The position vector of the centre of mass $\overrightarrow{\mathrm{r}} \mathrm{cm}$ of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:

(A) $\vec{r} \mathrm{~cm}=\frac{13}{8} L \hat{x}+\frac{5}{8} L \hat{y}$

(B) $\vec{r} \mathrm{~cm}=\frac{5}{8} \mathrm{~L} \hat{\mathrm{x}}+\frac{13}{8} \mathrm{~L} \hat{\mathrm{y}}$

(C) $\vec{r} \mathrm{~cm}=\frac{3}{8} L \hat{x}+\frac{11}{8} L \hat{y}$

(D) $\vec{r} \mathrm{~cm}=\frac{11}{8} L \hat{x}+\frac{3}{8} L \hat{y}$

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Conservation of Momentum

The position vector of the centre of mass

r cm

of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:

(A) $r cm = \frac{13}{8} L \overset{x}{^} + \frac{5}{8} L \overset{y}{^}$ ✓ Correct
(B) $r cm = \frac{5}{8} L \overset{x}{^} + \frac{13}{8} L \overset{y}{^}$
(C) $r cm = \frac{3}{8} L \overset{x}{^} + \frac{11}{8} L \overset{y}{^}$
(D) $r cm = \frac{11}{8} L \overset{x}{^} + \frac{3}{8} L \overset{y}{^}$

Answer: (A)

Solution

Break the bent bar into straight pieces and use the fact that each uniform piece has its mass at its own midpoint, with mass proportional to length. Taking the pieces of mass

2 m

m

and

m

located at

(L, L)

(2 L, \frac{L}{2})

and

(\frac{5 L}{2}, 0)

, the centre of mass is the mass-weighted average of these positions. The

x

-coordinate is

\frac{2 m ( L ) + m ( 2 L ) + m ( \frac{5 L}{2} )}{4 m} = \frac{13}{8} L

and the

y

-coordinate is

\frac{2 m ( L ) + m ( \frac{L}{2} ) + m ( 0 )}{4 m} = \frac{5}{8} L

, so

r_{cm} = \frac{13}{8} L \overset{x}{^} + \frac{5}{8} L \overset{y}{^}

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Conservation of Momentum

Solution

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