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(D) Consider a small elemental ring of radius $x$ and width $dx$ on the circular mop. The area of this ring is $dA = 2\pi x dx$. The total area of the

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$\frac{2}{3}\mu FR$

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(D) Consider a small elemental ring of radius $x$ and width $dx$ on the circular mop. The area of this ring is $dA = 2\pi x dx$. The total area of the mop is $A = \pi R^2$. Since the force $F$ is distributed uniformly,the normal force $dN$ on this elemental ring is $dN = (F/A) dA = (F/(\pi R^2)) 	imes 2\pi x dx = (2F/R^2) x dx$. The frictional force on this ring is $df = \mu dN = \mu (2F/R^2) x dx$. The torque $d	au$ due to this frictional force about the axis of rotation is $d	au = x df = x 	imes \mu (2F/R^2) x dx = (2\mu F/R^2) x^2 dx$. To find the total torque $	au$,we integrate $d	au$ from $x = 0$ to $x = R$: $	au = \int_0^R (2\mu F/R^2) x^2 dx = (2\mu F/R^2) [x^3/3]_0^R = (2\mu F/R^2) (R^3/3) = \frac{2}{3}\mu FR$.

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