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Rotational Motion

149762 Three particles are attached to a ring of mass $m$ and radius $R$ as shown in the figure. The centre of mass the ring has a speed $V_{0}$ and rolls without slipping on a horizontal surface. The kinetic energy of the system in the position shown in the figure is

1

6 m V_{0}^{2}

2

12 m V_{0}^{2}

3

2 m V_{0}^{2}

4

8 m V_{0}^{2}

Explanation:

A When body is pure rolling
$V_{0} = R ω$
Speed on the left side is:
$V_{1} = \sqrt{V_{0}^{2} + (R ω)^{2}}$
$V_{1} = \sqrt{V_{0}^{2} + V_{0}^{2}}$
$V_{1} = V_{0} \sqrt{2}$
Kinetic energy $(E_{1}) = \frac{1}{2} \times 2 m \times V_{1}^{2}$
$= m \times V_{0}^{2} \times 2$
$= 2 m V_{0}^{2}$
Speed on the top-
$V_{2} = V_{0} + R ω$
$V_{2} = V_{0} + V_{0} = 2 V_{0}$
$E_{2} = \frac{1}{2} {mV}_{2}^{2}$
$E_{2} = \frac{1}{2} m {(2 V_{0})}^{2}$
$E_{2} = 2 m V_{0}^{2}$
Speed on right side-
$V_{3} = \sqrt{V_{0}^{2} + (R ω)^{2}}$
$V_{3} = \sqrt{V_{0}^{2} + {(V_{0})}^{2}}$
$V_{3} = \sqrt{2} V_{0}$
$E_{3} = \frac{1}{2} m {(V_{3})}^{2} = \frac{1}{2} m \times {(\sqrt{2} V_{0})}^{2}$
$E_{3} = {mV}_{0}^{2}$
Kinetic energy of ring
$E_{ring} = \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} {mR}^{2} ω^{2}$
$= \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} m (R ω)^{2}$
$= \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} {mV}_{0}^{2}$
$E_{ring} = m V_{0}^{2}$
Total kinetic energy $(E) = E_{1} + E_{2} + E_{3} + E_{ring}$
$= 2 {mV}_{0}^{2} + 2 {mV}_{0}^{2} + {mV}_{0}^{2} + {mV}_{0}^{2}$
$= 6 {mV}_{0}^{2}$

AP EAMCET(Medical)-2016

Rotational Motion

149763 Two blocks of equals mass are released on two smooth sides of a double inclined plane with a fixed base as shown in the figure. If each angle of inclination is $45^{\circ}$ , the acceleration of the centre of mass of the system of the two blocks is (Acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

10 {ms}^{- 2}

vertically downward

2

10 {ms}^{- 2}

vertically upward

3

5 {ms}^{- 2}

vertically downward

4

5 {ms}^{- 2}

vertically upward

Explanation:

A Given that, Mass of two blocks are equal $m_{1} = m_{2} = m$

Acceleration of $A = a_{1}$
\& acceleration of $B = a_{2}$ , both are equal.
Gravity due to acceleration $(g) = - 10 m / s^{2}$
Acceleration of centre of mass of system
$a_{cm} = \frac{m_{1} a_{1} + m_{2} a_{2}}{m_{1} + m_{2}}$
$\Rightarrow \frac{m \times (- 10) + m \times (- 10)}{m + m}$
$a_{cm} = \frac{- 20 m}{2 m} \Rightarrow - 10 m / s^{2}$

AP EAMCET(Medical)-2016

Rotational Motion

1 (i)-(v), (ii)-(vii), (iii)-(vi), (iv)-(viii)

2 (i)-(vii), (ii)-(v), (iii)-(vi), (iv)-(viii)

3 (i)-(vii), (ii)-(v), (iii)-(viii), (iv)-(vi)

4 (i)-(viii), (ii)-(v), (iii)-(vi), (iv)-(vii)

AP EMCET(Medical)-2011

Rotational Motion

149765 Four particles, each of mass $1 kg$ are placed at the corners of a square $O A B C$ of side $1 m$ . ' $O$ ' is at the origin the co-ordinate system. $OA$ and $O C$ are aligned along positive $X$ -axis and positive $Y$ -axis respectively. The position vector of the centre of mass is (in ' $m$ '):

1

\hat{i} + \hat{j}

2

\frac{1}{2} (\hat{i} + \hat{j})

3

(\hat{i} - \hat{j})

4

\frac{1}{2} (\hat{i} - \hat{j})

Explanation:

B From the question given that, Four particles, each of mass $1 kg$
So, $m_{1} = m_{2} = m_{3} = m_{4} = 1 kg$
and $OA = AB = OC = BC = 1 m$
According to the question -

