Is the Angular Velocity of Pinned Rods Calculated Correctly?

[DrNG]

Homework Statement

As shown in the figure, two rods PQ and QR, pinned at Q and rod QR is hinged at R. P moves with a velocity v0 and acceleration a0 along the incline. We are required to find the angular velocities of P and Q, angular acceleration of PQ as well as QR.

Relevant Equations

ωPQ=v⊥/ℓ

My line of thinking is as follows:

$$\omega_{PQ} = \frac{v_{\perp}}{\ell} = \frac v\ell \frac{\sqrt3}{2}$$

Similarly for rod $QR$

$$\omega_{QR} = \frac{v_{\perp}}{\ell} = \frac v\ell \frac{\sqrt3}{2}$$

Is my reasoning correct?

 

[haruspex]

Your first equation appears to ignore that Q will move.
Say P is moving up the plane. Then Q moves closer to the plane, increasing the rate at which the rod rotates.

 

[DrNG]

Your first equation appears to ignore that Q will move.
Say P is moving up the plane. Then Q moves closer to the plane, increasing the rate at which the rod rotates.

Thanks for your reply!

I get it now.
Can you please suggest on which lines should I attempt further.. I'm not so comfy with rotating frames...

 

[haruspex]

Thanks for your reply!

I get it now.
Can you please suggest on which lines should I attempt further.. I'm not so comfy with rotating frames...

I would consider QR rotated through an angle $\phi$ and find expressions for the position of P and the angle of PQ in terms of that. Could be messy, though.
If it were only the angular velocity that is required, you could take $\phi$ as small and make suitable approximations, but the acceleration is a bit harder.

 

[Lnewqban]

For the instantaneous position of the mechanism, I would first try to calculate the tangential velocity and acceleration of Q (both horizontal vectors), which depend on Vo and ao.
You will need to create a system of coordinates aligned in a form that makes calculations easier.
Q is a common point for links RQ and PQ; therefore, you will have velocity values and directions for both ends of link QP and could calculate its center of rotation and angular velocity and acceleration.

 

[haruspex]

acceleration of Q (both horizontal vector

The acceleration of Q is not horizontal. That's why it it's a bit tougher.