What is the significance of a minus sign in kinematics equations?

[Bauxiet]

Homework Statement

I have chosen my axis as you can see on the picture. To the left, it is the positive 'r'-axis, up is the positive 'y'-axis. I have to calculate the angular speed. My solution is correct, but I'm getting into trouble at the end. I have to take the roots of the last equation to become the angular speed. Because my acceleration 'ar' is positive, I become a negatif number so I can't take the root. What am I doing wrong?

Homework Equations

The Attempt at a Solution

Thank you!

 

[Doc Al]

I have to calculate the angular speed.

Can you provide a statement of the problem?

 

[jtbell]

If the original statement is not in English, I suggest that you tell us both the original statement (in whatever language it's in), and then your attempt at translating it. That would be useful if someone reads this, who knows your language.

 

[Bauxiet]

Sorry thought the question was in the picture.

The question is, calculate the angular speed so that r = 0,72 m. Everything else that you need is in the picture.
The spring is 0,4 m long in rest. Spring constant is in the picture.

Thanks guys!

 

[Doc Al]

Shouldn't we have the mass?

 

[Doc Al]

Because my acceleration 'ar' is positive, I become a negatif number so I can't take the root. What am I doing wrong?

It's difficult to see what you're doing that leads you to try to take the square root of a negative number.

The force and the acceleration all point to the left (in your diagram). No need for any negative numbers. (Or if you use them, they will cancel out.)

 

[Bauxiet]

http://blob:http%3A//imgur.com/fe849a41-c1e6-4557-98ab-d2fc40a81c18

I have to put my acceleration, calculated out of the free body diagram, in the ar = ... equation. Since my ar is positive and r(double point) = 0, I can't take the root to find my angular acceleration. If something is not clear enough, just say me. Thanks!

EDIT: The mass is 3 kg.

 

[Doc Al]

I have to put my acceleration, calculated out of the free body diagram, in the ar = ... equation. Since my ar is positive and r(double point) = 0, I can't take the root to find my angular acceleration.

The negative sign in the equation $a_r = - r \dot{\theta}^2$ just indicates the direction of the acceleration, which is centripetal. When doing your calculation, you'll set the magnitude of the acceleration equal to $r \dot{\theta}^2$ and leave off the minus sign.

 

[Bauxiet]

The negative sign in the equation $a_r = - r \dot{\theta}^2$ just indicates the direction of the acceleration, which is centripetal. When doing your calculation, you'll set the magnitude of the acceleration equal to $r \dot{\theta}^2$ and leave off the minus sign.

Thank you. But how do I know this? Is 'r' a vector? I see the formulas just as a formula where I have to fill in my value of something without changing signs in the formula. How do I know when I can change a value? Maybe this is a weird question, but I want to get the bigger view. Thank you

 

[Doc Al]

Thank you. But how do I know this? Is 'r' a vector? I see the formulas just as a formula where I have to fill in my value of something without changing signs in the formula. How do I know when I can change a value? Maybe this is a weird question, but I want to get the bigger view. Thank you

You should not just plug numbers into a formula blindly without first understanding what it means. a_r is a component, which can be negative or positive depending on the direction. In this case, the negative sign tells you that the direction is negative (which is toward the axis).

 

[Bauxiet]

You should not just plug numbers into a formula blindly without first understanding what it means. a_r is a component, which can be negative or positive depending on the direction. In this case, the negative sign tells you that the direction is negative (which is toward the axis).

Yes indeed, I know that the acceleration sign is - or + if a result is negatif, it is poiting in the other direction. But in this case, my ar was positive. How did I know that the right part of the equation to calculate the angular speed must be positive and NOT negative? This is not the acceleration ... Do you understand me?

Or on another view. Because my ar is positive, the right part SHOULD be also positive so I can remove the minus sign? It sound weird to me.

 

[Doc Al]

Yes indeed, I know that the acceleration sign is - or + if a result is negatif, it is poiting in the other direction. But in this case, my ar was positive. How did I know that the right part of the equation to calculate the angular speed must be positive and NOT negative? This is not the acceleration ... Do you understand me?

The way I look at it is this. You solved for the acceleration using Newton's 2nd law and a force analysis. Then you used an expression for centripetal acceleration. I would have used $a_r = \omega^2 r$. The only reason your expression has a minus sign is that you are "borrowing" from a more elaborate treatment of cylindrical coordinates and the signs indicate direction. No real need for that if you know the formula for centripetal acceleration in terms of ω.

Whenever a minus sign appears in a formula, you need to understand what it means. Here it just shows direction, since we're discussing components of vectors. (By the way, if you consistently used a minus sign to indicate "left" then your value for force and acceleration would also have been negative, and thus the negative signs would have canceled out.)

 

[Doc Al]

Or on another view. Because my ar is positive, the right part SHOULD be also positive so I can remove the minus sign? It sound weird to me.

Realize (to repeat myself a bit more clearly) that your a_r is positive means that you are using a sign convention where left is positive. But the formula you grabbed for the centripetal acceleration uses the opposite sign convention.

 

[David Lewis]

Whenever a minus sign appears in a formula, you need to understand what it means. Here it just shows direction, since we're discussing components of vectors.

Plus or minus sign gives you the directional sense (or sense) of the vector. The direction of the vector's line of action can be spelled out in a diagram or represented by a unit vector in a Cartesian coordinate system.