0
$\begingroup$

You drop a pencil, point down, into the water. If you release the pencil from a height of 4.0 cm (measured from the point to the water level), how far will it dive? Treat the pencil as a cylinder of radius 0.40 cm, length 19 cm, and mass 4.0 g. Ignore the frictional and inertial resistance of the water, but take into account the buoyant force.

first, we find the initial velocity from the conservation of energy:

$$V_0 = \sqrt{2*9.8*0.04} = 0.885$$

next, if we draw A free body diagram we would see that when the pencil is underwater there is a force that accelerates the object up that force is the buoyant force minus the weight, we find the buoyance force is equal to pressure times area, and since pressure is density times height times acceleration due to earth gravity hence:

$$F_b = mg$$ $$A\rho y g - mg = ma$$

$$\frac{0.49y-0.0392}{0.004} = a$$

$$\frac{dv}{dy} = 122.5y-9.8$$ $$\int_{v_0}^v dv = \int (122.5y-9.8)dy$$

I don't know the limits of integration of the last integral. and how would you find how far the pencil would dive.

$\endgroup$
14
  • $\begingroup$ ???............. $\endgroup$ Commented May 14, 2023 at 2:31
  • $\begingroup$ you guys have been very quiet $\endgroup$ Commented May 14, 2023 at 3:05
  • $\begingroup$ I don't understand your initial velocity. If you hold a pencil then drop it, the velocity of the pencil when holding it should be $0$ $m/s$, correct? $\endgroup$ Commented May 14, 2023 at 3:20
  • $\begingroup$ We are talking about the velocity when the pencil enters the water because the buoyant force is on that range ""I think"" $\endgroup$ Commented May 14, 2023 at 3:25
  • 2
    $\begingroup$ The first thing I would do is to make sure that the signs of all my vertical directions, speeds, and forces agreed. Using $-9.8$ as the acceleration due to gravity implies that up is positive, but using $122.5y$ as the acceleration due to the water suggests that up is negative, since the buoyancy is less when the pencil is higher but $122.5y$ increases as $y$ increases. Also, $dv/dy$ is not acceleration. $\endgroup$
    – David K
    Commented May 14, 2023 at 4:08

1 Answer 1

0
$\begingroup$

One error in the question is the formula $$ \frac{\mathrm dv}{\mathrm dy} \stackrel?= 122.5y - 9.8. \tag1$$ The formula one line earlier is correct except for the direction of acceleration; if positive velocity is down then positive acceleration should also be down, and the correct formula is $$ a = \frac{0.0392 - 0.49y}{0.004} = 9.8 - 122.5y. \tag2$$

In addition to the sign error, Equation $(1)$ is incorrect because it substitutes $\dfrac{\mathrm dv}{\mathrm dy}$ for acceleration, but acceleration is $\dfrac{\mathrm dv}{\mathrm dt}$ instead. But we can use the chain rule to write $$ a = \frac{\mathrm dv}{\mathrm dt} = \frac{\mathrm dy}{\mathrm dt}\frac{\mathrm dv}{\mathrm dy} = v \frac{\mathrm dv}{\mathrm dy}. $$ So that's where the extra "$v$" comes from.

If we take the indefinite integral of both sides of Equation $(2)$ with respect to $y$, we get $$ \int a\,\mathrm dy = \int v \frac{\mathrm dv}{\mathrm dy}\,\mathrm dy = \int (9.8 - 122.5y)\,\mathrm dy $$ which is close to what you wrote in a comment; I just have not yet replaced $\displaystyle\int v \dfrac{\mathrm dv}{\mathrm dy}\,\mathrm dy$ by $\displaystyle\int v\,\mathrm dv,$ because I want to consider both sides as a function of $y$ for a little longer.

For limits of integration, it makes sense to start at $y=0$ (because that's when the acceleration $9.8 - 122.5y$ first applies) and to stop at some depth $y=y1$ where we want this integration to stop: $$ \int_0^{y_1} v \frac{\mathrm dv}{\mathrm dy}\,\mathrm dy = \int_0^{y_1} (9.8 - 122.5y)\,\mathrm dy $$

Now we can change variables from $y$ to $v$ on the left by specifying that $v_0 = 0.885438\ldots$ is the velocity at the instant the pencil hits the water and $v_1$ is the velocity when the pencil reaches depth $y_1$: $$ \int_{v_0}^{v_1} v\,\mathrm dv = \int_0^{y_1} (9.8 - 122.5y)\,\mathrm dy. $$

