SATHEE: Centre of Mass 1 Question 2

Centre of Mass 1 Question 2

2. A uniform rectangular thin sheet $A B C D$ of mass $M$ has length $a$ and breadth $b$ , as shown in the figure. If the shaded portion $H B G O$ is cut-off, the coordinates of the centre of mass of the remaining portion will be

(2019 Main, 8 April II)

(a) $\frac{2 a}{3}, \frac{2 b}{3}$

(b) $\frac{5 a}{12}, \frac{5 b}{12}$

(d) $\frac{5 a}{3}, \frac{5 b}{3}$

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Answer:

Correct Answer: 2. (b)

Solution:

The given rectangular thin sheet $A B C D$ can be drawn as shown in the figure below,

Here,

Area of complete lamina, $A_{1} = a b$

Area of shaded part of lamina $= \frac{a}{2} \times \frac{b}{2} = \frac{a b}{4}$

$(x_{1}, y_{1}) =$ coordinates of centre of mass of complete lamina

$= \frac{a}{2}, \frac{a}{2}$

$(x_{2}, y_{2}) =$ coordinates of centre of mass of shaded part of lamina $= \frac{3 a}{4}, \frac{3 a}{4}$

$∴$ Using formula for centre of mass, we have

$X_{C M} = \frac{A_{1} x_{1} - A_{2} x_{2}}{A_{1} - A_{2}}$

$= \frac{a b \frac{a}{2} - \frac{a b}{4} \frac{3 a}{4}}{a b - \frac{a b}{4}} = \frac{\frac{8 a^{2} b - 3 a^{2} b}{16}}{\frac{3 a b}{4}} = \frac{5 a}{12}$

Similarly, $Y_{C M} = \frac{A_{1} y_{1} - A_{2} y_{2}}{A_{1} - A_{2}}$

$= \frac{a b \frac{b}{2} - \frac{a b}{4} \frac{3 b}{4}}{a b - \frac{a b}{4}} = \frac{5 b}{12}$

The coordinate of the centre of mass is $\frac{5 a}{12}, \frac{5 b}{12}$ .

Alternate Solution

Let $m$ be the mass of entire rectangular lamina.

So, the mass of the shaded portion of lamina $= \frac{m}{4}$

Using the relation,

$\begin{aligned} X_{C M} & = \frac{m_{1} x_{1} - m_{2} x_{2}}{m_{1} - m_{2}}, we get \\ X_{C M} & = \frac{m \frac{a}{2} - \frac{m}{4} \frac{3 a}{4}}{m - \frac{m}{4}} = \frac{\frac{a}{2} - \frac{3 a}{16}}{\frac{3}{4}} = \frac{5 a}{12} \end{aligned}$

Similarly, $Y_{C M} = \frac{m_{1} y_{1} - m_{2} y_{2}}{m_{1} - m_{2}}$ , we get

$Y_{C M} = \frac{m \frac{a}{2} - \frac{m}{4} \frac{3 b}{4}}{m - \frac{m}{4}} = \frac{\frac{b}{2} - \frac{3 b}{16}}{\frac{3}{4}} = \frac{5 b}{12}$

$∴$ The coordinates of the centre of mass of remaining portion will be $\frac{5 a}{12}, \frac{5 b}{12}$ .

Centre of Mass 1 Question 2

2. A uniform rectangular thin sheet $A B C D$ of mass $M$ has length $a$ and breadth $b$ , as shown in the figure. If the shaded portion $H B G O$ is cut-off, the coordinates of the centre of mass of the remaining portion will be

Answer:

Alternate Solution

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Centre of Mass 1 Question 2

2. A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be

Answer:

Alternate Solution

Engineering Preparation

Medical Preparation

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Important Resources

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Syllabus

Test Platform

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NCERT Exercise Video Solution

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2. A uniform rectangular thin sheet $A B C D$ of mass $M$ has length $a$ and breadth $b$ , as shown in the figure. If the shaded portion $H B G O$ is cut-off, the coordinates of the centre of mass of the remaining portion will be