Angular velocity of a pendulum

[jisbon]

Homework Statement

There are two identical, uniform rods each of length 𝐿 and mass 𝑚 forming a
symmetrical T-shaped pendulum as shown.
The pendulum is released at some angle 𝜃 (less than 90D). Derive the angular velocity of the pendulum when the 𝜃 = 0.

Relevant Equations

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So this question has something probably due to the conservation of angular momentum. I am able to find the moment of inertia about the pivot, which is :
$$\dfrac {1}{3}mL^{2}+\left( \dfrac {1}{12}ML^{2}+mL^{2}\right) =\dfrac {17}{12}ml^{2} $$

Now to find angular velocity when angle =0, how am I able to apply the conservation of angular momentum law without knowing the initial angular speed and angle? (even though they tell me to release less than 90 --> what does this imply in this case?)

Thanks

 

[BvU]

So this question has something probably due to the conservation of angular momentum

Think again: at t=0, L=0

 

[jisbon]

Think again: at t=0, L=0

t as in time? If so, yes because of the pendulum is not released yet, so L =0
When theta is 0, how can I find what is the L though? There is no time provided etc.

 

[BvU]

The 'think again' was aimed at your assumption that you need conservation of angular momentum for this exercise. Under what circumstances is angular momentum conserved ? Does that apply here ?

 

[jisbon]

The 'think again' was aimed at your assumption that you need conservation of angular momentum for this exercise. Under what circumstances is angular momentum conserved ? Does that apply here ?

Angular momentum is conserved when there is no external forces ...

Thinking about it again, should I be using the energy method instead? GPE = GPE + KE ? Or is there other methods to use?

 

[haruspex]

Angular momentum is conserved when there is no external forces ...

And the Earth is not part of the system whose angular momentum you are considering, so...?

 

[BvU]

Thinking about it again, should I be using the energy method instead? GPE = GPE + KE

By all means, yes !
So now consider how you want to express KE...

 

[jisbon]

By all means, yes !
So now consider how you want to express KE...

Will there be any translational KE in this case?

 

[haruspex]

Will there be any translational KE in this case?

That depends on how you choose to consider the motion.
If you think of it as a rotation about the hinge then that expresses the whole motion, so you can treat it as entirely rotational, with the appropriate moment of inertia.
Or you think of it as a rotation about the mass centre, plus the linear velocity of the mass centre.
Either way you get the same total KE.

 

[jisbon]

Alright.. So if I use the energy method:

If I take the dot as the origin, is it safe to say that:

$$mg\left( \cos \theta \dfrac {L}{2}\right) =\dfrac {1}{2}I\left( \dfrac {v}{r}\right) ^{2} $$

where left hand side shows the GPE , no KE at initial angle, and RHS as when theta =0, and there is no gpe

 

[jbriggs444]

$$mg\left( \cos \theta \dfrac {L}{2}\right)$$

Remember that only differences in potential energy are physically meaningful. This oversight masks the fact that you also have a sign error here.

 

[BvU]

And that the lower rod also has mass and mechanical energy (potential and kinetic) ...

 

[haruspex]

Alright.. So if I use the energy method:

View attachment 252824

If I take the dot as the origin, is it safe to say that:

$$mg\left( \cos \theta \dfrac {L}{2}\right) =\dfrac {1}{2}I\left( \dfrac {v}{r}\right) ^{2} $$

where left hand side shows the GPE , no KE at initial angle, and RHS as when theta =0, and there is no gpe

Further to @jbriggs444's comment...
Which dot? I see one at the pivot and another half way down.
How are you defining I?
What is r?