Is the Angular Velocity of Pinned Rods Calculated Correctly?

AI Thread Summary
The discussion centers on the calculation of angular velocity for pinned rods PQ and QR, with the initial equations presented being critiqued for not accounting for the movement of point Q. As point P moves up the plane, point Q's position changes, affecting the rotation rate of the rod. Suggestions are made to analyze the system using a small angle approximation for angular velocity while noting that calculating acceleration is more complex due to Q's non-horizontal acceleration. The importance of establishing a suitable coordinate system to simplify calculations is emphasized. Overall, the conversation highlights the intricacies of dynamics in rotating frames and the need for careful consideration of all moving parts.
DrNG
Messages
2
Reaction score
0
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?
 

Attachments

  • 4arZm.png
    4arZm.png
    5.7 KB · Views: 131
Physics news on Phys.org
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.
 
haruspex said:
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...
 
DrNG said:
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.
 
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.
 
Lnewqban said:
acceleration of Q (both horizontal vector
The acceleration of Q is not horizontal. That's why it it's a bit tougher.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Back
Top