What is the angle between the acceleration and velocity when rotating?

[bolzano95]

Homework Statement

A body is circulating on a fixed circumference with radius R=1m. It moves with an angular velocity $ω = \frac {k} {\sqrt{φ}}$.

Relevant Equations

Relevant equations are listed below in solving process.

$ω = \frac {k} {\sqrt{φ}}$
What is the angle between acceleration and velocity after 1spin (2π radians)?

First I decided to find out what is the angular acceleration:
$α = \frac {dω} {dt} = \frac {dω} {dt} \frac {dφ} {dφ} = \frac {dω} {dφ} ω \implies$after integrating $\implies α = - \frac {k^2} {2φ^2}$
$a_t= αR = - \frac {k^2} {2φ^2} R$
$a_r= ω^2R= \frac {k^2} {φ} R$
$a= \sqrt{a_t^2+a_r^2}=...$

Because $\vec{v}$ is parallel to $\vec{a_t}$ we can use $tanθ = |\frac {a_r} {a_t}|\implies$ after evaluating for $φ=2π \implies θ=85,45°$

Is this correct?

 

[PeroK]

Did you draw a diagram?

 

[bolzano95]

I did. I'm just not sure about the result. Unfortunately I have no solution manual.

 

[PeroK]

I did. I'm just not sure about the result. Unfortunately I have no solution manual.

You're almost right, but you need to draw a diagram.

 

[kuruman]

I did. I'm just not sure about the result. Unfortunately I have no solution manual.

Perhaps you can post the diagram that you drew?

 

[bolzano95]

Posted above in the solving process.

 

[kuruman]

It helps. Is the angular speed increasing or decreasing as this thing goes around?

 

[bolzano95]

If you look at the equation $ω = \frac {k} {\sqrt{φ}}$ I would say the angular speed is decreasing.

 

[kuruman]

If you look at the equation $ω = \frac {k} {\sqrt{φ}}$ I would say the angular speed is decreasing.

Right which means that the linear speed is also decreasing. What is the direction of the tangential acceleration for that to happen?

 

[PeroK]

If you look at the equation $ω = \frac {k} {\sqrt{φ}}$ I would say the angular speed is decreasing.

Yes, and $\alpha$ was negative in your calculations.

 

[bolzano95]

If I increase the angle, then the angular velocity decreases but is still positive. So the velocity vector is OK as drawn.
The angular velocity is decreasing so I need a negative angular acceleration and if it is negative then is directed in the clockwise direction of circulation. The tangential acceleration is then parallel to the velocity but has an opposite direction.
The correct solution then is $$ ψ=π-θ= 94,55^\circ$$

 

[kuruman]

If I increase the angle, then the angular velocity decreases but is still positive. So the velocity vector is OK as drawn.
The angular velocity is decreasing so I need a negative angular acceleration and if it is negative then is directed in the clockwise direction of circulation. The tangential acceleration is then parallel to the velocity but has an opposite direction.
The correct solution then is $$ ψ=π-θ= 94,55^\circ$$

Flipping the angle from π - θ to π + θ will not give you the correct answer if θ is incorrect.
At this point I would suggest another line of attack that will give you the angle without a diagram. It's what I did. Write the linear acceleration vector $\vec a$ and linear velocity vector $\vec v$ in unit vector notation (polar coordinates) and then consider that the cosine of the angle $\psi$ between them is given by $$\cos\psi = \frac{\vec a \cdot \vec v}{av}.$$

 

[haruspex]

If I increase the angle, then the angular velocity decreases but is still positive. So the velocity vector is OK as drawn.
The angular velocity is decreasing so I need a negative angular acceleration and if it is negative then is directed in the clockwise direction of circulation. The tangential acceleration is then parallel to the velocity but has an opposite direction.
The correct solution then is $$ ψ=π-θ= 94,55^\circ$$

I get the same.

 

[kuruman]

I get the same.

I too get the same now. I had a sign error in my solution.