Physical pendulum with no fixed pole

[bznm]

Yes, that's right for the zero-momentum frame.

didn't we say "all inertial frames agree on ω for the rod"?

 

[TSny]

didn't we say "all inertial frames agree on ω for the rod"?

Yes, that's right. Do you see this as a problem?

 

[bznm]

I'd say that ωθ=0 is opposite to ω0

Yes, that's right *for the zero-momentum frame*.

*all inertial frames agree* on ω for the rod

so, ωθ=0 is opposite to ω0 in every inertial frame and
U_max = -2/3 ω_θ=0 R cosθ + w = -2/3 ω_θ=0 R + 2/3 ω₀ R=4/3 ω₀ R..
What about it?

[I was reading something about angular velocity: the property "it doesn't depend on an origin" is true for the rigid body (reference), but for a particle I need an origin.. isn't it?]

 

[bznm]

I tried to recap everything and I think I spotted some mistakes.. could you please read it and tell me if it sounds good, @TSny ? Please!

$d_{cm}=\frac{\frac{L}{2}\cdot2M +0 \cdot M}{3M}=\frac{L}{3}$

$v_{cm}= \omega d_{cm}$

At start the horizontal linear momentum is $3Mv_{cm_{0}}$, while at $\theta_{max}$ rod-and-sleeve is just translating at some velocity V, so the hor. linear momentum is $3MV$.

As the h. linear momentum conserves itself (there are no horizontal forces), it must be $V=v_{cm_{0}} = \omega_{0}\frac{L}{3}$.

There are no dissipative forces, so U + K conserves.

At first ($\theta=0$), instead the motion is entirely rotational: $E_{\theta=0}=\frac{1}{2}I\omega_0^2$, where I is the moment of inertia, calculated as $I=\frac{1}{3}(2M)L^2$.

At $\theta_{max}$, the motion is entirely translational, so $E_{\theta_{max}}$ is $\frac{1}{2}(3M) V^2+ L(2M)g(1-\cos(\theta_{max}))$ (where the second term is the expression for the potential energy.

From $E_{\theta_{max}}=E_{\theta=0}$ I get $\cos(\theta_{max})=1-\frac{5L\omega_0^2}{12g}$.

Then, using the following notation (measured in the zero-momentum frame):

U is the instantaneous velocity of the sleeve. Positive toward the right.

vx is the x-component of the center of mass of the rod. Positive x is to the right.

vy is the y-component of the center of mass of the rod. Positive y is upward.

θ is the angular displacement of rod from horizontal. Counterclockwise is positive.

R is the distance pivot-center of mass ($d_{cm}$)

ω is the angular velocity of the rod. Counterclockwise is positive.

vx=U+ωRcosθ

In the zero-momentum frame MU+2Mvx is constantly 0.
M(U)+2M(U+ωRcosθ)=0 hence $U = -2/3 ω R \cos(θ)$.

While the rod is swinging up it loses horizontal momentum, so also the sleeve is reducing its horizontal momentum (talking about abs values). While the rod swings down, instead, it gains horizontal momentum, and so does the sleeve. What we want to verify is that effectively the sleeve has its maximum KE at $\theta=0$ (we want to show that |U| never decreases as |θ| decreases).

Let's recap:

vx = -U/2 (from conservation of momentum)

$U = -\frac{2}{3}ωRcosθ$

vy = ωRsinθ

If |U| were to decrease as |θ| decreased, |vx| decreases, |ω| decreases, |vy| decreases.. that means that if |U| decreases when |θ| decreases, then KE of the system decreases, but this is not possible because also potential energy is decreasing and there are no dissipative forces. OK, we finally found out that max sleeve speed is when the rod comes back in the vertical position.

If θ max, with vx the velocity in the original frame:

- in the zero momentum frame the horizontal momentum is 3M*(vx - w) = 0 i.e. w=vx (w is the transformation factor between the frames).
I conclude that w is V.

$U_{max} = -2/3 ω_{θ=0} R cosθ + w$

Due to the symmetry of motion in the zero-momentum frame $ω_{θ=0}$ is opposite to ω0 in every inertial frame and

$U_{max} = -2/3 ω_{θ=0} R cosθ + w = -2/3 ω_{0} R + ω_0 R=1/3 ω_0 R$

 

[TSny]

so, ωθ=0 is opposite to ω0 in every inertial frame and
U_max = -2/3 ω_θ=0 R cosθ + w = -2/3 ω_θ=0 R + 2/3 ω₀ R=4/3 ω₀ R..
What about it?

I think that's correct. Most people would probably prefer to see the answer in terms of the length of the rod, L, instead of R.

[I was reading something about angular velocity: the property "it doesn't depend on an origin" is true for the rigid body (reference), but for a particle I need an origin.. isn't it?]

Yes, that's right.

 

[TSny]

I tried to recap everything and I think I spotted some mistakes.. could you please read it and tell me if it sounds good, @TSny ? Please!

