2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Mriganka Basu Roy Chowdhury Sep 1 '14 at 12:41
  • 1
    $\begingroup$ That looks right to me. $\endgroup$ – Floris Sep 1 '14 at 19:47
  • $\begingroup$ 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 $g=10$. $\endgroup$ – Mriganka Basu Roy Chowdhury Sep 2 '14 at 7:25
  • 1
    $\begingroup$ 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. $\endgroup$ – Floris Sep 2 '14 at 8:14
  • $\begingroup$ No. They give no hints. :( Thats the problem. +1 for that document. :D $\endgroup$ – Mriganka Basu Roy Chowdhury Sep 2 '14 at 13:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.