Co-ordinate of $O = (x_{1}, y_{1}) = (0, 0)$
Co-ordinate of $A = (x_{2}, y_{2}) = (1, 0)$
Co-ordinate of $B = (x_{3}, y_{3}) = (1, 1)$
Co-ordinate of $C = (x_{4}, y_{4}) = (0, 1)$
$x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$x_{cm} = \frac{1 \times 0 + 1 \times 1 + 1 \times 1 + 1 \times 0}{1 + 1 + 1 + 1} = \frac{2}{4} = \frac{1}{2} m$
$y_{cm} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$= \frac{1 \times 0 + 1 \times 0 + 1 \times 1 + 1 \times 1}{1 + 1 + 1 + 1} = \frac{2}{4} = \frac{1}{2} m$
The position vector of centre of mass $({\vec{r}}_{cm})$
$= x_{c m} \hat{i} + y_{c m} \hat{j}$
$\Rightarrow \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} \Rightarrow \frac{1}{2} (\hat{i} + \hat{j})$

AP EAMCET(Medical)-2006

Rotational Motion

149762 Three particles are attached to a ring of mass $m$ and radius $R$ as shown in the figure. The centre of mass the ring has a speed $V_{0}$ and rolls without slipping on a horizontal surface. The kinetic energy of the system in the position shown in the figure is

1

6 m V_{0}^{2}

2

12 m V_{0}^{2}

3

2 m V_{0}^{2}

4

8 m V_{0}^{2}

Explanation:

A When body is pure rolling
$V_{0} = R ω$
Speed on the left side is:
$V_{1} = \sqrt{V_{0}^{2} + (R ω)^{2}}$
$V_{1} = \sqrt{V_{0}^{2} + V_{0}^{2}}$
$V_{1} = V_{0} \sqrt{2}$
Kinetic energy $(E_{1}) = \frac{1}{2} \times 2 m \times V_{1}^{2}$
$= m \times V_{0}^{2} \times 2$
$= 2 m V_{0}^{2}$
Speed on the top-
$V_{2} = V_{0} + R ω$
$V_{2} = V_{0} + V_{0} = 2 V_{0}$
$E_{2} = \frac{1}{2} {mV}_{2}^{2}$
$E_{2} = \frac{1}{2} m {(2 V_{0})}^{2}$
$E_{2} = 2 m V_{0}^{2}$
Speed on right side-
$V_{3} = \sqrt{V_{0}^{2} + (R ω)^{2}}$
$V_{3} = \sqrt{V_{0}^{2} + {(V_{0})}^{2}}$
$V_{3} = \sqrt{2} V_{0}$
$E_{3} = \frac{1}{2} m {(V_{3})}^{2} = \frac{1}{2} m \times {(\sqrt{2} V_{0})}^{2}$
$E_{3} = {mV}_{0}^{2}$
Kinetic energy of ring
$E_{ring} = \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} {mR}^{2} ω^{2}$
$= \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} m (R ω)^{2}$
$= \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} {mV}_{0}^{2}$
$E_{ring} = m V_{0}^{2}$
Total kinetic energy $(E) = E_{1} + E_{2} + E_{3} + E_{ring}$
$= 2 {mV}_{0}^{2} + 2 {mV}_{0}^{2} + {mV}_{0}^{2} + {mV}_{0}^{2}$
$= 6 {mV}_{0}^{2}$

AP EAMCET(Medical)-2016

Rotational Motion

149763 Two blocks of equals mass are released on two smooth sides of a double inclined plane with a fixed base as shown in the figure. If each angle of inclination is $45^{\circ}$ , the acceleration of the centre of mass of the system of the two blocks is (Acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

10 {ms}^{- 2}

vertically downward

2

10 {ms}^{- 2}

vertically upward

3

5 {ms}^{- 2}

vertically downward

4

5 {ms}^{- 2}

vertically upward

Explanation:

A Given that, Mass of two blocks are equal $m_{1} = m_{2} = m$

Acceleration of $A = a_{1}$
\& acceleration of $B = a_{2}$ , both are equal.
Gravity due to acceleration $(g) = - 10 m / s^{2}$
Acceleration of centre of mass of system
$a_{cm} = \frac{m_{1} a_{1} + m_{2} a_{2}}{m_{1} + m_{2}}$
$\Rightarrow \frac{m \times (- 10) + m \times (- 10)}{m + m}$
$a_{cm} = \frac{- 20 m}{2 m} \Rightarrow - 10 m / s^{2}$

