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
5153 cm
B
5173 cm
C
5193 cm
D
7173 cm

Solution

## The correct option is B $$\displaystyle \dfrac{5\sqrt{17}}{3}$$ cm Total length of a rope, $$L=30cm$$Let mass density $$=\lambda$$$$\mu=0.5$$ coefficient of friction.Frictional force,$$\mu[\lambda(L-l)]g=\lambda lg$$$$\mu (L-l)=l$$$$\dfrac{l}{L}=\dfrac{\mu}{\mu +1}$$$$l=30\dfrac{0.5}{0.5+1}=10cm$$$$L-l=20cm$$The center of mass along x-axis is given by$$X_{CM}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}$$$$X_{CM}=\dfrac{20\lambda\times (-10)+10\lambda \times (0)}{20\lambda +10\lambda}$$$$X_{CM}=-\dfrac{20}{3}cm$$The center of mass along y-axis is given by$$Y_{CM}=\dfrac{m_1y_1+m_2y_2}{m_1+m_2}$$$$Y_{cm}=\dfrac{5\times 10\lambda+20\lambda \times 0}{20\lambda +10\lambda}$$$$Y_{CM}=-\dfrac{5}{3}cm$$We know that,$$CM=\sqrt{(X_{CM})^2+(Y_{CM})^2}$$$$CM=\sqrt{(\dfrac{-20}{3})^2+(\dfrac{-5}{3})^2}$$$$CM=\dfrac{5\sqrt{17}}{3}cm$$The correct option is B.Physics

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