A seemingly obvious place to stop the integration is at the lowest depth reached by the pencil, that is, $y_1$ is the depth where the velocity is zero. So we might try $$ \int_{v_0}^0 v\,\mathrm dv \stackrel?= \int_0^{y_1} (9.8 - 122.5y)\,\mathrm dy.\tag3 $$

On the left we have $$ \int_{v_0}^0 v\,\mathrm dv = \left[ \frac12 v^2 \right]_{v_0}^0 = -\frac12 v_0^2 = -\frac12(2\times 9.8 \times 0.04) = -0.392. $$

On the right we have $$ \int_0^{y_1} (9.8 - 122.5y)\,\mathrm dy = \bigg[ 9.8 y - 61.25 y^2 \bigg]_0^{y_1} = 9.8 y_1 - 61.25 y_1^2. $$

Equation $(3)$ therefore is equivalent to $$ 9.8 y_1 - 61.25 y_1^2 = -0.392, $$ which has solutions $ y_1 = -0.0331371\ldots$ and $y_1 = 0.193137\ldots . $

The first solution is obviously out of range, but there is also a problem with the second solution: $0.193137$ meters is longer than the length of the pencil. The right-hand integral can only be used for values of $y_1$ between $0$ and $0.19$, because once the bottom end of the pencil is more than $19$ cm below the surface, the water is pressing down on the top surface of the pencil as well as upward on the bottom surface and the acceleration equation changes.

So Equation $(3)$ is invalid. We cannot follow the pencil all the way to its deepest depth with this equation; we can only go to a depth equal to the pencil's length, $$ \int_{v_0}^{v_1} v\,\mathrm dv = \int_0^{0.19} (9.8 - 122.5y)\,\mathrm dy. \tag4 $$ The right-hand side of Equation $(4)$ is $$ 9.8 (0.19) - 61.25 (0.19)^2 = -0.349125 . $$ The left-hand side is $$ \int_{v_0}^{v_1} v\,\mathrm dv = \left[ \frac12 v^2 \right]_{v_0}^{v_1} = \frac12 v_1^2 - \frac12 v_0^2 = \frac12 v_1^2 - 0.392. $$ Therefore $\frac12 v_1^2 - 0.392 = -0.349125, $ which has solution $v_1 = \pm 0.292831\ldots,$ and we want the positive solution since the pencil is still descending.

To complete the exercise you need to find the acceleration $a_1$ of the pencil below the depth $0.19$ (hint: it is constant from that point downward) and solve the integral $$ \int_{v_1}^0 v\,\mathrm dv = \int_{0.19}^{y_2} a_1\,\mathrm dy. $$ where $v_1 = 0.292831\ldots$ and $y_2$ is the maximum depth reached by the pencil.


Personally, I would rather use potential energy to solve the problem. The potential energy at the deepest position of the pencil is equal to the potential energy at the start, because in both cases there is no velocity and so no kinetic energy. But in the deepest position of the pencil you have to account for the premise that the water displaced by the pencil has been raised to the surface.

$\endgroup$
4
  • $\begingroup$ eq 3 seems valid because the answer is -0.19 $\endgroup$ Commented May 22, 2023 at 15:10
  • $\begingroup$ why do you think the pencil cant go deeper than his length? $\endgroup$ Commented May 22, 2023 at 15:15
  • $\begingroup$ To the nearest $1$ cm, the answer is still $-0.19$. But the math says the pencil has not yet lost all downward speed at the instant it reaches exactly $19$ cm below the surface. It is almost stopped by then, but not quite. It seems a poorly-designed problem; if you were meant to use only Equation (3) then you should have been given a setup where the pencil never gets too near to its full length underwater. Instead, the problem seems to be set up in such a way as to teach you to ignore boundary conditions, which I think is bad teaching. $\endgroup$
    – David K
    Commented May 22, 2023 at 15:46
  • $\begingroup$ Perhaps I misunderstood the last comment: it says "can't go deeper" but I read "can go deeper". In fact the math says that the pencil will go deeper than its own length. Equation 3 demonstrates this because when we solve it to more than two digits we find that $y_1 > 0.19$ (which means "deeper" because we measured positive distance in the downward direction). Physically, this is not a problem at all; the pencil just goes underwater. Only the math is a bit tricky because the bouyant force stops increasing when the pencil is fully underwater. $\endgroup$
    – David K
    Commented May 22, 2023 at 22:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .