QBManager

Rotational Motion

149713 A body weighs $8 g$ when placed in one pan and $18 g$ when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty, then the true weight of the body is

1

15 g

2

13 g

3

10 g

4

12 g

Explanation:

D
$8 Pg = wq$

On equating equations (i) and (ii), we get
$\frac{wq}{8 g} = \frac{18 qg}{w}$
$w^{2} = 18 \times 8 \times g^{2}$
$w = \sqrt{18 \times 8} g$
$w = 12 g$

UPSEE - 2015

Rotational Motion

149714 Three identical spheres of mass $m$ each are placed at the corners of an equilateral triangle of side $2 m$ . Taking one of the corner as the origin, the position vector of the centre of mass is

1

\sqrt{3} (\hat{i} - \hat{j})

2

\frac{\hat{i}}{\sqrt{3}} + \hat{j}

3

\frac{\hat{i} + \hat{j}}{3}

4

\hat{i} + \frac{\hat{j}}{\sqrt{3}}

Explanation:

D
From figure,
$\sin 60^{\circ} = d / 2$
$\Rightarrow d = 2 \sin 60^{\circ}$
$= 2 \times \frac{\sqrt{3}}{2}$
$= \sqrt{3}$
The coordinates of centre of mass are-
$\overset{―}{x} = \frac{\sum m_{i} x_{i}}{\sum m_{i}} = \frac{m \times 0 + m \times 1 + m \times 2}{m + m + m} = 1$
$\overset{―}{y} = \frac{\sum m_{i} y_{i}}{\sum m_{i}} = \frac{m \times 0 + m (\sqrt{3}) + m \times 0}{m + m + m}$
$\overset{―}{y} = \frac{m \sqrt{3}}{3 m} = \frac{1}{\sqrt{3}}$
Hence, position vector of centre of mass is $(\hat{i} + \frac{\hat{j}}{\sqrt{3}})$

UPSEE - 2014

Rotational Motion

149715 Two blocks of masses $6 kg$ and $4 kg$ are placed on a frictionless surface and connected by a spring. If the heavier mass is given a velocity of $14 {ms}^{- 1}$ in the direction of lighter one, then the velocity gained by the centre of mass will be

1

7.4 {ms}^{- 1}

2

14 {ms}^{- 1}

3

8.4 {ms}^{- 1}

4

10.0 {ms}^{- 1}

Explanation:

C Given:
$m_{1} = 6 kg$
$m_{2} = 4 kg$
$v_{1} = 14 m / \sec .$
If the spring is massless, then no force will be exerted on the smaller mass i.e. Velocity of smaller mass will be zero.
Now, the velocity of center of mass (CM) is
$v_{CM} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$= \frac{6 \times 14 + 4 \times 0}{6 + 4}$
$v_{CM} = \frac{84}{10} = 8.4 m / \sec$
Hence, the velocity of COM is $8.4 m / \sec$

UPSEE - 2009

Rotational Motion

149717 The centre of mass of three particles of masses $1 kg, 2 kg$ and $3 kg$ is at $(3, 3, 3)$ with reference to a fixed coordinate system. Where should a fourth particle of mass $4 kg$ be placed, so that the centre of mass of the system of all particles shifts to a point $(1, 1, 1)$ ?

1

(- 1, - 1, - 1)

2

(- 2, - 2, - 2)

3

(2, 2, 2)

4

(1, 1, 1)

Explanation:

B Suppose particle of $4 kg$ mass is placed at point $(x, y, z)$ . Then the center of mass of $6 kg (1 kg +$ $2 kg + 3 kg$ ) mass and $4 kg$ mass must lie at the point (1, $1, 1)$ .
$x_{CM} = \frac{6 \times 3 + 4 \times x}{6 + 4} = 1$
$10 = 18 + 4 x$
$x = - 2 m$
Similarly,
$y_{C M} = \frac{6 \times 3 + 4 \times y}{6 + 4} = 1$
$10 = 18 + 4 y$
$y = - 2 m$
And for $z$ -coordinate,
$z_{C M} = \frac{6 \times 3 + 4 \times z}{6 + 4} = 1$
$10 = 18 + 4 z$
$z = - 2 m$
Hence, the fourth particle must be placed at the point $(- 2, - 2, - 2)$

JCECE-2010

Rotational Motion

149718 A small disc of radius $2 cm$ is cut from a disc of radius $6 cm$ . If the distance between their centers is $3.2 cm$ , what is the shift in the centre of mass of the disc?

