Kinematics Homework: Solving for Velocity and Acceleration

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The discussion revolves around solving kinematics homework related to velocity and acceleration. The user has derived expressions for velocity and acceleration but encounters an issue when trying to find the tangent of the angle for velocity, resulting in a negative cotangent. Responses suggest using a diagram to visualize the relationship between vectors and angles, emphasizing the geometric interpretation of the problem. Additionally, there is a clarification on the assumption that angular velocity (ω) is constant. The conversation highlights the importance of understanding vector relationships in kinematics.
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Homework Statement



In the attachment.

Homework Equations





The Attempt at a Solution



First of all, sorry for the English, I'm from Israel, speak Hebrew here.
Ok, I found v = (-w*R*sin(wt), w*R*cos(wt)), a = (-w^2 * R*cos(wt), -w^2*R*sin(wt)).
I did tg to ay / ax and found the angle of a.
The problem is when I do tg to v, I come out with tg = -ctg, how can I solve that?
Hope u got it, 10x in advance.
 

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Welcome to PF!

asi123 said:
First of all, sorry for the English, I'm from Israel, speak Hebrew here.

Ok, I found v = (-w*R*sin(wt), w*R*cos(wt)), a = (-w^2 * R*cos(wt), -w^2*R*sin(wt)).
I did tg to ay / ax and found the angle of a.
The problem is when I do tg to v, I come out with tg = -ctg, how can I solve that?

Shalom asi123! Welcome to PF! :smile:

(btw, we'd say "I used tg = ay/ax", or "I divided ay by ax to find the tg of the angle of a." :smile:)

Two ways to do it …

One … draw a diagram! … that's often a good idea anyway …

in this case, you can see that v is parallel to (-sinωt, cosωt), and a is parallel to (-cosωt, -sinωt), so from ordinary geometry the angle between them is … ?

Two … as you say, the tangent for a is the same as minus the cotangent for v.

Well, ctgθ = tg(what)?

And -tgθ = tg(what)? :smile:

(oh … this assumes that ω is constant … are you sure it is?)
 
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