$d_{cm}=\frac{\frac{L}{2}\cdot2M +0 \cdot M}{3M}=\frac{L}{3}$

$v_{cm}= \omega d_{cm}$

At start the horizontal linear momentum is $3Mv_{cm_{0}}$, while at $\theta_{max}$ rod-and-sleeve is just translating at some velocity V, so the hor. linear momentum is $3MV$.

OK. For me, it's easier to get the initial horizontal momentum of the system (in the original frame) by just calculating the momentum of the rod since the sleeve is at rest. No need to worry about the location of the CM of the entire system.

As the h. linear momentum conserves itself (there are no horizontal forces), it must be $V=v_{cm_{0}} = \omega_{0}\frac{L}{3}$.

There are no dissipative forces, so U + K conserves.

At first ($\theta=0$), instead the motion is entirely rotational: $E_{\theta=0}=\frac{1}{2}I\omega_0^2$, where I is the moment of inertia, calculated as $I=\frac{1}{3}(2M)L^2$.

At $\theta_{max}$, the motion is entirely translational, so $E_{\theta_{max}}$ is $\frac{1}{2}(3M) V^2+ L(2M)g(1-\cos(\theta_{max}))$ (where the second term is the expression for the potential energy.

The potential energy expression is not quite correct. The potential energy is determined by the location of the CM of the rod.

From $E_{\theta_{max}}=E_{\theta=0}$ I get $\cos(\theta_{max})=1-\frac{5L\omega_0^2}{12g}$

Once the PE is corrected, you will get a different answer. But your method is correct.

Let's recap:

vx = -U/2 (from conservation of momentum)

$U = -\frac{2}{3}ωRcosθ$

vy = ωRsinθ

If |U| were to decrease as |θ| decreased, |vx| decreases, |ω| decreases, |vy| decreases.. that means that if |U| decreases when |θ| decreases, then KE of the system decreases, but this is not possible because also potential energy is decreasing and there are no dissipative forces. OK, we finally found out that max sleeve speed is when the rod comes back in the vertical position.

OK

If θ max, with vx the velocity in the original frame:

- in the zero momentum frame the horizontal momentum is 3M*(vx - w) = 0 i.e. w=vx (w is the transformation factor between the frames).
I conclude that w is V.

OK

$U_{max} = -2/3 ω_{θ=0} R cosθ + w$

Due to the symmetry of motion in the zero-momentum frame $ω_{θ=0}$ is opposite to ω0 in every inertial frame and

$U_{max} = -2/3 ω_{θ=0} R cosθ + w = -2/3 ω_{0} R + ω_0 R=1/3 ω_0 R$

Does $ω_{θ=0} = ω_{0}$ or does $ω_{θ=0} = -ω_{0}$?

Finally, I thought you found earlier that $w = (2/3) ω_0 R = (1/3) ω_0 L$.

 

[bznm]

I originally wrote

In the original frame I have that px at start = px at θ max.
That is 2M ω R = 3M vx, so vx=2/3 ω R for θ max.
I conclude that w is 2/3 ω0 R.

but then I thought that I was wrong with the calculation of px at start. In fact if I consider the system, whose center of mass is located at R=L/3, then I have to calculate px with *3*M ω R = 3M vx, so vx=ω R=ω0 * L/3=w.
Now I see that with R you always considered L/2, while I thought at L/3.. we also considered two different centers of mass.. do the other things I wrote work even with this misunderstanding? :(
I corrected the calculation of Umax:
$U_{max} = -2/3 ω_{θ=0} R cosθ + w = 2/3 ω_{0} R + ω_0 R=ω_0 R$

While for the problem regarding the PE, I suppose that's enough to replace L with R..

 

[TSny]

In post #41 you wrote $v_x = U + \omega R \cos \theta$ which is true for $R = L/2$. So I always thought R stood for L/2.
Let's agree to never use the symbol R again. We'll either write L/2 or L/3 explicitly.

Anyway, I do agree that w = $\omega_0 L/3$. Would you mind rewriting your final calculation for U_max?

[EDIT: For the PE, the CM of the rod moves in an arc of radius L/2, not L.]

 

[bznm]

@TSny:
I corrected the calculation of Umax:
$U_{max} = -2/3 ω_{θ=0} L/2 cosθ + w =1/3 ω_{0} L + ω_0 L/3=2/3 L ω_0$

About PE:
At first ($\theta=0$), instead the motion is entirely rotational: $E_{\theta=0}=\frac{1}{2}I\omega_0^2$, where I is the moment of inertia, calculated as $I=\frac{1}{3}(2M)L^2$.

At $\theta_{max}$, the motion is entirely translational, so $E_{\theta_{max}}$ is $\frac{1}{2}(3M) V^2+ L/2 (2M)g(1-\cos(\theta_{max}))$ (where the second term is the expression for the potential energy.

From $E_{\theta_{max}}=E_{\theta=0}$ I get $\cos(\theta_{max})=1-\frac{L\omega_0^2}{6g}$.

What about it now?