1

0.4 cm

2

2.4 cm

3

1.8 cm

4

1.2 cm

Explanation:

A Given, $R = 6 cm, r = 2 cm$

Let the mass per unit area $= m$
$M =$ Mass of disc $= π R^{2} m = π (6)^{2} m = 36 π m$
$M_{1} =$ Mass of removed disc $= π (r)^{2} m = π (2)^{2} m = 4 π m$
$M_{2} =$ Mass of remaining portion $= M - M_{1}$
$= 36 π m - 4 π m = 32 π m$
Center of removed disc $r_{1} = 3.2 cm$
Center of remaining disc $= r_{2}$
Hence,
$M \times 0 = M_{1} r_{1} + M_{2} r_{2}$
$0 = M_{1} r_{1} + M_{2} r_{2}$
$r_{2} = [\frac{(- M_{1} r_{1})}{M_{2}}]$
$r_{2} = [\frac{(- 4 π m) \times 3.2}{32 π m}]$
$r_{2} = - 0.4 cm$
Negative sign indicate the shift from the centre.

JCECE-2007

Rotational Motion

149713 A body weighs $8 g$ when placed in one pan and $18 g$ when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty, then the true weight of the body is

1

15 g

2

13 g

3

10 g

4

12 g

Explanation:

D
$8 Pg = wq$

On equating equations (i) and (ii), we get
$\frac{wq}{8 g} = \frac{18 qg}{w}$
$w^{2} = 18 \times 8 \times g^{2}$
$w = \sqrt{18 \times 8} g$
$w = 12 g$

UPSEE - 2015

Rotational Motion

149714 Three identical spheres of mass $m$ each are placed at the corners of an equilateral triangle of side $2 m$ . Taking one of the corner as the origin, the position vector of the centre of mass is

1

\sqrt{3} (\hat{i} - \hat{j})

2

\frac{\hat{i}}{\sqrt{3}} + \hat{j}

3

\frac{\hat{i} + \hat{j}}{3}

4

\hat{i} + \frac{\hat{j}}{\sqrt{3}}

Explanation:

D
From figure,
$\sin 60^{\circ} = d / 2$
$\Rightarrow d = 2 \sin 60^{\circ}$
$= 2 \times \frac{\sqrt{3}}{2}$
$= \sqrt{3}$
The coordinates of centre of mass are-
$\overset{―}{x} = \frac{\sum m_{i} x_{i}}{\sum m_{i}} = \frac{m \times 0 + m \times 1 + m \times 2}{m + m + m} = 1$
$\overset{―}{y} = \frac{\sum m_{i} y_{i}}{\sum m_{i}} = \frac{m \times 0 + m (\sqrt{3}) + m \times 0}{m + m + m}$
$\overset{―}{y} = \frac{m \sqrt{3}}{3 m} = \frac{1}{\sqrt{3}}$
Hence, position vector of centre of mass is $(\hat{i} + \frac{\hat{j}}{\sqrt{3}})$

UPSEE - 2014

Rotational Motion

149715 Two blocks of masses $6 kg$ and $4 kg$ are placed on a frictionless surface and connected by a spring. If the heavier mass is given a velocity of $14 {ms}^{- 1}$ in the direction of lighter one, then the velocity gained by the centre of mass will be

1

7.4 {ms}^{- 1}

2

14 {ms}^{- 1}

3

8.4 {ms}^{- 1}

4

10.0 {ms}^{- 1}

Explanation:

C Given:
$m_{1} = 6 kg$
$m_{2} = 4 kg$
$v_{1} = 14 m / \sec .$
If the spring is massless, then no force will be exerted on the smaller mass i.e. Velocity of smaller mass will be zero.
Now, the velocity of center of mass (CM) is
$v_{CM} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$= \frac{6 \times 14 + 4 \times 0}{6 + 4}$
$v_{CM} = \frac{84}{10} = 8.4 m / \sec$
Hence, the velocity of COM is $8.4 m / \sec$

UPSEE - 2009

Rotational Motion

149717 The centre of mass of three particles of masses $1 kg, 2 kg$ and $3 kg$ is at $(3, 3, 3)$ with reference to a fixed coordinate system. Where should a fourth particle of mass $4 kg$ be placed, so that the centre of mass of the system of all particles shifts to a point $(1, 1, 1)$ ?

