Which statement is true? (Velocity and Acceleration of a Tennis Ball)

[paulimerci]

What happens during circular movement?
Could you have another acceleration vector that could be combined with the vector of centripetal acceleration in such a way that the resultant acceleration vector forms a constant angle with the velocity vector of the ball?

I understand that the velocity vector and acceleration vector are perpendicular to each other at every location on the circle in uniform circular motion at a constant speed because the acceleration vector points in the direction of the circle's center and the velocity is always tangent to the circle. Where should I draw "another acceleration vector" that you have pointed out in the above question?

 

[haruspex]

Where should I draw "another acceleration vector" that you have pointed out in the above question?

The centripetal acceleration is the acceleration component normal to the velocity, so any other acceleration must be parallel to it. That's the tangential component.

 

[haruspex]

Oh, I think I see a path forward now... by the Chain Rule:

$$ \frac{dv_x}{dt} = v_x\frac{dv_x}{dx} = a \cos ( \theta + \beta) \tag{10} $$

By (9):

$$v_x^2 = a \sin \beta \frac{\cos \theta(x)}{ \theta'(x)} \tag{9'}$$

Differentiate (9') w.r.t $x$ (dropping the function notation):

$$ 2 v_x \frac{dv_x}{dx} =- a \sin \beta \left( \sin \theta + \frac{ \cos \theta}{ \theta'^2} \theta'' \right)$$

Then sub (10) into the result

$$ 2 \cos ( \theta + \beta ) = - \sin \beta \left( \sin \theta + \frac{ \cos \theta}{ \theta'^2} \theta'' \right) \tag{11}$$

Now, we can solve for $\theta(x)$...Theoretically and find $y(x)$ from (1).

However, given the form of (11), I think I'm not going to bother.

It looks a bit nicer in $(s, \psi)$ coordinates. s is distance along the path, ψ is the angle the path makes to some fixed direction.
$\ddot s=a\cos(\theta)$
$\dot s\dot\psi=a\sin(\theta)$
That's not hard to solve; the hard part is figuring out what it looks like as a curve,

 

[paulimerci]

The centripetal acceleration is the acceleration component normal to the velocity, so any other acceleration must be parallel to it. That's the tangential component.

Does it look like this?

 

[haruspex]

Does it look like this?

Yes, though it does not have to be a circle. The diagram works for any section of the path short enough to be approximated as an arc of a circle.

 

[paulimerci]

Yes, though it does not have to be a circle. The diagram works for any section of the path short enough to be approximated as an arc of a circle.

Thank you, that makes sense!

 

[paulimerci]

Thank you, that makes sense!

Is there anything I missed or need to work on?

 

[Lnewqban]

I understand that the velocity vector and acceleration vector are perpendicular to each other at every location on the circle in uniform circular motion at a constant speed because the acceleration vector points in the direction of the circle's center and the velocity is always tangent to the circle. Where should I draw "another acceleration vector" that you have pointed out in the above question?

You understand correctly.
In accelerated circular movement there is a component of the acceleration that is tangential.

 

[paulimerci]

You understand correctly.
In accelerated circular movement there is a component of the acceleration that is tangential.

View attachment 321158

Oh, thank you so much. A lot is explained by these two diagrams. Worth it. I've learned something far more thoroughly than I anticipated.

 

[erobz]

It looks a bit nicer in $(s, \psi)$ coordinates. s is distance along the path, ψ is the angle the path makes to some fixed direction.

Indeed it does look quite tidy and succinct.

So $\psi$ is my $\theta$, and $\theta$ is my $\beta$. Is that correct?

 

[haruspex]

Indeed it does look quite tidy and succinct.

So $\psi$ is my $\theta$, and $\theta$ is my $\beta$. Is that correct?

Yes.
I don't know what the technical term is for $(s, \psi)$ coordinates. I was introduced to them briefly at high school, but I've not come across them since, and couldn't find a mention on the net.

