How Do You Calculate Acceleration in a Limacon-Shaped Path?

[Nikstykal]

Homework Statement

The rod OA rotates clockwise with a constant angular velocity of 6 rad/s. Two pin-connected slider blocks located at B move freely on OA and the curved rod whose shape is a limacon described by the equation r = 200*(2-cosθ) mm. Determine the magnitude of the acceleration of the slider blocks at the instant θ = 135°.

Homework Equations

α_r=r''-rθ'²
α_θ=rθ''+2r'θ'
α = √(α_r²+α_θ²)

The Attempt at a Solution

I am trying to solve for the magnitude using the normal and tangential acceleration components.
r = 200*(2-cosθ) --> r(135) = 541.42 mm θ' = 6 rad/s
r' = 200 sinθ * θ' --> r'(135) = 848.53 mm/s θ'' = 0
r'' = 200cosθ * θ'²= -5091.17 mm/s2

When I plug all those values in I keep getting the wrong answer. Can someone tell me where I am going wrong?

 

[TSny]

Careful with the signs. Note that the rod is rotating clockwise.

 

[Nikstykal]

Careful with the signs. Note that the rod is rotating clockwise.

Can you explain that more? Is this because there is a negative change in θ over time?

 

[Nikstykal]

I ended up getting the right answer, must have just been a miscalculation.

 

[TSny]

Can you explain that more? Is this because there is a negative change in θ over time?

Yes. In the figure, $\theta$ increases in the counterclockwise direction. Therefore, if the rod rotates clockwise, $\dot{\theta}$ is a negative number. You should find that $\dot{r}$ is negative when $\theta = 135^o$.

 

[Nikstykal]

Yes. In the figure, $\theta$ increases in the counterclockwise direction. Therefore, if the rod rotates clockwise, $\dot{\theta}$ is a negative number. You should find that $\dot{r}$ is negative when $\theta = 135^o$.

Okay I understand that. Thank you for explaining. I just go lucky in my calculations because they ended up cancelling the negatives out.