1

(- 1, - 1, - 1)

2

(- 2, - 2, - 2)

3

(2, 2, 2)

4

(1, 1, 1)

Explanation:

B Suppose particle of $4 kg$ mass is placed at point $(x, y, z)$ . Then the center of mass of $6 kg (1 kg +$ $2 kg + 3 kg$ ) mass and $4 kg$ mass must lie at the point (1, $1, 1)$ .
$x_{CM} = \frac{6 \times 3 + 4 \times x}{6 + 4} = 1$
$10 = 18 + 4 x$
$x = - 2 m$
Similarly,
$y_{C M} = \frac{6 \times 3 + 4 \times y}{6 + 4} = 1$
$10 = 18 + 4 y$
$y = - 2 m$
And for $z$ -coordinate,
$z_{C M} = \frac{6 \times 3 + 4 \times z}{6 + 4} = 1$
$10 = 18 + 4 z$
$z = - 2 m$
Hence, the fourth particle must be placed at the point $(- 2, - 2, - 2)$

JCECE-2010

Rotational Motion

149718 A small disc of radius $2 cm$ is cut from a disc of radius $6 cm$ . If the distance between their centers is $3.2 cm$ , what is the shift in the centre of mass of the disc?

1

0.4 cm

2

2.4 cm

3

1.8 cm

4

1.2 cm

Explanation:

A Given, $R = 6 cm, r = 2 cm$

Let the mass per unit area $= m$
$M =$ Mass of disc $= π R^{2} m = π (6)^{2} m = 36 π m$
$M_{1} =$ Mass of removed disc $= π (r)^{2} m = π (2)^{2} m = 4 π m$
$M_{2} =$ Mass of remaining portion $= M - M_{1}$
$= 36 π m - 4 π m = 32 π m$
Center of removed disc $r_{1} = 3.2 cm$
Center of remaining disc $= r_{2}$
Hence,
$M \times 0 = M_{1} r_{1} + M_{2} r_{2}$
$0 = M_{1} r_{1} + M_{2} r_{2}$
$r_{2} = [\frac{(- M_{1} r_{1})}{M_{2}}]$
$r_{2} = [\frac{(- 4 π m) \times 3.2}{32 π m}]$
$r_{2} = - 0.4 cm$
Negative sign indicate the shift from the centre.

JCECE-2007

Rotational Motion

149713 A body weighs $8 g$ when placed in one pan and $18 g$ when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty, then the true weight of the body is

1

15 g

2

13 g

3

10 g

4

12 g

Explanation:

D
$8 Pg = wq$

On equating equations (i) and (ii), we get
$\frac{wq}{8 g} = \frac{18 qg}{w}$
$w^{2} = 18 \times 8 \times g^{2}$
$w = \sqrt{18 \times 8} g$
$w = 12 g$

UPSEE - 2015

Rotational Motion

149714 Three identical spheres of mass $m$ each are placed at the corners of an equilateral triangle of side $2 m$ . Taking one of the corner as the origin, the position vector of the centre of mass is

1

\sqrt{3} (\hat{i} - \hat{j})

2

\frac{\hat{i}}{\sqrt{3}} + \hat{j}

3

\frac{\hat{i} + \hat{j}}{3}

4

\hat{i} + \frac{\hat{j}}{\sqrt{3}}

Explanation:

D
From figure,
$\sin 60^{\circ} = d / 2$
$\Rightarrow d = 2 \sin 60^{\circ}$
$= 2 \times \frac{\sqrt{3}}{2}$
$= \sqrt{3}$
The coordinates of centre of mass are-
$\overset{―}{x} = \frac{\sum m_{i} x_{i}}{\sum m_{i}} = \frac{m \times 0 + m \times 1 + m \times 2}{m + m + m} = 1$
$\overset{―}{y} = \frac{\sum m_{i} y_{i}}{\sum m_{i}} = \frac{m \times 0 + m (\sqrt{3}) + m \times 0}{m + m + m}$
$\overset{―}{y} = \frac{m \sqrt{3}}{3 m} = \frac{1}{\sqrt{3}}$
Hence, position vector of centre of mass is $(\hat{i} + \frac{\hat{j}}{\sqrt{3}})$

UPSEE - 2014

Rotational Motion

149715 Two blocks of masses $6 kg$ and $4 kg$ are placed on a frictionless surface and connected by a spring. If the heavier mass is given a velocity of $14 {ms}^{- 1}$ in the direction of lighter one, then the velocity gained by the centre of mass will be

1

7.4 {ms}^{- 1}

2

14 {ms}^{- 1}

3

8.4 {ms}^{- 1}

4

10.0 {ms}^{- 1}

Explanation:

