# Weight held by jet stream of water

Consider a bucket of weight $W$, held in air by a jet stream of water from the ground (moving vertically upwards) at height $h$. The water is fired from the ground at the rate of $\mu$ (units kg/s) with an initial velocity of $v_0$. Find the maximum possible value of $h$. Find conditions to achieve the same.

(To observers, this is a problem from Kleppner & Kollenkow: An Introduction to Mechanics).

What I did: Consider a differential mass element $dm$ near the bucket before hitting it. Its velocity is $v_f = \sqrt{v_0^2 - 2gh}$, so its momentum is $v_f dm$. Suppose after hitting the bucket it had a velocity of $v'$, and the time of contact was $\Delta t$. Since we must exert a force of $W$ upwards on the bucket, the rate-of-change of momentum of $dm$ is $W$. Clearly $dm(v' + v_f) = W\Delta t$, which in the limit $\Delta t \to 0$, becomes $W = \mu (v' + v_f)$. Thus $2gh = v_0^2 - \left(\frac{W}{\mu} - v'\right)^2$. This raises a question. Setting $v' = \frac{W}{\mu}$, we can attain $h = v_0^2 / 2g$, but at that height, the water would have no velocity, so no momentum change.

What are my errors, and how to attack such problems (with continuous masses)? ( Please excuse me for any silly mistakes. I'm only a beginner :( ). A detailed solution which I can study would be extremely helpful.

You correctly identify the residual velocity of the water after bouncing off the bucket as a critical parameter in the calculation. Where you go wrong is in assuming that you can assign any value you want to it.

If your bucket's bottom was shaped in such a way as to "turn around" the water jet hitting it, then you would have the maximum possible momentum exchange - namely $2\mu v$ where $v$ is the velocity at height $h$. Making the jet go faster than that would require you to add energy to it - imaging having a little propeller at the bottom of your bucket which sends the water down with a greater velocity than it arrived. You would end up with more kinetic energy in the water after this interaction than before - because the propeller did work on the water stream.

For a passive bucket the kinetic energy of the water cannot increase.

Do you see it now?

• Setting $v' = v_f$ in magnitude, gives me, $v_f = W/2\mu$, which in turn gives, $2gh = v_0^2 - \left(\frac{W}{2\mu}\right)^2$. Is this the right answer, with the conditions being a perfectly elastic collision? – Mriganka Basu Roy Chowdhury Sep 1 '14 at 12:41
• That looks right to me. – Floris Sep 1 '14 at 19:47
• But Kollenkow gives : (In SI units) For $v=20$,$\mu=0.5,W=10,h_\mathrm{max} \approx 17$. I didn't understand what was meant by a weight of $10\ \mathrm{kg}$. But putting in those values gives you $15$, with $g=10$. – Mriganka Basu Roy Chowdhury Sep 2 '14 at 7:25
• That's strange - according to scribd.com/doc/48248022/Impact-of-a-Jet-of-Water we are doing the right thing (and the bucket that turns around the water as we are proposing is called a "Pelton Bucket" - I learnt something new today). Does your book give any hints on what they are doing? They seem to be able to support the weight of 10 N with a mass flow of 0.5 kg/s and a jet velocity of < 10 m/s. I don't know how they do that. – Floris Sep 2 '14 at 8:14
• No. They give no hints. :( Thats the problem. +1 for that document. :D – Mriganka Basu Roy Chowdhury Sep 2 '14 at 13:52