Answer:
Explanation:
To find the angular velocity of the tank at which the bottom of the tank is exposed
From the information given:
At rest, the initial volume of the tank is:
[tex]V_i = \pi R^2 h_i --- (1)[/tex]
where;
height h which is the height for the free surface in a rotating tank is expressed as:
[tex]h = \dfrac{\omega^2 r^2}{2g} + C[/tex]
at the bottom surface of the tank;
r = 0, h = 0
∴
[tex]h = \dfrac{\omega^2 r^2}{2g} + C[/tex]
0 = 0 + C
C = 0
Thus; the free surface height in a rotating tank is:
[tex]h=\dfrac{\omega^2 r^2}{2g} --- (2)[/tex]
Now; the volume of the water when the tank is rotating is:
dV = 2π × r × h × dr
Taking the integral on both sides;
[tex]\int \limits ^{V_f}_{0} \ dV = \int \limits ^R_0 \times 2 \pi \times r \times h \ dr[/tex]
replacing the value of h in equation (2); we have:
[tex]V_f} = \int \limits ^R_0 \times 2 \pi \times r \times ( \dfrac{\omega ^2 r^2}{2g} ) \ dr[/tex]
[tex]V_f = \dfrac{ \pi \omega ^2}{g} \int \limits ^R_0 \ r^3 \ dr[/tex]
[tex]V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{r^4}{4} \Big]^R_0[/tex]
[tex]V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{R^4}{4} \Big] --- (3)[/tex]
Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.
Then [tex]V_f = V_i[/tex]
Replacing equation (1) and (3)
[tex]\dfrac{\pi \omega^2}{g}( \dfrac{R^4}{4}) = \pi R^2 h_i[/tex]
[tex]\omega^2 = \dfrac{4g \times h_i }{R^2}[/tex]
[tex]\omega =\sqrt{ \dfrac{4g \times h_i }{R^2}}[/tex]
[tex]\omega = \sqrt{\dfrac{4 \times 9.81 \ m/s^2 \times 0.7 \ m}{(0.5)^2} }[/tex]
[tex]\omega = \sqrt{109.87 }[/tex]
[tex]\mathbf{\omega = 10.48 \ rad/s}[/tex]
Finally, the angular velocity of the tank at which the bottom of the tank is exposed = 10.48 rad/s