C Given:
$m_{1} = 6 kg$
$m_{2} = 4 kg$
$v_{1} = 14 m / \sec .$
If the spring is massless, then no force will be exerted on the smaller mass i.e. Velocity of smaller mass will be zero.
Now, the velocity of center of mass (CM) is
$v_{CM} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$= \frac{6 \times 14 + 4 \times 0}{6 + 4}$
$v_{CM} = \frac{84}{10} = 8.4 m / \sec$
Hence, the velocity of COM is $8.4 m / \sec$

UPSEE - 2009

Rotational Motion

149717 The centre of mass of three particles of masses $1 kg, 2 kg$ and $3 kg$ is at $(3, 3, 3)$ with reference to a fixed coordinate system. Where should a fourth particle of mass $4 kg$ be placed, so that the centre of mass of the system of all particles shifts to a point $(1, 1, 1)$ ?

1

(- 1, - 1, - 1)

2

(- 2, - 2, - 2)

3

(2, 2, 2)

4

(1, 1, 1)

Explanation:

B Suppose particle of $4 kg$ mass is placed at point $(x, y, z)$ . Then the center of mass of $6 kg (1 kg +$ $2 kg + 3 kg$ ) mass and $4 kg$ mass must lie at the point (1, $1, 1)$ .
$x_{CM} = \frac{6 \times 3 + 4 \times x}{6 + 4} = 1$
$10 = 18 + 4 x$
$x = - 2 m$
Similarly,
$y_{C M} = \frac{6 \times 3 + 4 \times y}{6 + 4} = 1$
$10 = 18 + 4 y$
$y = - 2 m$
And for $z$ -coordinate,
$z_{C M} = \frac{6 \times 3 + 4 \times z}{6 + 4} = 1$
$10 = 18 + 4 z$
$z = - 2 m$
Hence, the fourth particle must be placed at the point $(- 2, - 2, - 2)$

JCECE-2010

Rotational Motion

149718 A small disc of radius $2 cm$ is cut from a disc of radius $6 cm$ . If the distance between their centers is $3.2 cm$ , what is the shift in the centre of mass of the disc?

1

0.4 cm

2

2.4 cm

3

1.8 cm

4

1.2 cm

Explanation:

A Given, $R = 6 cm, r = 2 cm$

Let the mass per unit area $= m$
$M =$ Mass of disc $= π R^{2} m = π (6)^{2} m = 36 π m$
$M_{1} =$ Mass of removed disc $= π (r)^{2} m = π (2)^{2} m = 4 π m$
$M_{2} =$ Mass of remaining portion $= M - M_{1}$
$= 36 π m - 4 π m = 32 π m$
Center of removed disc $r_{1} = 3.2 cm$
Center of remaining disc $= r_{2}$
Hence,
$M \times 0 = M_{1} r_{1} + M_{2} r_{2}$
$0 = M_{1} r_{1} + M_{2} r_{2}$
$r_{2} = [\frac{(- M_{1} r_{1})}{M_{2}}]$
$r_{2} = [\frac{(- 4 π m) \times 3.2}{32 π m}]$
$r_{2} = - 0.4 cm$
Negative sign indicate the shift from the centre.

JCECE-2007

Rotational Motion

149713 A body weighs $8 g$ when placed in one pan and $18 g$ when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty, then the true weight of the body is

1

15 g

2

13 g

3

10 g

4

12 g

Explanation:

D
$8 Pg = wq$

On equating equations (i) and (ii), we get
$\frac{wq}{8 g} = \frac{18 qg}{w}$
$w^{2} = 18 \times 8 \times g^{2}$
$w = \sqrt{18 \times 8} g$
$w = 12 g$

UPSEE - 2015

Rotational Motion

149714 Three identical spheres of mass $m$ each are placed at the corners of an equilateral triangle of side $2 m$ . Taking one of the corner as the origin, the position vector of the centre of mass is

1

\sqrt{3} (\hat{i} - \hat{j})

2

\frac{\hat{i}}{\sqrt{3}} + \hat{j}

3

\frac{\hat{i} + \hat{j}}{3}

4

\hat{i} + \frac{\hat{j}}{\sqrt{3}}

Explanation:

D
From figure,
$\sin 60^{\circ} = d / 2$
$\Rightarrow d = 2 \sin 60^{\circ}$
$= 2 \times \frac{\sqrt{3}}{2}$
$= \sqrt{3}$
The coordinates of centre of mass are-
$\overset{―}{x} = \frac{\sum m_{i} x_{i}}{\sum m_{i}} = \frac{m \times 0 + m \times 1 + m \times 2}{m + m + m} = 1$
$\overset{―}{y} = \frac{\sum m_{i} y_{i}}{\sum m_{i}} = \frac{m \times 0 + m (\sqrt{3}) + m \times 0}{m + m + m}$
$\overset{―}{y} = \frac{m \sqrt{3}}{3 m} = \frac{1}{\sqrt{3}}$
Hence, position vector of centre of mass is $(\hat{i} + \frac{\hat{j}}{\sqrt{3}})$

