_{AB} \mathbf{k} \times \mathbf{r}_{AB} \, \tfrac{\text{m}}{\text{s}}
Substituting:
\begin{aligned} \mathbf{v}_B & = (0.96 \, \mathbf{i} + 0.72 \, \mathbf{j}) + \omega_{AB} \, \mathbf{k} \times (0.316 \, \mathbf{i} - 0.096 \, \mathbf{j}) \, \tfrac{\text{m}}{\text{s}} \\ & = (0.96 \, \mathbf{i} + 0.72 \, \mathbf{j}) + \omega_{AB} (0.096 \, \mathbf{i} + 0.316 \, \mathbf{j}) \, \tfrac{\text{m}}{\text{s}} \\ & = (0.96 + 0.096 \, \omega_{AB}) \, \mathbf{i} + (0.72 + 0.316 \, \omega_{AB}) \, \mathbf{j} \, \tfrac{\text{m}}{\text{s}} \end{aligned}
It is possible to apply the constraints: since \mathbf{v}_B can only move horizontally, its \mathbf{j}-component must be zero:
0.72 + 0.316 \, \omega_{AB} = 0
Solving for \omega_{AB}:
\omega_{AB} = -\frac{0.72}{0.316} = -2.27 \, \tfrac{\text{rad}}{\text{s}}
Substitute \omega_{AB} = -2.27 into the \mathbf{i}-component of \mathbf{v}_B:
v_B = 0.96 + 0.096 \cdot (-2.27) \, \tfrac{\text{m}}{\text{s}} = 0.742\, \tfrac{\text{m}}{\text{s}} The direction is known and therefore:
\begin{aligned} \omega_{AB} & = -2.27 \, \tfrac{\text{rad}}{\text{s}}\\ \mathbf{v}_B & = 0.742 \, \mathbf{i} \, \tfrac{\text{m}}{\text{s}} \end{aligned}
To compute the acceleration of A, we use the formula for the acceleration of a rigid body:
\mathbf{a}_A = \mathbf{a}_O + \mathbf{\alpha}_{OA} \mathbf k \times \mathbf{r}_{OA} - \omega_{OA}^2 \mathbf{r}_{OA}
where:
Using these values:
\begin{aligned} \mathbf{a}_A & = \mathbf{a}_O + \mathbf{\alpha}_{OA} \mathbf k \times \mathbf{r}_{OA} - \omega_{OA}^2 \mathbf{r}_{OA} \\ & = 0 + 5\, \mathbf k \times \left( -0.072 \, \mathbf{i} + 0.096 \, \mathbf{j} \right)\tfrac{\text{m}}{\text{s}^2} - \omega_{OA}^2 \left(-0.072 \, \mathbf{i} + 0.096 \, \mathbf{j}\right)\tfrac{\text{m}}{\text{s}^2} \\ & = \left(-0.48 \, \mathbf{i} - 0.36 \, \mathbf{j} \right) \tfrac{\text{m}}{\text{s}^2} - 100 \left(-0.072 \, \mathbf{i} + 0.096 \, \mathbf{j}\right)\tfrac{\text{m}}{\text{s}^2} \\ & = \left(-0.48 \, \mathbf{i} - 0.36 \, \mathbf{j} \right) \tfrac{\text{m}}{\text{s}^2} + \left(7.2 \, \mathbf{i} - 9.6 \, \mathbf{j}\right)\tfrac{\text{m}}{\text{s}^2} \\ & = 6.72 \, \mathbf{i} - 9.96 \, \mathbf{j} \tfrac{\text{m}}{\text{s}^2} \end{aligned}
The acceleration of B is computed using the relative acceleration formula:
\mathbf{a}_B = \mathbf{a}_A + \alpha_{AB} \mathbf{k} \times \mathbf{r}_{AB} - \omega_{AB}^2 \mathbf{r}_{AB}.
