Find speed of CoM after collision between ball and "square structure"

[songoku]

You know the horizontal velocity of the mass centre. What, instantaneously, is the horizontal velocity of the particle? Of the horizontal rod? What about the vertical rod?

The instantaneous horizontal velocity of the particle, horizontal rod and vertical rod will also be $\frac{1}{3}v_o$. But if I take origin as the axis, wouldn't the orbital angular momentum after collision is zero since there is no perpendicular distance between the motion and origin?

Thanks

 

[haruspex]

The instantaneous horizontal velocity of the particle, horizontal rod

Yes

and vertical rod

No. The rotation will affect it. I should have made clear that I mean the horizontal velocity of the mass centre of each.

 

[songoku]

No. The rotation will affect it. I should have made clear that I mean the horizontal velocity of the mass centre of each.

Since the question does not give information about the vertical impulse by the table, how to incorporate the rotation to find the horizontal velocity of vertical rod after collision?

Is it correct to say the total angular momentum at origin after collision will be contributed by three things: spin angular momentum of vertical rod, spin angular momentum of horizontal rod and orbital angular momentum of center of mass of vertical rod, and sum all of them?

Thanks

 

[haruspex]

Here’s what looks to me to be the simplest approach…

Forget about the velocity you found for the mass centre of the system. It is not helpful.
Let the velocity after impact of the ball (and rod joint) be $u$, and the angular velocity of the rods be $\omega$.
For the linear momentum after impact, you have the horizontal velocities of the ball and horizontal rod as $u$. What is the horizontal velocity of the mass centre of the vertical rod? What equation does conservation give you?

For the angular momentum about the (static) point of impact, you have:

• 0 for the ball
• angular momentum of the horizontal rod about the point of impact due to $\omega$. There is no angular momentum due to its linear motion. What is its moment of inertia about that axis?
• angular momentum of the vertical rod about the point of impact

For angular momentum of the vertical rod about the point of impact consider its motion as the sum of the linear motion of its lower end ($u$) and its rotation ($\omega$).
What is its moment of inertia about that axis? What is its angular momentum due to

• its linear motion?
• its rotation?

 

[Lnewqban]

 

[haruspex]

View attachment 354921

View attachment 354928

Nice pictures, but it is not clear what point you are making. The second picture is somewhat irrelevant since we are only concerned with the motion the instant after impact. The frame has not moved yet.

 

[Lnewqban]

@haruspex
I believe that I was not trying to make any point when I posted those diagrams.
I made them for myself at first, while trying to understand this problem.
Perhaps those could help @songoku clarify the ideas he has shared in his posts 29, 31 and 33.

The exaggerated second diagram and the animated Judo movement show how the common center of mass must move up some for the pivoting point (x-y origin) to try to slide horizontally under it.

Assuming that the frame is rigid and that the horizontal and vertical members remain perpendicular to each other, it seems to me that the collision should simultaneously induce that vertical movement, a horizontal movement, and a rotation in the combined CoM.

 

[haruspex]

@haruspex
I believe that I was not trying to make any point when I posted those diagrams.
I made them for myself at first, while trying to understand this problem.
Perhaps those could help @songoku clarify the ideas he has shared in his posts 29, 31 and 33.

The exaggerated second diagram and the animated Judo movement show how the common center of mass must move up some for the pivoting point (x-y origin) to try to slide horizontally under it.

Assuming that the frame is rigid and that the horizontal and vertical members remain perpendicular to each other, it seems to me that the collision should simultaneously induce that vertical movement, a horizontal movement, and a rotation in the combined CoM.

Ok, but I worry that students might think you are implying an error in their working. Perhaps when just posting helpful diagrams you could mention the purpose?
Thanks.

 

[songoku]

Here’s what looks to me to be the simplest approach…

Forget about the velocity you found for the mass centre of the system. It is not helpful.
Let the velocity after impact of the ball (and rod joint) be $u$, and the angular velocity of the rods be $\omega$.
For the linear momentum after impact, you have the horizontal velocities of the ball and horizontal rod as $u$. What is the horizontal velocity of the mass centre of the vertical rod? What equation does conservation give you?

Since there is vertical impulse, I can't use conservation of linear momentum so the approach will be using conservation of angular momentum.

