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Assuming energy is conserved of course, no friction and similar mass. looking at this as a collision I know the mass relation plays a part, but how? Also how is having 1/4 of a circle is different from 1/5 or 1/3?

edit: no gravity and ball starting velocity is $V$. Ideally, mass of ball $M_{ball}$ and mass of tube $M_{tube}$ are different, but if its easier to solve for $M_{ball} = M_{tube}$, that will do too.

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    $\begingroup$ What details are given in the question? Is the ball thrown at a fixed velocity? Is there a gravitational field? What do you mean by "similar mass"? $\endgroup$
    – Lili FN
    Jul 18, 2021 at 20:36
  • $\begingroup$ no gravitation. fixed velocity's. and equal mass if its easier, even tho i'd want to generalize it to M1 and M2. $\endgroup$
    – DAcheese
    Jul 18, 2021 at 20:42
  • $\begingroup$ This reminds me a lot of the well known question about a frictionless swastika-shaped (sorry!!) rotating lawn sprinkler that sucks water in, instead of spraying it out - the question being, does it move clockwise, counterclockwise or stay still. Many of the same issues are in play. $\endgroup$
    – Stilez
    Jul 18, 2021 at 21:15
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    $\begingroup$ A good place to start, if we assume perfect friction free and idealised experiment, is conservation of momentum. Meaning linear momentum X and Y directions, and angular momentum (rotation). That should help you a lot...... $\endgroup$
    – Stilez
    Jul 18, 2021 at 21:18
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    $\begingroup$ I don't think this is really a collision problem. It looks like the tube is just the same size as the ball. The ball simply slides down the tube and emerges from the other end. The ball goes the direction of the tube. So the question is how much does the direction change as the ball slides around a quarter circle? $\endgroup$
    – mmesser314
    Jul 18, 2021 at 23:13

2 Answers 2

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The thing that will simplify calculation is the assumption that the cross-sectional diameter of the tube is equal to that of the ball. Because then, at the exit point, the horizontal velocity of the ball must be equal to the horizontal velocity of the tube. Why? Because then, at any point of its journey, the ball will not be able to move radially without causing the same radial movement of the tube. At exit, the radial direction happens to be horizontal.

Let's call the final horizontal velocity of both the tube and ball at exit, $v_x (= v_{bx} = v_{tx})$, final vertical velocity of tube, $v_{ty}$, that of the ball, $v_{by}$. For brevity, let's call $M_{ball}$ $m$, and $M_{tube}$, $M$ So, we have: $$ mV = Mv_x + mv_x \implies v_x = \frac{m}{M+m}V \tag{1} $$ $$ Mv_{ty} = mv_{by} \implies v_{ty} = \frac{m}{M}v_{by} \tag{2} $$ $(1)$ and $(2)$ are by conservation of momentum. By conservation of energy we have: $$ \begin{align} \frac{1}{2}mV^2 &= \frac{1}{2}Mv_t^2 + \frac{1}{2}mv_b^2 \\ \implies mV^2 &= M(v_x^2 + v_{ty}^2) + m(v_x^2 + v_{by}^2) \\ &= M(v_x^2 + \frac{m^2}{M^2}v_{by}^2) + m(v_x^2 + v_{by}^2) \\ &= (m+M)v_x^2 + \frac{m}{M}(m+M) v_{by}^2 \\ \implies \frac{m}{M}v_{by}^2 &= \frac{mV^2}{M+m} - v_x^2 \\ &= \frac{mV^2}{M+m} - \frac{m^2V^2}{(M+m)^2} \\ &= \frac{mM}{(M+m)^2}V^2 \\ \implies v_{by} &= \frac{M}{(M+m)}V \tag{3} \end{align} $$ By $(1)$ and $(3)$, $$ \begin{align} \boxed{\frac{v_{by}}{v_{bx}} = \frac{M}{m}} \end{align} $$ This ratio gives the direction of the final velocity of the ball. If $M \gg m$, the final direction will be almost upward as expected.

$\frac{1}{4}$ of the circle maximizes the final vertical velocities, anything else than that will cause smaller final vertical velocities.