Using:
\begin{aligned} \mathbf{a}_B & = \mathbf{a}_A + \mathbf{\alpha}_{AB} \mathbf k \times \mathbf{r}_{AB} - \omega_{AB}^2 \mathbf{r}_{AB} \\ & = \left( 6.72 \, \mathbf{i} - 9.96 \, \mathbf{j} \right)\tfrac{\text{m}}{\text{s}^2} + \mathbf{\alpha}_{AB} \mathbf k \times \left(0.316 \, \mathbf{i} - 0.096 \, \mathbf{j} \right)\tfrac{\text{m}}{\text{s}^2} - \left(-2.27\right)^2 \left( 0.316 \, \mathbf{i} - 0.096 \, \mathbf{j}\right)\tfrac{\text{m}}{\text{s}^2} \\ & = \left( 6.72 \, \mathbf{i} - 9.96 \, \mathbf{j} \right)\tfrac{\text{m}}{\text{s}^2} + \alpha_{AB} (0.096 \, \mathbf{i} + 0.316 \, \mathbf{j}) + \left(-1.629 \, \mathbf{i} + 0.4947 \, \mathbf{j}\right)\tfrac{\text{m}}{\text{s}^2} \end{aligned}
Since \mathbf{a}_B is only in the \mathbf{i}-direction, the \mathbf{j}-component of \mathbf{a}_B must equal 0. We can use this to solve for \alpha_{AB} \mathbf{k}, then substitute to find a_B. The \mathbf{j}-component of \mathbf{a}_B must be zero:
-9.96 + 0.4947 + 0.316 \alpha_{AB} = 0
Solving for \alpha_{AB}:
\alpha_{AB} = \frac{9.96 - 0.4947}{0.316} \approx 29.95 \, \tfrac{\text{rad}}{\text{s}^2}
And finally:
a_B = 6.72 - 1.629 + 0.096 \cdot 29.95 \approx 7.966 \, \tfrac{\text{m}}{\text{s}^2}
The direction is known and therefore:
\begin{aligned} \alpha_{AB} & = 29.95 \, \tfrac{\text{rad}}{\text{s}^2}\\ \mathbf{a}_B & = 0.7966 \, \mathbf{i} \, \tfrac{\text{m}}{\text{s}^2} \end{aligned}
Let’s consider a wheel rolling (without slipping) on a fixed horizontal surface, and \omega = \dot{\theta} is the angular velocity and \alpha = \ddot{\theta} is the angular acceleration.
In the coordinate system \mathbf{i} (x-direction) points to the right, \mathbf{j} (y-direction) points downward, and \mathbf{k} (z-direction) points into the plane of motion.
The instantaneous center of rotation for a rolling wheel without slipping is the point of contact P with the ground. At any instant, this point has zero velocity.
To find the velocity of the geometric center C of the wheel, we use the relative velocity equation:
\mathbf{v}_C = \mathbf{v}_P + \mathbf{\omega} \times \mathbf{r}_{CP}
Given that \mathbf{v}_P = \mathbf{0} (since P is the instantaneous center), the equation simplifies to:
\mathbf{v}_C = \mathbf{\omega} \times \mathbf{r}_{CP}
Here, \mathbf{r}_{CP} is the position vector from P to C. Since C is directly above P at a distance r, we have:
\mathbf{r}_{PC} = -r \mathbf{j}
The angular velocity vector is:
\mathbf{\omega} = \omega \mathbf{k}
Calculating the cross product:
\mathbf{\omega} \times \mathbf{r}_{CP} = \omega \mathbf{k} \times (-r \mathbf{j}) = \omega r (\mathbf{k} \times \mathbf{j}) = \omega r (-\mathbf{i}) = -\omega r \mathbf{i}
Considering the direction of rolling (positive in the clockwise direction), the velocity of the center C becomes:
\mathbf{v}_C = \omega r \mathbf{i}
To determine the acceleration \mathbf{a}_C of the geometric center, we differentiate the velocity \mathbf{v}_C with respect to time:
\mathbf{a}_C = \frac{d\mathbf{v}_C}{dt} = \frac{d}{dt} (\omega r \mathbf{i}) = r \frac{d\omega}{dt} \mathbf{i} = r \alpha \mathbf{i}
Thus, the acceleration of the geometric center C is:
\mathbf{a}_C = r \alpha \mathbf{i}
This acceleration is directed solely along the x-axis.
Next, we determine the acceleration \mathbf{a}_P of the instantaneous center of rotation P. Using the acceleration relationship for a rigid body:
\begin{aligned} \mathbf{a}_P & = \mathbf{a}_C + \alpha \mathbf{k} \times \mathbf{r}_{PC} - \omega^2 \mathbf{r}_{PC} \\ & = r \alpha \mathbf{i} + \alpha \mathbf{k} \times r \mathbf{j} - \omega^2 r \mathbf{j} \\ & = r \alpha \mathbf{i} - r \alpha \mathbf{i} - r\omega^2 \mathbf{j} \\ & = - r\omega^2 \mathbf{j} \end{aligned}
While the instantaneous center of rotation P has zero velocity at the instant considered, it possesses a non-zero acceleration directed upward.
This upward acceleration is a consequence of the rotational motion of the wheel, specifically due to the centripetal acceleration arising from the wheel’s angular velocity \omega.
Let’s consider a wheel rolling (without slipping) on a curved surface, and \omega = \dot{\theta} is the angular velocity and \alpha = \ddot{\theta} is the angular acceleration.
In the coordinate system \mathbf{i} (x-direction) points to the right, \mathbf{j} (y-direction) points downward, and \mathbf{k} (z-direction) points into the plane of motion.
We’ll use normal and tangential coordinates, as these are well-suited for the wheel’s motion along a curved path.
The relative velocity equation, will be used to determine the velocity and acceleration at various points of interest. We start with the velocity of the center of the wheel, using the point of contact as our reference point.
The velocity of the center of the wheel is expressed as follows:
\mathbf{v}_C = \mathbf{v}_P + \omega \mathbf{k} \times \mathbf{r}_{PC}
Substituting the known terms, the velocity at the instantaneous center is zero (\mathbf{v}_P = 0), and the vector \mathbf{r}_{CP} is the radius r in the \mathbf{e}_n direction:
\mathbf{v}_C = 0 + \omega \mathbf{k} \times (r \mathbf{e}_n)
Crossing \mathbf{k} with \mathbf{e}_n results in the tangential direction \mathbf{e}_t:
\mathbf{v}_C = r \omega \mathbf{e}_t
The velocity is tangential to the path, and its magnitude, r \omega, represents the speed of the wheel’s center.
The acceleration of the center of the wheel can be found by taking the time derivative of \mathbf{v}_C. The expression becomes:
\mathbf{a}_C = \frac{\mathrm d}{\mathrm dt} (r \omega \mathbf{e}_t)
Since r is constant, the derivative expands as:
\mathbf{a}_C = r \alpha \mathbf{e}_t + r \omega \frac{\mathrm d\mathbf{e}_t}{\mathrm dt}
The derivative \frac{\mathrm d\mathbf{e}_t}{\mathrm dt} is related to the curvature of the path:
\frac{\mathrm d\mathbf{e}_t}{\mathrm dt} = \frac{\dot{s}}{\rho} \mathbf{e}_n
where \dot{s}, the magnitude of velocity, equals r \dot{\theta} = r \omega, and \rho is the radius of curvature. Substituting, the acceleration becomes:
\mathbf{a}_C = r \alpha \mathbf{e}_t + r \omega \left( \frac{r \omega}{\rho} \mathbf{e}_n \right)
Simplifying:
\mathbf{a}_C = r \alpha \mathbf{e}_t + \frac{r \omega^2}{\rho} \mathbf{e}_n
This result reveals that the center of the wheel experiences a tangential acceleration r \alpha = r \ddot{\theta} and a normal (radial) acceleration \frac{r \omega^2}{\rho} = \frac{r \dot{\theta}^2}{\rho}, directed inward.
Finally, we determine the acceleration at the point of contact \mathbf{P}. Using the relative acceleration equation:
\mathbf{a}_P = \mathbf{a}_C + \alpha \mathbf{k} \times \mathbf{r}_{CP} - \omega^2 \mathbf{r}_{CP}
The vector \mathbf{r}_{CP} is -r \mathbf{e}_n, so:
\mathbf{a}_P = \mathbf{a}_C + \alpha \mathbf{k} \times (-r \mathbf{e}_n) - \omega^2 (-r \mathbf{e}_n)
Substituting \mathbf{a}_C:
\mathbf{a}_P = \left( r \alpha \mathbf{e}_t + \frac{r \omega^2}{\rho} \mathbf{e}_n \right) - r \alpha \mathbf{e}_t + r \omega^2 \mathbf{e}_n
Simplifying, the tangential terms cancel, and the result becomes:
\mathbf{a}_P = \left( 1 + \frac{r}{\rho} \right) r \omega^2 \mathbf{e}_n = \left( 1 + \frac{r}{\rho} \right) r \dot{\theta}^2 \mathbf{e}_n
This expression describes the acceleration at the point of contact for a wheel rolling on a curved path. It accounts for the curvature of the path and combines the effects of tangential and normal accelerations.
It worth noting that as \rho \to \infty, the fixed plane curve relationship approach the fixed straight line conditions:
\begin{aligned} \mathbf{a}_C & = r \alpha \mathbf{e}_t + \frac{r \omega^2}{\rho} \mathbf{e}_n \underset{\rho \to \infty}{=} r \alpha \mathbf{e}_t = r \alpha \, \mathbf{i} \\[5pt] \mathbf{a}_P & = \left( 1 + \frac{r}{\rho} \right) r \omega^2 \mathbf{e}_n \underset{\rho \to \infty}{=} r \omega^2 \mathbf{e}_n = -r \omega^2 \, \mathbf{j} \end{aligned}