AP EAMCET(Medical)-2016

Rotational Motion

1 (i)-(v), (ii)-(vii), (iii)-(vi), (iv)-(viii)

2 (i)-(vii), (ii)-(v), (iii)-(vi), (iv)-(viii)

3 (i)-(vii), (ii)-(v), (iii)-(viii), (iv)-(vi)

4 (i)-(viii), (ii)-(v), (iii)-(vi), (iv)-(vii)

AP EMCET(Medical)-2011

Rotational Motion

149765 Four particles, each of mass $1 kg$ are placed at the corners of a square $O A B C$ of side $1 m$ . ' $O$ ' is at the origin the co-ordinate system. $OA$ and $O C$ are aligned along positive $X$ -axis and positive $Y$ -axis respectively. The position vector of the centre of mass is (in ' $m$ '):

1

\hat{i} + \hat{j}

2

\frac{1}{2} (\hat{i} + \hat{j})

3

(\hat{i} - \hat{j})

4

\frac{1}{2} (\hat{i} - \hat{j})

Explanation:

B From the question given that, Four particles, each of mass $1 kg$
So, $m_{1} = m_{2} = m_{3} = m_{4} = 1 kg$
and $OA = AB = OC = BC = 1 m$
According to the question -

Co-ordinate of $O = (x_{1}, y_{1}) = (0, 0)$
Co-ordinate of $A = (x_{2}, y_{2}) = (1, 0)$
Co-ordinate of $B = (x_{3}, y_{3}) = (1, 1)$
Co-ordinate of $C = (x_{4}, y_{4}) = (0, 1)$
$x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$x_{cm} = \frac{1 \times 0 + 1 \times 1 + 1 \times 1 + 1 \times 0}{1 + 1 + 1 + 1} = \frac{2}{4} = \frac{1}{2} m$
$y_{cm} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$= \frac{1 \times 0 + 1 \times 0 + 1 \times 1 + 1 \times 1}{1 + 1 + 1 + 1} = \frac{2}{4} = \frac{1}{2} m$
The position vector of centre of mass $({\vec{r}}_{cm})$
$= x_{c m} \hat{i} + y_{c m} \hat{j}$
$\Rightarrow \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} \Rightarrow \frac{1}{2} (\hat{i} + \hat{j})$

AP EAMCET(Medical)-2006

Rotational Motion

149762 Three particles are attached to a ring of mass $m$ and radius $R$ as shown in the figure. The centre of mass the ring has a speed $V_{0}$ and rolls without slipping on a horizontal surface. The kinetic energy of the system in the position shown in the figure is

1

6 m V_{0}^{2}

2

12 m V_{0}^{2}

3

2 m V_{0}^{2}

4

8 m V_{0}^{2}

Explanation:

A When body is pure rolling
$V_{0} = R ω$
Speed on the left side is:
$V_{1} = \sqrt{V_{0}^{2} + (R ω)^{2}}$
$V_{1} = \sqrt{V_{0}^{2} + V_{0}^{2}}$
$V_{1} = V_{0} \sqrt{2}$
Kinetic energy $(E_{1}) = \frac{1}{2} \times 2 m \times V_{1}^{2}$
$= m \times V_{0}^{2} \times 2$
$= 2 m V_{0}^{2}$
Speed on the top-
$V_{2} = V_{0} + R ω$
$V_{2} = V_{0} + V_{0} = 2 V_{0}$
$E_{2} = \frac{1}{2} {mV}_{2}^{2}$
$E_{2} = \frac{1}{2} m {(2 V_{0})}^{2}$
$E_{2} = 2 m V_{0}^{2}$
Speed on right side-
$V_{3} = \sqrt{V_{0}^{2} + (R ω)^{2}}$
$V_{3} = \sqrt{V_{0}^{2} + {(V_{0})}^{2}}$
$V_{3} = \sqrt{2} V_{0}$
$E_{3} = \frac{1}{2} m {(V_{3})}^{2} = \frac{1}{2} m \times {(\sqrt{2} V_{0})}^{2}$
$E_{3} = {mV}_{0}^{2}$
Kinetic energy of ring
$E_{ring} = \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} {mR}^{2} ω^{2}$
$= \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} m (R ω)^{2}$
$= \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} {mV}_{0}^{2}$
$E_{ring} = m V_{0}^{2}$
Total kinetic energy $(E) = E_{1} + E_{2} + E_{3} + E_{ring}$
$= 2 {mV}_{0}^{2} + 2 {mV}_{0}^{2} + {mV}_{0}^{2} + {mV}_{0}^{2}$
$= 6 {mV}_{0}^{2}$

