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A spherical shot of W gm weight and radius r cm, lies at the bottom of cylindrical bucket of radius R cm. The bucket is filled with water up to a depth of $h\,\mathrm{cm}$ $\left(h > 2r\right).$ Show that the minimum amount of work done in lifting the shot just clear of the water must be$$ [W(h-4r^3/3R^2)+W'(r-h+2r^3/3R^2)] \,.$$ W'gm is the weight of water displaced by the shot.

I tried to solve this question by assuming that Work is generally defined as force times distance. The depth is the distance in this case (or however far you lift the object) and the force is the apparent weight of the submerged object so W(app) = mg − Fb = mg − ρ(fluid)gV(obj)

but i dont know weather to add drag force or not and how to use w'( wt of water displaced by shot)

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    $\begingroup$ The energy needed to overcome "drag force" usually is a function of speed. The faster you try to move whatever it is, the more the "drag force" opposes the movement. Since the question asks you to compute the minimum amount of work needed, you probably can assume that the shot is lifted so slowly that the drag force can be ignored. But of course, only your instructor can say for certain. $\endgroup$ – Solomon Slow Jun 11 '18 at 15:15
  • $\begingroup$ Welcome to the site! We expect mathematics to be typeset using MathJax, as in the edit by @Nat. If you want to further edit your question, you're expected to build on top of the existing improvements, instead of rolling them back. $\endgroup$ – Emilio Pisanty Jun 11 '18 at 15:15
  • $\begingroup$ Furthermore, please read carefully our guidelines for homework and exercise questions. $\endgroup$ – Emilio Pisanty Jun 11 '18 at 15:16
  • $\begingroup$ Hi Rahul and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Jun 11 '18 at 15:17
  • $\begingroup$ @jameslarge thanxx.. though i neglect drag force but still how to use W' and how to reach W′(r−h+2r^3/3R^2)] . $\endgroup$ – Rahul Shukla Jun 11 '18 at 15:18
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If you assume that the ball was lifted slowly, you can ignore the drag losses and any changes in the kinetic energy, so the problem is reduced to the changes in the potential energy of the ball and the water displaced by the ball.

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The water level will decrease by $d$ and you can find $d$ by equating the volume of the ball with the volume of the cylinder of height $d$.

You can calculate the increase of the potential energy of the ball knowing that its weight is $W$ and its center of gravity has increased from $r$ to $h-d+r$.

You can calculate the decrease of the potential energy of water displaced by the ball knowing that its weight is $W'$ and its center of gravity has decreased from $h-\frac d 2$ to $r$.

The minimal work needed to lift the ball to the level just above the new level of water will be equal to the sum of changes of the potential energy of the ball and the potential energy of the water displaced by the ball.

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  • $\begingroup$ @ V.F. thanxx.. i get you but from where you get d/2 . should't we have to take only d? $\endgroup$ – Rahul Shukla Jun 12 '18 at 12:47
  • $\begingroup$ @RahulShukla $h-d/2$ is the height of the center of gravity of the slice of the displaced water ($d$, of course, is the height or the thickness of the slice). $\endgroup$ – V.F. Jun 12 '18 at 19:18

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