Physical pendulum with no fixed pole

[bznm]

Yes

Here's where using the zero-momentum reference frame is useful. The animation you posted earlier is similar to being in the zero-momentum frame. What is the orientation of the rod when the sleeve has maximum speed?

Uhmm. it isn't very easy to see unless you already know the answer :P
To me it seems it is with the rod in vertical position.. But I'm not sure.. Please, help me!

 

[TSny]

Uhmm. it isn't very easy to see unless you already know the answer :P
To me it seems it is with the rod in vertical position.. But I'm not sure..

OK, fair enough. It seems "intuitive" that the maximum speed of the sleeve in the zero-momentum frame will occur when the rod is vertical. But "intuitive" is not a good reason. I tried to come up with a good argument without any math, but so far with no luck. But with conservation of momentum and a couple of kinematic relations you can prove the result.

 

[bznm]

OK, fair enough. It seems "intuitive" that the maximum speed of the sleeve in the zero-momentum frame will occur when the rod is vertical. But "intuitive" is not a good reason. I tried to come up with a good argument without any math, but so far with no luck. But with conservation of momentum and a couple of kinematic relations you can prove the result.

Could you please show me how to extract max sleeve velocity and how to transformer it with respect to the inertial frame of the figure? This is the first exercise of this kind that I was given.. It isn't that easy to get the results! I would be very grateful!

 

[bznm]

I still didn't get to the result. Can anybody help me with finding the max sleeve speed?

 

[TSny]

We'll try to show that the maximum speed of the sleeve in the zero-momentum frame of reference occurs when the rod is vertical. It will then be easy to show that the maximum speed of the sleeve in the original frame also occurs when the rod is vertical.

Notation: [edited]

CM stands for center of mass.

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

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

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

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

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

Goal: show that U takes on its max value when θ = 0.

Step 1. Use conservation of momentum to derive a relation between U and v_x

 

[bznm]

We'll try to show that the maximum speed of the sleeve in the zero-momentum frame of reference occurs when the rod is vertical. It will then be easy to show that the maximum speed of the sleeve in the original frame also occurs when the rod is vertical.

Notation:

CM stands for center of mass.

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

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

v_y 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.

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

Goal: show that U takes on its max value when θ = 0.

Step 1. Use conservation of momentum to derive a relation between U and v_x

Uhm OK, thanks @TSny for the help. I assume that you mean "v_x is the x-component of the *(velocity of)* center of mass of the rod".

v_x=U+ωRcosθ where R is the distance between the pole and CM.
due to the conservation of horizontal momentum I have:
MU+2Mv_x= const (in the zero-momentum frame is =0).
If I substitute I get
M(U+w)+2M(U+ωRcosθ+w)=0 (where w is the zero momentum frame velocity, i.e. the transformation constant between the frames)
U = -2/3 ω R cosθ - w. This is in the zero momentum frame. In the original frame U= -2/3 ω R cosθ - 2w.
w is a constant, so I don't care about it.
U is maximum in absolute value when theta is 0 and ω is maximum (this also occurs when theta is zero because potential energy is 0).
Is it correct?

 

[bznm]

Furthermore, when θ max, rod-and-sleeve are moving with the same velocity.

This means that if θ max
- in the zero momentum frame the horizontal momentum is 3M*(vx + w) = 0 i.e. w=-v_x
- In the original frame I have that p_x at start = p_x at θ max.
That is 2M ω L = 3M v_x, so v_x=2/3 ω L for θ max.
I have found that the value of w is = -2/3 ω L.

is it correct?
What do I have to do now?

 

[TSny]

Uhm OK, thanks @TSny for the help. I assume that you mean "v_x is the x-component of the *(velocity of)* center of mass of the rod".

Yes, thanks for catching that.

v_x=U+ωRcosθ where R is the distance between the pole and CM.

Yes, good. (You've already worked out my "Step 2" that was coming next!)

due to the conservation of horizontal momentum I have:
MU+2Mv_x= const (in the zero-momentum frame is =0).

Yes. I should have made it clearer that we are going to first work out the maximum speed of the sleeve for the zero-momentum frame. All the symbols in the notation list that I posted should be considered as measured in the zero-momentum frame. So, the right hand side of the above equation is zero.

If I substitute I get
M(U+w)+2M(U+ωRcosθ+w)=0 (where w is the zero momentum frame velocity, i.e. the transformation constant between the frames)
U = -2/3 ω R cosθ - w. This is in the zero momentum frame.

Since all symbols are considered measured relative to the zero-momentum frame, there is no need to introduce w. So, let w = 0. We will get back to the original frame later.