Total angular momentum before collision = total angular momentum after collision
$$0=-\frac{2}{3}L . v_{CM, vertical rod}+I_{L-rod}.\omega$$

For the angular momentum about the (static) point of impact, you have:

• 0 for the ball
• angular momentum of the horizontal rod about the point of impact due to $\omega$. There is no angular momentum due to its linear motion. What is its moment of inertia about that axis?

$$I_o=\int_0^L r^2 dm$$
$$=\int_0^L x^2 \frac{\alpha_o}{L}x ~ dx$$
$$=\frac{\alpha_o.L^3}{4}$$

Because the axis of rotation of horizontal rod is at darker region, by using parallel axis theorem:
$$I=I_o+ML^2$$
$$=\frac{\alpha_o.L^3}{4}+\int_0^L \frac{\alpha_o}{L}x~dx . L^2$$
$$=\frac{3}{4}\alpha_o.L^3$$

For angular momentum of the vertical rod about the point of impact consider its motion as the sum of the linear motion of its lower end ($u$) and its rotation ($\omega$).
What is its moment of inertia about that axis? What is its angular momentum due to

• its linear motion?
• its rotation?

Moment of inertia = $I_o = \frac{\alpha_o.L^3}{4}$

Angular momentum due to velocity $u$ is zero because there is no perpendicular distance to the origin.

Angular momentum due to its rotation is $I_o.\omega = \frac{\alpha_o.L^3}{4} . \omega$

 

[haruspex]

Since there is vertical impulse, I can't use conservation of linear momentum

You only can’t use conservation of the vertical component of the linear momentum. The horizontal component is conserved.

by using parallel axis theorem:

Your integral used distance from the lighter end of the rod, not distance from the mass centre of the rod. The $I_o$ you computed is the MoI about that end of the rod, so it gave the right value but for the wrong rod. No need for the parallel axis theorem.

For the horizontal rod, use the same integral but with the mass density as $\alpha_0(1-\frac xL)$.

Once you have both of those, as a check, you can use the parallel axis theorem in reverse to find the MoI about the mass centre in two ways:
$I_{heavy end}-M(\frac L3)^2=I_c =I_{light end}-M(\frac {2L}3)^2$

Angular momentum due to velocity $u$ is zero because there is no perpendicular distance to the origin.

No.
We are treating the motion of the vertical rod as it rotating about its lower end while moving as a whole linearly to the right at speed $u$. In this view, the whole rod has that linear velocity, so for finding the contribution to its angular momentum we need to take its mass as being at the rod's mass centre.

 

[songoku]

You only can’t use conservation of the vertical component of the linear momentum. The horizontal component is conserved.

Wouldn't the horizontal velocity of center of mass of vertical rod also be $u$?

For the horizontal rod, use the same integral but with the mass density as $\alpha_0(1-\frac xL)$.

Sorry I don't understand why the mass density is this for horizontal rod

 

[haruspex]

Wouldn't the horizontal velocity of center of mass of vertical rod also be $u$?

No. The lower end of the rod is moving to the right with velocity $u$, but the rod is rotating anticlockwise, so the mass centre’s horizontal velocity must be less, or even negative.
However, for the purpose of calculating its angular momentum, I am suggesting to consider the motion of that rod as a linear motion speed $u$ for the rod as a whole, plus an anticlockwise rotation rate $\omega$ about the lower end.
What is the angular momentum contribution from each?

Sorry I don't understand why the mass density is this for horizontal rod

Because the density is $\alpha_0$ at $x=0$ and $0$ at $x=L$.
Don’t confuse the $x$ in the diagram for question a) with the $x$ in the diagrams for the other parts.

 

[songoku]

No. The lower end of the rod is moving to the right with velocity $u$, but the rod is rotating anticlockwise, so the mass centre’s horizontal velocity must be less, or even negative.
However, for the purpose of calculating its angular momentum, I am suggesting to consider the motion of that rod as a linear motion speed $u$ for the rod as a whole, plus an anticlockwise rotation rate $\omega$ about the lower end.
What is the angular momentum contribution from each?

Angular momentum contribution from:
(1) linear motion = $\frac{2}{3} MuL$

(2) rotational motion = $\frac{1}{4} \alpha_o L^3 \omega$