AP EAMCET(Medical)-2016

Rotational Motion

149763 Two blocks of equals mass are released on two smooth sides of a double inclined plane with a fixed base as shown in the figure. If each angle of inclination is $45^{\circ}$ , the acceleration of the centre of mass of the system of the two blocks is (Acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

10 {ms}^{- 2}

vertically downward

2

10 {ms}^{- 2}

vertically upward

3

5 {ms}^{- 2}

vertically downward

4

5 {ms}^{- 2}

vertically upward

Explanation:

A Given that, Mass of two blocks are equal $m_{1} = m_{2} = m$

Acceleration of $A = a_{1}$
\& acceleration of $B = a_{2}$ , both are equal.
Gravity due to acceleration $(g) = - 10 m / s^{2}$
Acceleration of centre of mass of system
$a_{cm} = \frac{m_{1} a_{1} + m_{2} a_{2}}{m_{1} + m_{2}}$
$\Rightarrow \frac{m \times (- 10) + m \times (- 10)}{m + m}$
$a_{cm} = \frac{- 20 m}{2 m} \Rightarrow - 10 m / s^{2}$

AP EAMCET(Medical)-2016

Rotational Motion

1 (i)-(v), (ii)-(vii), (iii)-(vi), (iv)-(viii)

2 (i)-(vii), (ii)-(v), (iii)-(vi), (iv)-(viii)

3 (i)-(vii), (ii)-(v), (iii)-(viii), (iv)-(vi)

4 (i)-(viii), (ii)-(v), (iii)-(vi), (iv)-(vii)

AP EMCET(Medical)-2011

Rotational Motion

149765 Four particles, each of mass $1 kg$ are placed at the corners of a square $O A B C$ of side $1 m$ . ' $O$ ' is at the origin the co-ordinate system. $OA$ and $O C$ are aligned along positive $X$ -axis and positive $Y$ -axis respectively. The position vector of the centre of mass is (in ' $m$ '):

1

\hat{i} + \hat{j}

2

\frac{1}{2} (\hat{i} + \hat{j})

3

(\hat{i} - \hat{j})

4

\frac{1}{2} (\hat{i} - \hat{j})

Explanation:

B From the question given that, Four particles, each of mass $1 kg$
So, $m_{1} = m_{2} = m_{3} = m_{4} = 1 kg$
and $OA = AB = OC = BC = 1 m$
According to the question -

Co-ordinate of $O = (x_{1}, y_{1}) = (0, 0)$
Co-ordinate of $A = (x_{2}, y_{2}) = (1, 0)$
Co-ordinate of $B = (x_{3}, y_{3}) = (1, 1)$
Co-ordinate of $C = (x_{4}, y_{4}) = (0, 1)$
$x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$x_{cm} = \frac{1 \times 0 + 1 \times 1 + 1 \times 1 + 1 \times 0}{1 + 1 + 1 + 1} = \frac{2}{4} = \frac{1}{2} m$
$y_{cm} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$= \frac{1 \times 0 + 1 \times 0 + 1 \times 1 + 1 \times 1}{1 + 1 + 1 + 1} = \frac{2}{4} = \frac{1}{2} m$
The position vector of centre of mass $({\vec{r}}_{cm})$
$= x_{c m} \hat{i} + y_{c m} \hat{j}$
$\Rightarrow \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} \Rightarrow \frac{1}{2} (\hat{i} + \hat{j})$

AP EAMCET(Medical)-2006

Rotational Motion

149762 Three particles are attached to a ring of mass $m$ and radius $R$ as shown in the figure. The centre of mass the ring has a speed $V_{0}$ and rolls without slipping on a horizontal surface. The kinetic energy of the system in the position shown in the figure is

1

6 m V_{0}^{2}

2

12 m V_{0}^{2}

3

2 m V_{0}^{2}

4

8 m V_{0}^{2}

Explanation:

A When body is pure rolling
$V_{0} = R ω$
Speed on the left side is:
$V_{1} = \sqrt{V_{0}^{2} + (R ω)^{2}}$
$V_{1} = \sqrt{V_{0}^{2} + V_{0}^{2}}$
$V_{1} = V_{0} \sqrt{2}$
Kinetic energy $(E_{1}) = \frac{1}{2} \times 2 m \times V_{1}^{2}$
$= m \times V_{0}^{2} \times 2$
$= 2 m V_{0}^{2}$
Speed on the top-
$V_{2} = V_{0} + R ω$
$V_{2} = V_{0} + V_{0} = 2 V_{0}$
$E_{2} = \frac{1}{2} {mV}_{2}^{2}$
$E_{2} = \frac{1}{2} m {(2 V_{0})}^{2}$
$E_{2} = 2 m V_{0}^{2}$
Speed on right side-
$V_{3} = \sqrt{V_{0}^{2} + (R ω)^{2}}$
$V_{3} = \sqrt{V_{0}^{2} + {(V_{0})}^{2}}$
$V_{3} = \sqrt{2} V_{0}$
$E_{3} = \frac{1}{2} m {(V_{3})}^{2} = \frac{1}{2} m \times {(\sqrt{2} V_{0})}^{2}$
$E_{3} = {mV}_{0}^{2}$
Kinetic energy of ring
$E_{ring} = \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} {mR}^{2} ω^{2}$
$= \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} m (R ω)^{2}$
$= \frac{1}{2} {mV}_{0}^{2} + \frac{1}{2} {mV}_{0}^{2}$
$E_{ring} = m V_{0}^{2}$
Total kinetic energy $(E) = E_{1} + E_{2} + E_{3} + E_{ring}$
$= 2 {mV}_{0}^{2} + 2 {mV}_{0}^{2} + {mV}_{0}^{2} + {mV}_{0}^{2}$
$= 6 {mV}_{0}^{2}$

AP EAMCET(Medical)-2016

Rotational Motion

149763 Two blocks of equals mass are released on two smooth sides of a double inclined plane with a fixed base as shown in the figure. If each angle of inclination is $45^{\circ}$ , the acceleration of the centre of mass of the system of the two blocks is (Acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

10 {ms}^{- 2}

vertically downward

2

10 {ms}^{- 2}

vertically upward

3

5 {ms}^{- 2}

vertically downward

4

5 {ms}^{- 2}

vertically upward

Explanation:

A Given that, Mass of two blocks are equal $m_{1} = m_{2} = m$

Acceleration of $A = a_{1}$
\& acceleration of $B = a_{2}$ , both are equal.
Gravity due to acceleration $(g) = - 10 m / s^{2}$
Acceleration of centre of mass of system
$a_{cm} = \frac{m_{1} a_{1} + m_{2} a_{2}}{m_{1} + m_{2}}$
$\Rightarrow \frac{m \times (- 10) + m \times (- 10)}{m + m}$
$a_{cm} = \frac{- 20 m}{2 m} \Rightarrow - 10 m / s^{2}$

AP EAMCET(Medical)-2016

Rotational Motion

1 (i)-(v), (ii)-(vii), (iii)-(vi), (iv)-(viii)

2 (i)-(vii), (ii)-(v), (iii)-(vi), (iv)-(viii)

3 (i)-(vii), (ii)-(v), (iii)-(viii), (iv)-(vi)

4 (i)-(viii), (ii)-(v), (iii)-(vi), (iv)-(vii)

AP EMCET(Medical)-2011

Rotational Motion

149765 Four particles, each of mass $1 kg$ are placed at the corners of a square $O A B C$ of side $1 m$ . ' $O$ ' is at the origin the co-ordinate system. $OA$ and $O C$ are aligned along positive $X$ -axis and positive $Y$ -axis respectively. The position vector of the centre of mass is (in ' $m$ '):

1

\hat{i} + \hat{j}

2

\frac{1}{2} (\hat{i} + \hat{j})

3

(\hat{i} - \hat{j})

4

\frac{1}{2} (\hat{i} - \hat{j})

Explanation:

B From the question given that, Four particles, each of mass $1 kg$
So, $m_{1} = m_{2} = m_{3} = m_{4} = 1 kg$
and $OA = AB = OC = BC = 1 m$
According to the question -

Co-ordinate of $O = (x_{1}, y_{1}) = (0, 0)$
Co-ordinate of $A = (x_{2}, y_{2}) = (1, 0)$
Co-ordinate of $B = (x_{3}, y_{3}) = (1, 1)$
Co-ordinate of $C = (x_{4}, y_{4}) = (0, 1)$
$x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$x_{cm} = \frac{1 \times 0 + 1 \times 1 + 1 \times 1 + 1 \times 0}{1 + 1 + 1 + 1} = \frac{2}{4} = \frac{1}{2} m$
$y_{cm} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$= \frac{1 \times 0 + 1 \times 0 + 1 \times 1 + 1 \times 1}{1 + 1 + 1 + 1} = \frac{2}{4} = \frac{1}{2} m$
The position vector of centre of mass $({\vec{r}}_{cm})$
$= x_{c m} \hat{i} + y_{c m} \hat{j}$
$\Rightarrow \frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} \Rightarrow \frac{1}{2} (\hat{i} + \hat{j})$

AP EAMCET(Medical)-2006

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