In the original frame U= -2/3 ω R cosθ - 2w.
w is a constant, so I don't care about it.

OK. w = 0 in the zero-momentum frame.

U is maximum in absolute value when theta is 0 and ω is maximum (this also occurs when theta is zero because potential energy is 0).
Is it correct?

Well, this is where it's tricky. The expression U= -2/3 ω R cosθ says that U is maximum when the product ωcosθ is a maximum. Even though cosθ takes on a maximum at θ = 0, that doesn't necessarily imply that the product ωcosθ takes on a maximum at θ = 0. We know that the system as a whole has maximum total KE when θ = 0 because the potential energy of the system is minimum at θ = 0. The total KE of the system is shared between the sleeve and the rod. Just because θ = 0 is the place where the total KE is maximum, it doesn't logically follow from this that the sleeve has its maximum KE at this point. For example, as the rod swings down, v_y increases and then decreases. So, v_y reaches a maximum before the rod becomes vertical. Could U also reach a maximum before the rod becomes vertical? I found it helpful to also have an expression for v_y in terms of ω and θ.

 

[TSny]

Furthermore, when θ max, rod-and-sleeve are moving with the same velocity.

This means that if θ max
- in the zero momentum frame the horizontal momentum is 3M*(vx + w) = 0 i.e. w=-v_x

w is zero in the zero-momentum frame.

- In the original frame I have that p_x at start = p_x at θ max.
That is 2M ω L = 3M v_x

The left hand side is not quite correct. The center of mass of the rod is not located a distance L from the pivot.

 

[bznm]

Yes, thanks for catching that.
The expression U= -2/3 ω R cosθ says that U is maximum when the product ωcosθ is a maximum. Even though cosθ takes on a maximum at θ = 0, that doesn't necessarily imply that the product ωcosθ takes on a maximum at θ = 0. We know that the system as a whole has maximum total KE when θ = 0 because the potential energy of the system is minimum at θ = 0. The total KE of the system is shared between the sleeve and the rod. Just because θ = 0 is the place where the total KE is maximum, it doesn't logically follow from this that the sleeve has its maximum KE at this point.

At first I thought about this problem, but I simply forgot while I was writing the message :D Is the assumption "ω increases while swinging down" is totally wrong?

For example, as the rod swings down, v_y increases and the then decreases. So, v_y reaches a maximum before the rod becomes vertical. Could U also reach a maximum before the rod becomes vertical? I found it helpful to also have an expression for v_y in terms of ω and θ.

Uhmmm..
Isn't v_y=- ω R sinθ? Let's assume 0 <= θ_max < pi/2.. But when θ goes from θ_max to 0, ω increases, while θ (and so sin θ) decreases... But I can't get θ such as v_y is max in abs value.. :/

 

[bznm]

w is zero in the zero-momentum frame.

The left hand side is not quite correct. The center of mass of the rod is not located a distance L from the pivot.

There was a typo.. L should be replaced by R.

 

[TSny]

At first I thought about this problem, but I simply forgot while I was writing the message :D Is the assumption "ω increases while swinging down" is totally wrong?

"ω increases while swinging down" is something we would like to prove (even though intuitively it seems true).

Isn't v_y=- ω R sinθ?

Yes, except if y is positive upward and counterclockwise is positive for ω and θ, then I don't believe there should be a minus sign on the right side.

Let's assume 0 <= θ_max < pi/2.. But when θ goes from θ_max to 0, ω increases, while θ (and so sin θ) decreases... But I can't get θ such as v_y is max in abs value.. :/

At this point, I think you have enough equations to prove that U is max at θ = 0. Let's recap the equations:

v_x = -U/2 (from conservation of momentum) ... Eq. 1

U = -(2ωRcosθ)/3 ..........Eq. 2

v_y = ωRsinθ ...........Eq. 3With these and the fact that the total KE of the system must increase as θ decreases, I think you can prove that U takes on a maximum at θ = 0.

 

[bznm]

"ω increases while swinging down" is something we would like to prove (even though intuitively it seems true).Yes, except if y is positive upward and counterclockwise is positive for ω and θ, then I don't believe there should be a minus sign on the right side.

At this point, I think you have enough equations to prove that U is max at θ = 0. Let's recap the equations:

v_x = -U/2 (from conservation of momentum) ... Eq. 1

U = -(2ωRcosθ)/3 ..........Eq. 2

v_y = ωRsinθ ...........Eq. 3With these and the fact that the total KE of the system must increase as θ decreases, I think you can prove that U takes on a maximum at θ = 0.

