A particle travels along a plane curve (Polar coordinates)

[Alexanddros81]

Homework Statement

13.24 A particle travels along a plane curve. At a certain instant, the polar
components of the velocity and acceleration are v_R=90mm/s, v_θ=60mm/s,
a_R=-50mm/s², and a_θ=20mm/s². Determine the component of acceleration that is tangent to the path of the particle at this instant.

Homework Equations

The Attempt at a Solution

I have tried somethings but didn't get the result which is -30.5mm/s²
Any hints?

 

[rude man]

Homework Statement

13.24 A particle travels along a plane curve. At a certain instant, the polar
components of the velocity and acceleration are v_R=90mm/s, v_θ=60mm/s,
a_R=-50mm/s², and a_θ=20mm/s². Determine the component of acceleration that is tangent to the path of the particle at this instant.

The Attempt at a Solution

I have tried somethings but didn't get the result which is -30.5mm/s²
Any hints?

What have you tried? What is a? What is v? Show your work.
PS answer is correct.

 

[Chestermiller]

In terms of the velocity components in the radial and theta directions and the unit vectors in the radial and theta directions, what is the velocity vector. What is the equation for a unit vector in the direction of the velocity? In terms of the acceleration components in the radial and theta directions, and the unit vectors in the radial and theta directions, what is the equation for the acceleration vector?

 

[Alexanddros81]

Here is my work:

 

[Chestermiller]

Here is my work:

View attachment 211139

Good start, but it's much simpler than this. You've shown that:
$$\vec{v}=90\vec{e_r}+60\vec{e_{\theta}}$$
$$\vec{a}=-50\vec{e_r}+20\vec{e_{\theta}}$$and
$$v=108.16$$
So, the unit vector in the direction of the velocity vector is:
$$\vec{i_v}=\frac{90\vec{e_r}+60\vec{e_{\theta}}}{108.16}=0.8321\vec{e_r}+0.5547\vec{e_{\theta}}$$
The tangential component of the acceleration is the same as its component in the direction of the velocity vector. This is equal to the dot product of the acceleration vector with the unit vector in the direction of the velocity vector. The dot product of two vectors is a distributive property. So, for two vectors, say $\vec{A}=A_r\vec{e_r}+A_{\theta}\vec{e_{\theta}}$ and $\vec{B}=B_r\vec{e_r}+B_{\theta}\vec{e_{\theta}}$, their dot product is given by:
$$\vec{A}\centerdot \vec{B}=(A_r\vec{e_r}+A_{\theta}\vec{e_{\theta}})\centerdot (B_r\vec{e_r}+B_{\theta}\vec{e_{\theta}})=A_rB_r+A_{\theta}B_{\theta}$$

 

[rude man]

Actually, the problem as stated is muddled.
In polar coordinates a vector has an r component and a θ component. The velocity vector should thus be v r + θ_v θ with bold being the unit vectors. From your data v is √(90² + 60²) = 108.17 mm/s while θ_v = tan^-1(60/90) = 33.69 deg. Same idea for the a vector: a = √(50 + 20) = 53.85 mm/s while θ_a = tan^-1(20/-50) = 180 - 21.80 = 158.2 deg.

We have all implicitly chosen to ignore polar coordinates "r" and "θ" and interpreted the data as cartesian "x" and "y" respectively. That's how the given answer is derived. It does not excuse stating that the given v and a data is given in polar coordinates when it isn't.
I hasten to add that, given polar coordinates i would perform dot-products by converting to cartesian first. I know there's another way but I'm too old to learn it.

 

[Chestermiller]

Actually, the problem as stated is muddled.
In polar coordinates a vector has an r component and a θ component. The velocity vector should thus be v r + θ_v θ with bold being the unit vectors. From your data v is √(90² + 60²) = 108.17 mm/s while θ_v = tan^-1(60/90) = 33.69 deg. Same idea for the a vector: a = √(50.+ 20) = 53.85 mm/s while θ_a = tan^-1(20/-50) = -21.80 deg.

We have all implicitly chosen to ignore polar coordinates "r" and "θ" and interpreted the data as cartesian "x" and "y" respectively. That's how the given answer is derived. It does not excuse stating that the given v and a data is given in polar coordinates when it isn't.

Hi rude man,

It seemed clear to me from the problem statement that what they meant was exactly as you described it above.

It also seemed clear that there was no need to determine the angles, since all that was really needed was to take the dot product of the acceleration vector with a unit vector in the direction of the velocity vector.

Chet