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

[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/