Ok @TSny , I know that K_tot= 1/2 M U^2 + 1/2 I ω^2
with a couple of substitions and calculus of moment of inertia, I get
U^2 = 1/M * 2K_tot * (R^2)/(R^2+3L^2) * cos^2 θ
let's group constants:
U^2 = c*K_tot * cos^2 θ
with c positive const.

when I go from θmax to 0,
K_tot increases (due to loss of potential energy) and cos^2 θ increases, hence U^2 increases.
In the domain [0,θmax] 0 is the maximum.
Is it correct?

 

[TSny]

I know that K_tot= 1/2 M U^2 + 1/2 I ω^2

The KE of the rod is not expressed correctly. For example, suppose ω = 0 at some instant when the sleeve is moving. Would the KE of the rod be zero?

But you have the right idea.

 

[bznm]

The KE of the rod is not expressed correctly. For example, suppose ω = 0 at some instant when the sleeve is moving. Would the KE of the rod be zero?

But you have the right idea.

haa sorry! what about Ktot= 1/2 (M+2M) U^2 + 1/2 I ω^2?

 

[TSny]

I don't think so. The KE of any rigid body can be expressed as the sum of the KE due to motion of the CM and the KE of rotation about the CM.

KE = (1/2)mV_CM²+ (1/2)I_cmω².

You can use this for the rod.

 

[bznm]

@TSny Ok, I'd want to express v_cm with respect to ω, v_cm= ωR, but that ω is the one that describes the rotation around the pivot, while the ω^2 in the rotational energy is referred to the cm (otherwise It wouldn't make sense to calculate I respect to CM). I realize now that probably I didn't fully understand your earlier comment " angular speed. There is no need to say "with respect to the pole". "..

Probably I missed something with angular kinematics. I mean, let's suppose I have a point rotating on a circumference with constant speed, in this case ω = v/R with respect to the center and every arc (that correspond to some time) cirresponds to the same angle, but what if choose as angular reference another , outer, point? In this case a defined piece of circumference doesn't always correspond to the same angle (when the arc is nearer to the point the angle is bigger)... Probably I'm completely wrong, I need your explanation! :(

 

[TSny]

You don't need to specify a reference point for defining ω. The angle Δθ through which an object rotates during a time Δt is defined independent of a point of reference and independent of inertial frame of reference. (Of course if you allow rotating reference frames, Δθ could depend on the frame of reference. But, we are not using rotating frames here.) So, all inertial frames agree on ω for the rod and, also, there is no need to specify an origin for defining ω.

The best way to determine v_cm is to use the expressions for v_x and v_y. You should find ultimately that you can express the KE of the rod due to the motion of the CM of the rod in terms of just U and θ. Likewise, for the rotational KE of the rod relative to the CM of the rod. So, the total KE of the system can be expressed in terms of just U and θ.

If you just want to show that U is maximum at θ = 0, you don't need to go through all of this. |U| will be a maximum at θ = 0 if you can show that |U| never decreases as |θ| decreases. If at some θ, |U| were to decrease as |θ| decreased, what would happen to |v_x|, |ω|, and |v_y| according to equations 1, 2 and 3, respectively? I'm using absolute values here since we are only interested in magnitudes.

 

[bznm]

Ahhh ok, @TSny I hope I got it:
|vx| proportional to U, proportional to ω cos θ
|U| proportional to ω cos θ
|vy| proportional to ω sin θ

when θ decreases, cos θ increases, sin θ decreases
if |U| decreases it must be that |ω| decreases
|vx| increases (because it is proportional to U
|vy| decreases (because |ω| decreases and sin θ decreases).

I guess that you'd like me to find a contradiction, but I still can't see it because all of these things could happen at certain moments during the swing-down. Isn't it?

 

[TSny]

when θ decreases, cos θ increases

OK

sin θ decreases

OK

if |U| decreases it must be that |ω| decreases

OK

|vx| increases (because it is proportional to U

Check this result for |vx|. We are only looking at absolute values.

|vy| decreases (because |ω| decreases and sin θ decreases)

OK

 

[bznm]

Check this result for |vx|. We are only looking at absolute values.

vx = -U/2, |vx| proportional to |U|.. if |U| decreases |vx| decreases... I can't see the error :(

 

[TSny]

vx = -U/2, |vx| proportional to |U|.. if |U| decreases |vx| decreases... I can't see the error :(

I thought you said |vx| increases.

 

[bznm]

I thought you said |vx| increases.

Oh my gosh, this problem is driving me crazy. I'm sorry for the misunderstanding :/
What should I do now? How can I conclude that |U| can't decrease?

Thanks again for your precious help!

 

[TSny]

Oh my gosh, this problem is driving me crazy. I'm sorry for the misunderstanding :/
What should I do now? How can I conclude that |U| can't decrease?

