What is the Correct Tangent Angle for Calculating Force on a Cylinder?

[Bauxiet]

Homework Statement

Calculate the force on the cilinder. You need the angle of the tangent and r.

Homework Equations

Down under here. The solution for the angle is 30°. But why does the formula I used did not work out?
My solution is 40,9°, why isn't this correct for this excercise?

The Attempt at a Solution

 

[Dr. Courtney]

I do not understand the problem statement. It does not look like English to me.

 

[S.G. Janssens]

I do not understand the problem statement. It does not look like English to me.

It is the native language of your 8th president
To the OP: It is polite (and, I believe, according to the rules here) to translate your question entirely.

 

[TSny]

Homework Statement

The solution for the angle is 30°. But why does the formula I used did not work out?
My solution is 40,9°, why isn't this correct for this excercise?

Why do you have a factor of $\dot{\theta}$ in the denominator?

 

[Bauxiet]

It is the native language of your 8th president
To the OP: It is polite (and, I believe, according to the rules here) to translate your question entirely.

Sorry guys, I didn't want to be inpolite. This is the translation:

The cylinder C can only move in the slot. The movement is described by r = 0,6*cos(theta) m. The lever OA turns left (counter clockwise) with an angular speed of 2 rad/s and has a angular acceleration of 0,8 rad/s^2 at the moment when theta = 30°. What is the force on the cylinder C at that moment. The cylinder touches only one side of the slot (without friction). The movement is horizontal.

I was solving this question. And I needed the angle between the lever and the tangentline of the cylinder. The formula is on my paper. I needed this to find the angles for my forces. But the fomula seems not to be correct. What did or do I wrong? Thanks guys! And sorry again, i did not want to be inpolite!

 

[Bauxiet]

Why do you have a factor of $\dot{\theta}$ in the denominator?
View attachment 104737

Just derivate of the term above. Chain rule for derivates...?

 

[TSny]

Note that tanΨ is dimensionless. Is the right hand side of your expression for tanΨ dimensionless?

 

[Bauxiet]

Why do you have a factor of $\dot{\theta}$ in the denominator?
View attachment 104737

Just derivate of the term above. Chain rule for derivates...?

EDIT: Okay, this is my fault. The second term doesn't need to be there.

I was confused because of the image above. I think that the derivate of above is dr/dt and that's why Theta has to be also be derived. (chain rule?). In this excercise it must be the derivate to theta and dsin(theta) /d(theta) = cos(theta). Because we have to derive to theta, when we apply the chainrule to theta, it is just a 1?

I think I got it, thank you guys very much!

 

[S.G. Janssens]

And sorry again, i did not want to be inpolite!

Geen zorgen (= don't worry), you fixed it just fine, nobody got hurt