 

[TSny]

$U_{max} = -2/3 ω_{θ=0} L/2 cosθ + w =1/3 ω_{0} L + ω_0 L/3=2/3 L ω_0$

About PE:
At first ($\theta=0$), instead the motion is entirely rotational: $E_{\theta=0}=\frac{1}{2}I\omega_0^2$, where I is the moment of inertia, calculated as $I=\frac{1}{3}(2M)L^2$.

At $\theta_{max}$, the motion is entirely translational, so $E_{\theta_{max}}$ is $\frac{1}{2}(3M) V^2+ L/2 (2M)g(1-\cos(\theta_{max}))$ (where the second term is the expression for the potential energy.

From $E_{\theta_{max}}=E_{\theta=0}$ I get $\cos(\theta_{max})=1-\frac{L\omega_0^2}{6g}$.

Looks great! Good work.

 

[bznm]

Ohhh, I'm happy!
but I tried writing the conservation of energy, without using special frames, between t=0 and the position in which the rod has come back in vertical position:
$1/2 I \omega_0^2 = 1/2 I \omega_{\theta=0}^2 +1/2 M U^2+1/2 (2M) v_{cm}^2$/
but if $\omega_0=-\omega_{\theta=0}$ then there's something wrong... what am I missing? :/

Please @TSny, keep in mind that this problem is very difficult with respect to my course! :(

 

[TSny]

Ohhh, I'm happy!

Good!

but I tried writing the conservation of energy, without using special frames, between t=0 and the position in which the rod has come back in vertical position:
$1/2 I \omega_0^2 = 1/2 I \omega_{\theta=0}^2 +1/2 M U^2+1/2 (2M) v_{cm}^2$/
but if $\omega_0=-\omega_{\theta=0}$ then there's something wrong... what am I missing? :/

Does the symbol $I$ on the left side stand for the same quantity as the symbol $I$ on the right side? On the left side, $I$ is about what point? On the right, $I$ is about what point?

 

[bznm]

Good!

Does the symbol $I$ on the left side stand for the same quantity as the symbol $I$ on the right side? On the left side, $I$ is about what point? On the right, $I$ is about what point?

Uhmm, I think that the I that appears in rotational energy depends on axis of rotation.. isn't it?
On my notes I read:
$KE = \sum 1/2m_i v_i^2 = \sum 1/2 m_i r_i^2 \omega^2 = 1/2 (\sum m_i r_i^2) \omega^2= 1/2 I \omega^2$

 

[TSny]

At the initial instant, the rod may be considered as rotating about the top end because the top end is instantaneously at rest. So, the KE of the rod may be written simply as $(1/2)I_\textrm{ end}ω^2$. But when the rod returns to vertical, the top end of the rod is not instantaneously at rest. Now you are expressing the KE of the rod as $(1/2)(2M)v_\textrm{cm}^2 + (1/2)I_\textrm{cm}ω^2$.

 

[bznm]

At the initial instant, the rod may be considered as rotating about the top end because the top end is instantaneously at rest. So, the KE of the rod may be written simply as $(1/2)I_\textrm{ end}ω^2$. But when the rod returns to vertical, the top end of the rod is not instantaneously at rest. Now you are expressing the KE of the rod as $(1/2)(2M)v_\textrm{cm}^2 + (1/2)I_\textrm{cm}ω^2$.

OOk, thanks.. you're right as always!
I also tried writing the momentum in the general frame:
M(U)+2M(U+ω L/2 cosθ)=2M ω0 L/2 (px at start)
so calculating for θ=0 I get
M(U)+2M(U-ω0 L/2 cosθ)=M ω0 L
U=1/3 (ω0) (L cos (θ)+ L)
... which gives Umax= 2/3 ω0 L that we got also with the use of special frame!

Could you tell me something more about the concept "due to the symmetry in zero momentum frame, |ω| = |ω0| for theta=0"?
It isn't very natural to me (at start only pendulum is moving, when the rod comes back also the sleeve is moving, why the hell should it come with the same angular velocity? ), even if I have seen that it just works

 

[TSny]

In the center of mass frame (zero-momentum frame), the motion is symmetric between swinging up and swinging back down. When the rod is vertical and swinging counterclockwise, the sleeve is moving to the left. When the rod reaches max angle, the rod and sleeve momentarily stop. Then, the rod and sleeve have the "reverse" motion as the rod swings back down. When the rod reaches vertical and moving clockwise, it will have the same magnitude of angular velocity as initially and the sleeve will have the same speed as initially.

 

[bznm]

In the center of mass frame (zero-momentum frame), the motion is symmetric between swinging up and swinging back down. When the rod is vertical and swinging counterclockwise, the sleeve is moving to the left. When the rod reaches max angle, the rod and sleeve momentarily stop. Then, the rod and sleeve have the "reverse" motion as the rod swings back down. When the rod reaches vertical and moving clockwise, it will have the same magnitude of angular velocity as initially and the sleeve will have the same speed as initially.

Ok, I got it.
Could you check that my last calculation is error free?

 

[TSny]

Could you check that my last calculation is error free?

It looks very good.