A tube of length 1 m is filled completely with an ideal liquid of mass 2 M , and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is F then angular velocity of the tube is…

Question

A tube of length 1 m is filled completely with an ideal liquid of mass 2 M , and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is F then angular velocity of the tube is $\sqrt{\frac{\mathrm{F}}{\alpha \mathrm{M}}}$ in SI unit. The value of $\alpha$ is __________.

Aicharya · Accepted Answer

Step 1: Consider the Force at a Distance $x$
When the tube rotates with angular velocity $\omega$, each element of the liquid experiences a centrifugal force.
The force at a distance $x$ from the pivot can be found by considering the differential force due to an element of liquid.
$d F=\rho A d x \omega^2 x$
where $\rho$ is the volumetric mass density of the liquid.
Step 2: Find the Volumetric Mass Density
Since the mass of the liquid is $2 M$, the length of the tube is 1 m, and let the area of cross-section is $A$, the volumetric mass density is:
$\rho=\frac{2 M}{A L}=\frac{2 M}{A}$
Step 3: Calculate the Total Force at the Other End
The force at $x=L$ is obtained by integrating:
$\begin{aligned}
& F=\int_0^L \rho x \omega^2 d x \\
& F=2 M \omega^2 \int_0^1 x d x \\
& F=\frac{2 M \omega^2}{2}=M \omega^2
\end{aligned}$
Step 4: Solve for $\omega$
$\begin{aligned}
& \omega^2=\frac{F}{M} \\
& \omega=\sqrt{\frac{F}{M}}
\end{aligned}$

A tube of length 1 m is filled completely with an ideal liquid of mass 2 M , and closed at both ends. The…

Solution