Well, based on what you have found, if there is some place in the motion where |U| decreases when |θ| decreases, all of the other quantities |v_x|, |v_y|, and |ω| would also decrease when |θ| decreased. Then what would happen to the total KE of the system as |θ| decreased?

 

[bznm]

Well, based on what you have found, if there is some place in the motion where |U| decreases when |θ| decreases, all of the other quantities |v_x|, |v_y|, and |ω| would also decrease when |θ| decreased. Then what would happen to the total KE of the system as |θ| decreased?

The conclusion was clear, once I corrected the typo and reviewed what I wrote.
If all that velocities decrease, then I can conclude that total KE decreases, but this is not possibile 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 horizontal position. Could you now guide me toward the final answer (i.e. max sleeve speed in original frame)?

 

[TSny]

Can you find the speed, w, of the zero-momentum frame relative to the original frame?

 

[bznm]

Can you find the speed, w, of the zero-momentum frame relative to the original frame?

could you point out what was conceptually wrong in message #42 ?

 

[TSny]

Furthermore, when θ max, rod-and-sleeve are moving with the same velocity.

This means that if θ max
- in the zero momentum frame the horizontal momentum is 3M*(vx + w) = 0 i.e. w=-v_x

OK, here vx is the velocity of the system at θmax in the original frame.

- In the original frame I have that p_x at start = p_x at θ max.
That is 2M ω L = 3M v_x, so v_x=2/3 ω L for θ max.

You didn't express the distance from the pivot to the CM of the rod correctly.

You are getting a negative value for w. It seems more natural to me to let w be the positive velocity that the zero-momentum frame is moving relative to the original frame. Then in the zero momentum frame you would have the equation 3m*(vx - w) = 0 and you will end up with a positive value of w. But that's just a matter of personal choice.

 

[bznm]

OK, here vx is the velocity of the system at θmax in the original frame.

You didn't express the distance from the pivot to the CM of the rod correctly.

You are getting a negative value for w. It seems more natural to me to let w be the positive velocity that the zero-momentum frame is moving relative to the original frame. Then in the zero momentum frame you would have the equation 3m*(vx - w) = 0 and you will end up with a positive value of w. But that's just a matter of personal choice.

1) 3m*(vx - w) = 0 hence w=vx
2) 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.
U_max = -2/3 ω_θ=0 R cosθ + w = -2/3 ω_θ=0 R cosθ + 2/3 ω₀ R
where R is the distance pivot-center of mass.. it should be L/3.

Is it correct?

 

[TSny]

1) 3m*(vx - w) = 0 hence w=vx

OK

2) 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.

OK

I conclude that w is 2/3 ω₀ R.

OK

U_max = -2/3 ω_θ=0 R cosθ + w = -2/3 ω_θ=0 R cosθ + 2/3 ω₀ R

Since you are looking at the time when θ = 0, you can simplify cosθ. Can you express ω_θ=0 in terms of ω₀?

where R is the distance pivot-center of mass.. it should be L/3.

Did you mean to say R = L/2?

 

[bznm]

OK
OK
OK

Since you are looking at the time when θ = 0, you can simplify cosθ. Can you express ω_θ=0 in terms of ω₀?

Should I use the fact delta K_{rodrotational} = - (K_sleeve + K_{rodtranslational})?

Did you mean to say R = L/2?

uhm but the rigid body rod-and-sleeve has mass 3M, with M in the pivot and 2M distributed in the L rod..
so the center of mass should be nearer to the pivot than L/2..
L/2 is only-rod center of mass, isn't it?

 

[TSny]

ω₀ is the initial counterclockwise angular speed of the rod (in both frames).
ω_θ=0 is the angular speed when the rod returns to vertical (moving clockwise).
Form the symmetry of the motion in the zero-momentum frame, it should be easy to see how ω_θ=0 is related to ω₀.

 

[bznm]

ω₀ is the initial counterclockwise angular speed of the rod (in both frames).
ω_θ=0 is the angular speed when the rod returns to vertical (moving clockwise).
Form the symmetry of the motion in the zero-momentum frame, it should be easy to see how ω_θ=0 is related to ω₀.

i watched again the animation... I'd say that ω_θ=0 is opposite to ω₀, but I think this is wrong because at start only the rod is moving, while when theta is 0 again also the sleeve is moving!

 

[TSny]

The animation is an approximation to the motion in the zero momentum frame. In this frame, the sleeve is moving except when the rod is at |θ|_max.

 

[TSny]

i watched again the animation... I'd say that ω_θ=0 is opposite to ω₀

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