Inner product between velocity and acceleration is zero (parametric)

  • #1
Lambda96
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Homework Statement
I should show the following ##\bigl\langle \dot{\textbf{r}}'(t) ,\ddot{\textbf{r}}'(t) \bigr\rangle=0##
Relevant Equations
none
Hi,

I am having problems with task b

Bildschirmfoto 2023-12-02 um 15.20.27.png

I then defined the velocity vector and the acceleration vector as follows

##dot{\textbf{r}}'(t) = \frac{1}{||\dot{\textbf{r}}(t)||} \left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right)##

and

##ddot{\textbf{r}}'(t) = \frac{1}{||\dot{\textbf{r}}(t)||} \frac{d}{dt}\left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right)##

Then I calculated the following:

##bigl\langle \dot{\textbf{r}}'(t) ,\ddot{\textbf{r}}'(t) \bigr\rangle=\dot{\textbf{r}}'(t) = \frac{1}{||\dot{\textbf{r}}(t)||^2} \left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right) \cdot \frac{d}{dt}\left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right)= \frac{1}{||\dot{\textbf{r}}(t)||^2} \Bigl( \dot{r_1}(t) \frac{d}{dt} \dot{r_1}(t) + \dot{r_2}(t) \frac{d}{dt} \dot{r_2}(t) \Bigr)=\frac{1}{||\dot{\textbf{r}}(t)||^2} \Bigl( \dot{r_1}(t) \ddot{r_1}(t) + \dot{r_2}(t) \ddot{r_2}(t) \Bigr)##

Unfortunately, I can't get any further. The following must apply ##\bigl\langle \dot{\textbf{r}}'(t) ,\ddot{\textbf{r}}'(t) \bigr\rangle=0## so that this is the case, the expression ##\dot{r_1}(t) \ddot{r_1}(t) + \dot{r_2}(t) \ddot{r_2}(t)=0##, but why is this the case?

My derivation is probably already wrong, but I don't know how else to solve the problem.
 
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  • #2
I don't know why my formula with Latex is not displayed correctly, does anyone know why?

In overleaf it is displayed correctly
 
  • #3
In $$dot{\textbf{r}}'(t) = \frac{1}{||\dot{\textbf{r}}(t)||} \left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right)$$ a "\" is missing in the beginning: $$\dot{\textbf{r}}'(t) = \frac{1}{||\dot{\textbf{r}}(t)||} \left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right)$$
 
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  • #4
Lambda96 said:
ddotr′(t)=1||r˙(t)||ddt(r1˙(t)r2˙(t))
no. The normalized acceleration vector should be the derivative of the normalized velocity vector.
 
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  • #5
To add a hint to that, what you are looking for is a property of any vector of constant magnitude. Don’t get bogged down in the particular expression in terms of a normalized vector.
 
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  • #6
Thank you Hill and Orodruin for your help 👍👍

I have now proceeded as follows

##\bigl\langle \dot{r}', \dot{r}' \bigr\rangle=\dot{r}' \cdot \dot{r}' =1##

Now I have simply formed the derivative ##\frac{d}{dt}##

##\ddot{r}' \cdot \dot{r}' + \dot{r}' \cdot \ddot{r}' =0##
##2 \dot{r}' \cdot \ddot{r}' =0##
##\dot{r}' \cdot \ddot{r}' =0##
 
  • #7
Lambda96 said:
Thank you Hill and Orodruin for your help 👍👍

I have now proceeded as follows

##\bigl\langle \dot{r}', \dot{r}' \bigr\rangle=\dot{r}' \cdot \dot{r}' =1##

Now I have simply formed the derivative ##\frac{d}{dt}##

##\ddot{r}' \cdot \dot{r}' + \dot{r}' \cdot \ddot{r}' =0##
##2 \dot{r}' \cdot \ddot{r}' =0##
##\dot{r}' \cdot \ddot{r}' =0##
Indeed.
 
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FAQ: Inner product between velocity and acceleration is zero (parametric)

What does it mean for the inner product between velocity and acceleration to be zero?

The inner product (or dot product) between velocity and acceleration being zero means that the velocity and acceleration vectors are orthogonal (perpendicular) to each other. This implies that the motion of the object is constrained in such a way that its speed remains constant, and any change in direction does not affect the magnitude of the velocity.

How can we mathematically express the condition that the inner product between velocity and acceleration is zero?

Mathematically, if **v(t)** is the velocity vector and **a(t)** is the acceleration vector, the condition can be expressed as **v(t) · a(t) = 0**. Here, "·" denotes the dot product. This equation signifies that the cosine of the angle between the velocity and acceleration vectors is zero, indicating orthogonality.

What are the implications of the inner product between velocity and acceleration being zero for the motion of a particle?

When the inner product between velocity and acceleration is zero, it implies that the particle's speed is constant. The acceleration vector only changes the direction of the velocity vector, not its magnitude. This typically describes a scenario where the particle is moving along a curved path at a constant speed.

Can you provide an example of a motion where the inner product between velocity and acceleration is zero?

A common example is uniform circular motion. In uniform circular motion, the velocity vector is always tangent to the circle, while the acceleration vector (centripetal acceleration) points towards the center of the circle. Since these two vectors are perpendicular to each other, their dot product is zero.

How can we verify if the inner product between velocity and acceleration is zero in a given parametric motion?

To verify this, first determine the parametric equations for the position vector **r(t)** of the particle. Then, compute the velocity vector **v(t) = dr(t)/dt** and the acceleration vector **a(t) = dv(t)/dt**. Finally, calculate the dot product **v(t) · a(t)**. If this dot product is zero for all values of t, then the velocity and acceleration are orthogonal.

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