Question

A cylindrical container with 20 cm diameter and 60 cm height is partially filled with 50 cm high liquid whose density is 850 kg/m³. Now the cylinder is rotated at a constant speed. The rotational speed at which the liquid will start spilling from the edges of the container is found to be 4n rad/sec. Find n in nearest integer.

Answer 1

n = 5

Answer 2

1. The free surface of a rotating liquid with angular speed ω is given by

   $z = a + \frac{\omega^2}{2g} r^2$,

   where $a$ is the depth at the center and $r$ is the radial distance.

2. For a cylindrical container (radius $R = 0.10$ m) that just starts spilling at the rim (i.e. $z(R)=0.60$ m),

   $a = 0.60 - \frac{\omega^2R^2}{2g}$.

3. The volume of liquid (which is constant) is

   $V = \pi R^2 \times 0.50 = 0.005\pi$ m³.

   Under rotation, the volume becomes

   $V = \int_0^R \left(a+\frac{\omega^2}{2g}r^2\right)2\pi r\,dr = \pi aR^2 + \frac{\pi\omega^2 R^4}{4g}$.

4. Equate volume:

   $$\pi aR^2 + \frac{\pi\omega^2 R^4}{4g} = 0.005\pi.$$

   Cancel π and substitute $a$:

   $$\left(0.60 - \frac{\omega^2R^2}{2g}\right)R^2 + \frac{\omega^2R^4}{4g} = 0.005.$$

   With $R=0.10$ m and $g=9.81$ m/s²:

   $$0.60(0.01) - \frac{\omega^2(0.0001)}{2(9.81)} + \frac{\omega^2(0.0001)}{4(9.81)} = 0.005.$$

   That is,

   $$0.006 - \frac{\omega^2(0.0001)}{19.62} + \frac{\omega^2(0.0001)}{39.24} = 0.005.$$

   Combine the ω² terms:

   $$0.006 - \left(\frac{1}{19.62}-\frac{1}{39.24}\right)\omega^2(0.0001) = 0.005.$$

   Since

   $$\frac{1}{19.62}-\frac{1}{39.24} = \frac{2-1}{39.24} = \frac{1}{39.24},$$

   then

   $$0.006 - \frac{\omega^2(0.0001)}{39.24} = 0.005.$$

   So,

   $$\frac{\omega^2(0.0001)}{39.24} = 0.006-0.005 = 0.001.$$

   Thus,

   $$\omega^2 = \frac{0.001 \times 39.24}{0.0001} = 392.4.$$

   Therefore,

   $$\omega \approx \sqrt{392.4} \approx 19.8\ \text{rad/sec}.$$

5. The problem states the speed is $4n$ rad/sec, so

   $$4n = 19.8 \quad \Rightarrow \quad n = \frac{19.8}{4} \approx 4.95.$$

   Rounded to the nearest integer, $n = 5$.