Question

Show that if an object accelerates in the sense that \(d^{2} s / d t^{2}>0\) and \(\kappa \neq 0,\) then the acceleration vector lies between \(\mathbf{T}\) and \(\mathbf{N}\) in the plane of \(\mathbf{T}\) and \(\mathbf{N}\). If an object decelerates in the sense that \(d^{2} s / d t^{2}<0\), then the acceleration vector lies in the plane of \(\mathbf{T}\) and \(\mathbf{N}\), but not between \(\mathbf{T}\) and \(\mathbf{N}\).

Short answer

What is the position of the acceleration vector with respect to the tangent and normal vectors for an accelerating object with non-zero curvature? Answer: The acceleration vector lies in the plane formed by the tangent vector (T) and the normal vector (N) for both accelerating and decelerating objects. For an object accelerating with non-zero curvature, the acceleration vector lies between the tangent and normal vectors.

Step-by-step solution

Express the acceleration vector in terms of T and N vectors

We know that acceleration is the derivative of velocity with respect to time: \(\mathbf{a} = \frac{d\mathbf{v}}{dt}\). The velocity vector can be written as the product of the speed (\(ds/dt\)) and the unit tangent vector \(\mathbf{T}\): \(\mathbf{v} = \frac{ds}{dt}\mathbf{T}\). Differentiate \(\mathbf{v}\) with respect to time to find the acceleration vector: \(\mathbf{a} = \frac{d\mathbf{v}}{dt}\). Remember that \(\mathbf{T}\) points in the direction of the curve, and \(\mathbf{N}\) is perpendicular to that direction.

Differentiate the velocity vector and find the acceleration vector

Differentiate the velocity vector \(\mathbf{v} = \frac{ds}{dt}\mathbf{T}\) with respect to time. Using the chain rule, we have: $$\mathbf{a} = \frac{d^2s}{dt^2}\mathbf{T} + \frac{ds}{dt} \frac{d\mathbf{T}}{dt}$$ Now, we need to further differentiate \(\frac{d\mathbf{T}}{dt}.\) We know that \(\frac{d\mathbf{T}}{dt}=\kappa \mathbf{N}\), where \(\kappa\) is the curvature. So, substitute it to get: $$\mathbf{a} = \frac{d^2s}{dt^2}\mathbf{T} + \frac{ds}{dt}(\kappa \mathbf{N})$$

Analyze the given conditions and determine the location of the acceleration vector

Now that we have the acceleration vector, let's analyze the cases given in the problem: 1. If \(d^2s/dt^2 > 0\) and \(\kappa \neq 0\): This means the object is accelerating, and the curvature is non-zero. Since both terms in the acceleration vector are positive, the acceleration vector lies in the plane of \(\mathbf{T}\) and \(\mathbf{N}\), and between them. 2. If \(d^2s/dt^2 < 0\): This means the object is decelerating. The expression for the acceleration vector now becomes: $$\mathbf{a} = -|\frac{d^2s}{dt^2}|\mathbf{T} + \frac{ds}{dt}(\kappa \mathbf{N})$$ Which means a part of the acceleration vector is in the opposite direction of the \(\mathbf{T}\). In this case, the acceleration vector still lies in the plane of \(\mathbf{T}\) and \(\mathbf{N}\), but not between them. This demonstrates that in both cases, the acceleration vector lies in the plane of \(\mathbf{T}\) and \(\mathbf{N}\). In the case of acceleration with non-zero curvature, the vector also lies between \(\mathbf{T}\) and \(\mathbf{N}\).

Key concepts

Tangent Vector (\mathbf{T})

The tangent vector, denoted as \(\mathbf{T}\), is a fundamental concept in understanding the motion of objects along a path. It is a unit vector that points in the direction of the curve at a particular point. Consider a curve represented parametrically, where each point on the path can be described by a position vector. The derivative of this position vector with respect to the parameter, usually time \(t\), gives us the velocity vector, which is crucial for defining \(\mathbf{T}\).

• The tangent vector \(\mathbf{T}\) is calculated as \(\mathbf{T} = \frac{\mathbf{v}}{\left| \mathbf{v} \right|}\), where \(\mathbf{v}\) is the velocity and \(\left| \mathbf{v} \right|\) is its magnitude, a.k.a. the speed.
• This normalization ensures that \(\mathbf{T}\) always has a magnitude (length) of 1, making it a unit vector.

Since \(\mathbf{T}\) aligns with the path's direction, it is perpendicular to other significant vectors, such as the normal vector \(\mathbf{N}\). Understanding \(\mathbf{T}\) is essential as it helps in differentiating concepts like acceleration into components that lie tangent and normal to the motion, which in turn plays a key role in the study of curved motion systems.

Normal Vector (\mathbf{N})

The normal vector, represented as \(\mathbf{N}\), is another critical element in the study of motion on curves. While the tangent vector \(\mathbf{T}\) points along the path, the normal vector \(\mathbf{N}\) points perpendicular to it, towards the center of curvature. Its primary role is to describe how sharply a path is curving at a point.

• \(\mathbf{N}\) is derived from the derivative of the tangent vector: \(\mathbf{N} = \frac{\frac{d\mathbf{T}}{dt}}{\left| \frac{d\mathbf{T}}{dt} \right|}\).
• The process involves normalizing the derivative to ensure that \(\mathbf{N}\) remains a unit vector.

This vector is crucial because it is directly involved in defining the curvature and is indicative of the radial component of acceleration. \(\mathbf{N}\) helps split the acceleration into radial and tangential components, providing insight into how an object moves around a curve. When looking at the acceleration vector, \(\mathbf{N}\) contributes to understanding the direction of the change in velocity tangential to the curve.

Curvature (\kappa)

Curvature, denoted as \(\kappa\), is a measure of how quickly a curve changes direction at a particular point. This concept is fundamental in describing the dynamics of curved paths and is intrinsically linked to the normal vector \(\mathbf{N}\). The larger the curvature, the tighter or more pronounced the bend in the curve is.

• Mathematically, curvature is defined as \(\kappa = \left| \frac{d\mathbf{T}}{ds} \right| \), where \(s\) is the arc length.
• It can also be expressed as the magnitude of the change in the tangent vector per unit length of the curve.

Curvature plays a significant role in determining the centripetal (or normal) component of acceleration because \(\frac{d\mathbf{T}}{dt} = \kappa \mathbf{N}\). In the context of the acceleration vector, curvature means that part of the acceleration is directed along \(\mathbf{N}\), affecting how the object maintains its path. Understanding \(\kappa\) gives us insights into how forces act on an object moving in a curved path, which is pivotal in physics and engineering applications.