 

[paulimerci]

You understand correctly.
In accelerated circular movement there is a component of the acceleration that is tangential.

View attachment 321158

Therefore, an object undergoing uniform acceleration is accelerating even though its speed is constant. How does the acceleration vs time graph look?

 

[jbriggs444]

Therefore, an object undergoing uniform acceleration is accelerating even though its speed is constant. How does the acceleration vs time graph look?

You are talking about uniform circular motion, right? So an acceleration versus time graph would involve two dimensions of acceleration versus one dimension of time. It is hard to convey this sort of three dimensional graph on a two dimensional page. However, a quick trip to Google finds this graph with the two dimensions superimposed. Both dimensions of acceleration are simple sine waves -- with a 90 degree offset from each other.

Viewed as vectors rather than as component pairs, the position vector will be moving around in a circular trajectory. The velocity vector will also trace out a circular path in velocity space. The acceleration vector will trace out a circular path in acceleration space. The "jerk" vector will trace out a circular path in "jerk" space. And so on.

Uniform circular motion has an unusual property -- the functions for position, velocity, acceleration, "jerk", "snap", "crackle", "pop" and all further derivatives all have graphs that look exactly the same. Each further derivative is shifted 90 degrees from the previous. For instance, the graph for "snap" will match the graph for position in terms of phase. [The amplitudes may differ, but a careful choice of units can make the amplitudes match as well].

 

[erobz]

Yes.
I don't know what the technical term is for $(s, \psi)$ coordinates. I was introduced to them briefly at high school, but I've not come across them since, and couldn't find a mention on the net.

The first seems pretty obvious as $\dot s$ is tangential to $s$

To derive the second equation, I think what is being said:

$$ \tan ( \Delta \psi ) = \frac{a \sin \theta \Delta t }{ \dot s } $$

such that when we take the limit as $\Delta t \to 0$ we get

$$ \dot s \dot \psi = a \sin \theta $$

Just making sure I understand the justification.

 

[haruspex]

The first seems pretty obvious as $\dot s$ is tangential to $s$

To derive the second equation, I think what is being said:

$$ \tan ( \Delta \psi ) = \frac{a \sin \theta \Delta t }{ \dot s } $$

such that when we take the limit as $\Delta t \to 0$ we get

$$ \dot s \dot \psi = a \sin \theta $$

Just making sure I get the justification.

Yes, that's how I derived it. I also checked it gave the right result for uniform circular motion.

 

[Lnewqban]

Therefore, an object undergoing uniform acceleration is accelerating even though its speed is constant. How does the acceleration vs time graph look?

Sorry, I don’t understand your question.
Could you explain it a little further?

 

[paulimerci]

You are talking about uniform circular motion, right? So an acceleration versus time graph would involve two dimensions of acceleration versus one dimension of time. It is hard to convey this sort of three dimensional graph on a two dimensional page. However, a quick trip to Google finds this graph with the two dimensions superimposed. Both dimensions of acceleration are simple sine waves -- with a 90 degree offset from each other.
View attachment 321203
Viewed as vectors rather than as component pairs, the position vector will be moving around in a circular trajectory. The velocity vector will also trace out a circular path in velocity space. The acceleration vector will trace out a circular path in acceleration space. The "jerk" vector will trace out a circular path in "jerk" space. And so on.

Uniform circular motion has an unusual property -- the functions for position, velocity, acceleration, "jerk", "snap", "crackle", "pop" and all further derivatives all have graphs that look exactly the same. Each further derivative is shifted 90 degrees from the previous. For instance, the graph for "snap" will match the graph for position in terms of phase. [The amplitudes may differ, but a careful choice of units can make the amplitudes match as well].

Thank you!

 

[paulimerci]

Sorry, I don’t understand your question.

jbriggs answered my question. Thank you.

Could you explain it a little further?

 

[kuruman]

For those interested I posted a solution to the trajectory problem here
https://www.physicsforums.com/threa...le-between-acceleration-and-velocity.1049440/