A tube of length L is filled completely with an in-compressible liquid of mass M and closed at both

Q: Why can't we just use the formula $F = M R \omega^2$ directly?

That formula works if all the mass is concentrated at a single distance $R$. Here, the liquid is spread out from $x=0$ to $x=L$. The parts closer to the axis experience less centrifugal force than the parts further away, so we must integrate.

Q: Why do we use the rotating frame instead of the ground frame?

In the ground frame, we would have to analyze the pressure gradient providing the centripetal acceleration for each fluid element. While perfectly valid, the rotating frame allows us to visualize the liquid 'pushing' outwards due to centrifugal force, which directly gives us the force on the outer wall.

Q: What if the tube was only half-filled?

If the tube was half-filled and rotating, the liquid would pool at the outer end. You would integrate from $x = L/2$ to $x = L$ instead of $0$ to $L$, assuming the liquid occupies the outer half.

Q: Does the cross-sectional area of the tube matter?

Interestingly, no! The area cancels out if you work with pressure, or it never appears if you work directly with linear mass density $\lambda = M/L$. The total force depends only on the total mass, length, and angular velocity.

Q: Why is the force exactly half of $M L \omega^2$?

Because the centrifugal force increases linearly with distance $x$. The average distance of the mass from the axis is $L/2$. So, the total force is equivalent to placing the entire mass at the center of mass of the tube ($L/2$).

Question

A tube of length $L$ is filled completely with an in-compressible liquid of mass $M$ and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $\omega$. The force exerted by the liquid at the other end is

Options:

(A) $\frac{M \omega^2 L}{2}$

(B) $M \omega^2 L$

(C) $\frac{M \omega^2 L}{4}$

(D) $\frac{M \omega^2 L^2}{2}$

Sigma Percentile · Accepted Answer

### The Setup: A Spinning Tube of Liquid

Imagine a tube completely filled with an incompressible liquid, rotating smoothly around one of its ends in a horizontal plane.

The question asks us to find the force this liquid exerts on the other, closed end of the tube.

To visualize this, think about what happens when you spin a bucket of water. The water wants to fly outwards, right? This is due to inertia, but in the rotating frame of the tube, we experience this as an outward **centrifugal force**.

### Slicing the Problem: The Calculus Approach

We cannot simply use the basic formula $F = M R \omega^2$ because the liquid is not concentrated at a single point. It is spread out along the entire length $L$.

The parts of the liquid closer to the axis of rotation experience a smaller centrifugal force, while the parts further away experience a larger force.

To handle this continuous distribution, we must use calculus. Let's pick a tiny slice of this liquid. We will take an element of length $dx$, located at a distance $x$ from the axis of rotation.

Since the liquid is uniformly distributed along the length $L$, the mass of our tiny element, $dm$, will simply be the total mass $M$ divided by $L$, multiplied by $dx$:

$dm = \frac{M}{L} dx$

### The Centrifugal Force on the Element

Now, what is the centrifugal force on this specific element?

The formula for centrifugal force is mass times distance times angular velocity squared. So, the force $dF$ on our element is:

$dF = (dm) x \omega^2$

Let's plug in the value of $dm$ we found earlier:

$dF = \left(\frac{M}{L} dx\right) x \omega^2$

Rearranging this a bit, we get a neat expression for $dF$:

$dF = \frac{M \omega^2}{L} x dx$

### Integrating to Find the Total Force

Here is the crucial step. The liquid is pushing outwards, and all these tiny forces from every single element add up.

The total force $F$ exerted on the outer end is the integral of $dF$ from the axis to the very end, that is, from $x = 0$ to $x = L$:

$F = \int_0^L dF$

Let's set up the integral. We can pull out the constants—$M$, $\omega^2$, and $L$—outside the integral sign:

$F = \frac{M \omega^2}{L} \int_0^L x dx$

The integral of $x$ is simply $\frac{x^2}{2}$. Now we just need to apply our limits, from $0$ to $L$:

$F = \frac{M \omega^2}{L} \left[ \frac{x^2}{2} \right]_0^L$

Substituting the upper limit $L$, we get:

$F = \frac{M \omega^2}{L} \left( \frac{L^2}{2} \right)$

One $L$ cancels out from the numerator and denominator. And we arrive at our final elegant result:

$F = \frac{M \omega^2 L}{2}$

### The Center of Mass Shortcut

Did it click? There is actually a brilliant shortcut to solve this problem in a single line!

If we look closely at our integral, we are essentially calculating $\int x \omega^2 dm$.

We can pull $\omega^2$ out, leaving us with $\omega^2 \int x dm$.

By the definition of the center of mass, $\int x dm$ is simply the total mass $M$ multiplied by the position of the center of mass $x_{cm}$.

Since the liquid is uniform, its center of mass is exactly in the middle, at $x_{cm} = \frac{L}{2}$.

So, the total force is just:

$F = M \omega^2 x_{cm} = M \omega^2 \left(\frac{L}{2}\right) = \frac{M \omega^2 L}{2}$

Boom! Solved in one line. This shows the profound elegance of physics when you understand the deeper principles at play.

Select Answer:

Visualized Solution

System Setup

The Rotating Frame

Choosing an Element

Mass of the Element

Centrifugal Force on Element

Substituting Mass

Total Force Setup

Integration

Evaluating the Integral

Final Answer

The Center of Mass Shortcut

The Sigma Insight: Dynamics of Circular Motion

The Setup

Slicing the Problem

The Centrifugal Force on the Element

Integrating to Find the Total Force

The Center of Mass Shortcut

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Select Answer:

Visualized Solution

System Setup

The Rotating Frame

Choosing an Element

Mass of the Element

Centrifugal Force on Element

Substituting Mass

Total Force Setup

Integration

Evaluating the Integral

Final Answer

The Center of Mass Shortcut

The Sigma Insight: Dynamics of Circular Motion

The Setup

Slicing the Problem

The Centrifugal Force on the Element

Integrating to Find the Total Force

The Center of Mass Shortcut

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