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    $\begingroup$ Can you clarify "at any point of its journey, the ball will not be able to move radially without causing the same radial movement of the tube" why would that be? also why the "radial direction happens to be horizontal"? $\endgroup$
    – DAcheese
    Jul 19, 2021 at 6:42
  • $\begingroup$ Radial direction is the direction along the radius of curvature of the tube. The ball will have no freedom of movement in that direction if its diameter is equal to that of the cross-section of the tube. As far as that direction is concerned, the ball and the tube are just one body for as long as the ball is inside the tube. Now, at the topmost part of the tube, this radial direction is the horizontal direction, i.e. the line from the center of curvature of the tube to this point is horizontal. $\endgroup$
    – manisar
    Jul 19, 2021 at 7:07
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    $\begingroup$ vty and vby would not be equal as you have considered different masses which will give false result as momentum is not conserved in y direction $\endgroup$ Jul 19, 2021 at 8:46
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    $\begingroup$ Yes, many thanks, I was fixing this mistake in the meanwhile. Just posted with correction. $\endgroup$
    – manisar
    Jul 19, 2021 at 8:55
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    $\begingroup$ And now our results have completely matched with two very different approaches :D $\endgroup$ Jul 19, 2021 at 10:29
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I am going to solve this question with the minimum amount of assumptions.

Now, one assumption I will make is that the ball is entering the tube by just touching the bottom half of the tube. I am also assuming that the size of the ball is less than that of the tube as shown in the figure. I would also assume that the moment of Inertia of the tube about its center is $I$. We can consider the rightward direction as X and Upward as Y.

enter image description here

Now, I would calculate the answer completely from the basics. Let's go into the frame of reference of the tube. In this frame of reference, you can clearly see that the Normal reaction on the ball will always be perpendicular to its path and hence it will only change the direction of the ball and to its speed with respect to the tube which will be $V$.

The normal reaction on the ball $N=\frac{{M_{ball}}V^2}{d}$

The distance between the centre of the ball and the centre of curvature of the tube is $ d = R+2r-2t$

Now there will be a torque acting on the tube which you can clearly solve and about the torque as $\tau= NdSin{\phi} = {M_{ball}V^2Sin{\phi}}$.

Hence, $$\int_{0}^{\omega} Id{\omega}=\int_{-\theta}^{\theta}{M_{ball}}V^2Sin{\phi}dt$$

$$\omega = 0$$

But this could have been derived just by symmetry. Hence, the tube won't have a rotational motion at the end of motion.

Now from symmetry, we can see that after rotation the tube would end up with the same angle from the X and Y-direction.

The direction of the velocity of the ball at the end is $2\theta$ from the X-axis. And the speed of the ball would be $V_x=V{Cos{2\theta}}+V_{x-tube}$

And $V_y=V{Sin{2\theta}}-V_{y-tube}$ considering that $V_y$ is towards +Y and $V_{y-tube}$ towards -Y

From Momentum conservation, we can see that $$M_{ball}V_y = M_{tube}V_{y-tube}$$ $$M_{ball}V=M_{ball}V_{x} + M_{tube}V_{x-tube}$$

After calculation this leads to a result of $$V_{x}=\frac {M_{tube}VCos{2\theta}+M_{ball}V}{M_{tube}+M_{ball}}$$ $$V_{y}=\frac {M_{tube}VSin{2\theta}}{M_{tube}+M_{ball}}$$ $$V_{x-tube}=\frac {2M_{ball}V{(Sin{\theta})^2}}{M_{ball}+M_{tube}}$$ $$V_{y-tube}=\frac {M_{tube}VSin{2\theta}}{M_{ball}+M_{tube}}$$

Now, this result will only be possible in 2 cases, If the r=t or when the velocity of the ball along x-direction at every point is equal to the velocity of the tube along the x-direction, why? because that will lead to the ball colliding to the upper surface of the tube which leads to multiple collisions inside the tube due to which we won't be able to calculate velocities at final conditions. This leads to the result $\theta = 45°$

But, If r=t then the above result would be applicable at all values of $\theta$.

Of course, You can derive this by other methods such as energy conservation, etc.

If you are considering quarter of a circle then $V_{x}= \frac {M_{ball}V}{M_{ball}+M_{tube}}$ and $V_{y}= \frac {M_{tube}V}{M_{ball}+M_{tube}}$

If we consider $M_{tube} >> M_{ball}$ then $V_{x}=0$ and $V_{y}=V$ which is would be an expected result.

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    $\begingroup$ Yours is a daring approach which I couldn't gather the courage for! I've just one worry - if the ball is smaller than the cross section of the tube, and tube happens to be of quite smaller mass than that of the ball, then somewhere during the ball's stay inside the tube, it may happen that tube attains more $v_x$ than ball's $v_x$. In that case, the ball will continuously collide back and forth with the walls of the tube. And that will leave the exit part very uncertain if the arc is not ending exactly vertically upward (in which case the collisions may die out). Not sure. $\endgroup$
    – manisar
    Jul 19, 2021 at 9:17
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    $\begingroup$ Yes, you are absolutely right, I have made the appropriate changes. After calculation, it leads to the result that the only value of theta which satisfies is 45 if the radius of tube and ball are unequal but the result would be applicated at all theta at r=t. $\endgroup$ Jul 19, 2021 at 10:26

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