To mop-clean a floor, a cleaning machine presses a circular mop of radius R vertically down with a t

Q: Why can't we just assume the total frictional force acts at the center of mass of the disc?

The center of mass of a uniform disc is at its geometric center (radius = 0). If you assumed all friction acted there, the torque would be zero! Friction acts all over the surface, and torque depends on the distance from the axis ($x$). Since $x$ varies, we must integrate.

Q: How did we get the area of the elemental ring as $2\pi x dx$?

Imagine taking the thin ring of radius $x$ and thickness $dx$ and cutting it open to lay it flat. It forms a thin rectangle. The length of this rectangle is the circumference of the ring ($2\pi x$), and its width is $dx$. Therefore, the area is length $\times$ width = $2\pi x dx$.

Q: Why is the normal force $dN$ proportional to the area?

The problem states that the total downward force $F$ is 'distributed uniformly' over the mop. This means the pressure (Force/Area) is constant everywhere. Since Pressure = $dN / dA$, a larger area $dA$ will experience a proportionally larger normal force $dN$.

Q: What if the mop was a thin circular ring instead of a solid disc?

If it were a thin ring of radius $R$, all the frictional force would act at the exact same distance $R$ from the axis. The total friction would be $\mu F$, and the total torque would simply be $\tau = (\mu F) \times R$. No integration needed!

Q: Does the angular speed $\omega$ affect the value of the frictional torque?

No, it does not! Kinetic friction depends only on the normal force and the coefficient of kinetic friction ($\mu$). It is independent of the relative velocity between the surfaces. Therefore, whether the mop spins fast or slow, the torque required to overcome friction remains the same.

Question

To mop-clean a floor, a cleaning machine presses a circular mop of radius $R$ vertically down with a total force $F$ and rotates it with a constant angular speed about its axis. If the force $F$ is distributed uniformly over the mop and if coefficient of friction between the mop and the floor is $\mu$, the torque applied by the machine on the mop in (N-m) is

Options:

(A) $\frac{2}{3}\mu FR$

(B) $\frac{\mu FR}{6}$

(C) $\frac{\mu FR}{3}$

(D) $\frac{\mu FR}{2}$

Sigma Percentile · Accepted Answer

### Visualizing the Rotating Mop

Imagine a heavy-duty cleaning machine pressing a circular mop against the floor. The mop has a radius $R$ and is rotating with a constant angular speed $\omega$. A total downward force $F$ is uniformly distributed over its entire area.

Our goal is to find the total torque required by the machine to overcome the kinetic friction between the mop and the floor.

Why can't we just use the simple formula $	au = 	ext{Force} 	imes 	ext{Radius}$? Because the mop is a continuous 2D surface! The frictional force acts at varying distances from the center. The outer edges sweep a larger circle and contribute more to the torque than the inner parts. To solve this, we must summon the power of calculus.

### The Elemental Ring Approach

Let's slice the mop into infinitely thin concentric rings. Consider a small elemental ring of radius $x$ and an infinitesimally small thickness $dx$.

The area of this tiny ring, $dA$, can be found by imagining we cut the ring and lay it flat. It forms a thin rectangle with a length equal to its circumference ($2\pi x$) and a width of $dx$.

Therefore, the area is:
$$dA = 2\pi x \, dx$$

### Calculating the Frictional Torque

Since the total force $F$ is uniformly distributed, the pressure $P$ (force per unit area) is constant everywhere on the mop:
$$P = \frac{F}{\pi R^2}$$

The normal force $dN$ acting just on our elemental ring will be this pressure multiplied by the ring's area $dA$:
$$dN = P \cdot dA = \left( \frac{F}{\pi R^2} ight) (2\pi x \, dx)$$

Simplifying this, we get:
$$dN = \frac{2F}{R^2} x \, dx$$

Now, the kinetic friction acting on this ring opposes its rotation. This frictional force $dF$ is simply the coefficient of friction $\mu$ times the normal force $dN$:
$$dF = \mu \cdot dN = \frac{2\mu F}{R^2} x \, dx$$

The tiny torque $d	au$ produced by this friction is the force $dF$ multiplied by its perpendicular distance from the center, which is $x$:
$$d	au = x \cdot dF = x \cdot \left( \frac{2\mu F}{R^2} x \, dx ight)$$

$$d	au = \frac{2\mu F}{R^2} x^2 \, dx$$

### The Final Integration

To find the total torque applied by the machine on the entire mop, we must integrate this elemental torque from the center (where $x = 0$) to the outer edge (where $x = R$).

$$	au = \int_0^R d	au = \int_0^R \frac{2\mu F}{R^2} x^2 \, dx$$

The constant terms $\frac{2\mu F}{R^2}$ can be pulled outside the integral:
$$	au = \frac{2\mu F}{R^2} \int_0^R x^2 \, dx$$

The integral of $x^2$ with respect to $x$ is simply $\frac{x^3}{3}$. Applying our limits from $0$ to $R$:
$$	au = \frac{2\mu F}{R^2} \left[ \frac{x^3}{3} ight]_0^R$$

$$	au = \frac{2\mu F}{R^2} \left( \frac{R^3}{3} - 0 ight)$$

Finally, the $R^2$ in the denominator cancels out two powers of $R$ in the numerator. We are left with the final expression for the total torque:

**$$	au = \frac{2}{3} \mu F R$$**

This is the torque the machine must overcome to keep the mop rotating. Notice how it is exactly $2/3$ of what the torque would be if all the friction acted purely at the outer rim!

Select Answer:

Visualized Solution

$Visualizing the Mop$

$Elemental Ring$

$Normal Force on Ring$

$Substituting Values$

$\text{Frictional Force & Torque}$

$Elemental Torque Expression$

$Integrating for Total Torque$

$Evaluating the Integral$

$Final Torque$

$Food for Thought$

The Sigma Insight: Dynamics of Rigid Body Rotation

Visualizing the Rotating Mop

The Elemental Ring Approach

Calculating the Frictional Torque

The Final Integration

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Mop

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Elemental Ring

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Normal Force on Ring

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Frictional Force & Torque}

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Elemental Torque Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integrating for Total Torque

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating the Integral

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Torque

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Food for Thought