Question

A rope of length 30 cm is on a horizontal table with maximum length hanging from edge A of the table. The coefficient of friction between the rope and table is 0.5. The distance of centre of mass of the rope from A is:

• A. $\frac{5\sqrt{15}}{3}$ cm
• B. $\frac{5\sqrt{17}}{3}$ cm
• C. $\frac{5\sqrt{19}}{3}$ cm
• D. $\frac{7\sqrt{17}}{3}$ cm

Answer 1

The correct option is B $\frac{5\sqrt{17}}{3}$ cm

Total length of a rope, $L=30cm$

Let mass density $=\lambda$

$\mu =0.5$ coefficient of friction.

Frictional force,

$\mu \left[\lambda \right(L-l\left)\right]g=\lambda lg$

$\mu (L-l)=l$

$\frac{l}{L}=\frac{\mu }{\mu +1}$

$l=30\frac{0.5}{0.5+1}=10cm$

$L-l=20cm$

The center of mass along x-axis is given by

$X_{CM}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$

$X_{CM}=\frac{20\lambda \times (-10)+10\lambda \times \left(0\right)}{20\lambda +10\lambda }$

$X_{CM}=-\frac{20}{3}cm$

The center of mass along y-axis is given by

$Y_{CM}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$

$Y_{cm}=\frac{5\times 10\lambda +20\lambda \times 0}{20\lambda +10\lambda }$

$Y_{CM}=-\frac{5}{3}cm$

We know that,

$CM=\sqrt{(X_{CM}{)}^2+(Y_{CM}{)}^2}$

$CM=\sqrt{(\frac{-20}{3}{)}^2+(\frac{-5}{3}{)}^2}$

$CM=\frac{5\sqrt{17}}{3}cm$

The correct option is B.