UPSEE - 2014

Rotational Motion

149715 Two blocks of masses $6 kg$ and $4 kg$ are placed on a frictionless surface and connected by a spring. If the heavier mass is given a velocity of $14 {ms}^{- 1}$ in the direction of lighter one, then the velocity gained by the centre of mass will be

1

7.4 {ms}^{- 1}

2

14 {ms}^{- 1}

3

8.4 {ms}^{- 1}

4

10.0 {ms}^{- 1}

Explanation:

C Given:
$m_{1} = 6 kg$
$m_{2} = 4 kg$
$v_{1} = 14 m / \sec .$
If the spring is massless, then no force will be exerted on the smaller mass i.e. Velocity of smaller mass will be zero.
Now, the velocity of center of mass (CM) is
$v_{CM} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$= \frac{6 \times 14 + 4 \times 0}{6 + 4}$
$v_{CM} = \frac{84}{10} = 8.4 m / \sec$
Hence, the velocity of COM is $8.4 m / \sec$

UPSEE - 2009

Rotational Motion

149717 The centre of mass of three particles of masses $1 kg, 2 kg$ and $3 kg$ is at $(3, 3, 3)$ with reference to a fixed coordinate system. Where should a fourth particle of mass $4 kg$ be placed, so that the centre of mass of the system of all particles shifts to a point $(1, 1, 1)$ ?

1

(- 1, - 1, - 1)

2

(- 2, - 2, - 2)

3

(2, 2, 2)

4

(1, 1, 1)

Explanation:

B Suppose particle of $4 kg$ mass is placed at point $(x, y, z)$ . Then the center of mass of $6 kg (1 kg +$ $2 kg + 3 kg$ ) mass and $4 kg$ mass must lie at the point (1, $1, 1)$ .
$x_{CM} = \frac{6 \times 3 + 4 \times x}{6 + 4} = 1$
$10 = 18 + 4 x$
$x = - 2 m$
Similarly,
$y_{C M} = \frac{6 \times 3 + 4 \times y}{6 + 4} = 1$
$10 = 18 + 4 y$
$y = - 2 m$
And for $z$ -coordinate,
$z_{C M} = \frac{6 \times 3 + 4 \times z}{6 + 4} = 1$
$10 = 18 + 4 z$
$z = - 2 m$
Hence, the fourth particle must be placed at the point $(- 2, - 2, - 2)$

JCECE-2010

Rotational Motion

149718 A small disc of radius $2 cm$ is cut from a disc of radius $6 cm$ . If the distance between their centers is $3.2 cm$ , what is the shift in the centre of mass of the disc?

1

0.4 cm

2

2.4 cm

3

1.8 cm

4

1.2 cm

Explanation:

A Given, $R = 6 cm, r = 2 cm$

Let the mass per unit area $= m$
$M =$ Mass of disc $= π R^{2} m = π (6)^{2} m = 36 π m$
$M_{1} =$ Mass of removed disc $= π (r)^{2} m = π (2)^{2} m = 4 π m$
$M_{2} =$ Mass of remaining portion $= M - M_{1}$
$= 36 π m - 4 π m = 32 π m$
Center of removed disc $r_{1} = 3.2 cm$
Center of remaining disc $= r_{2}$
Hence,
$M \times 0 = M_{1} r_{1} + M_{2} r_{2}$
$0 = M_{1} r_{1} + M_{2} r_{2}$
$r_{2} = [\frac{(- M_{1} r_{1})}{M_{2}}]$
$r_{2} = [\frac{(- 4 π m) \times 3.2}{32 π m}]$
$r_{2} = - 0.4 cm$
Negative sign indicate the shift from the centre.

JCECE-2007

Rotational Motion

149713 A body weighs $8 g$ when placed in one pan and $18 g$ when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty, then the true weight of the body is

1

15 g

2

13 g

3

10 g

4

12 g

Explanation:

D
$8 Pg = wq$

On equating equations (i) and (ii), we get
$\frac{wq}{8 g} = \frac{18 qg}{w}$
$w^{2} = 18 \times 8 \times g^{2}$
$w = \sqrt{18 \times 8} g$
$w = 12 g$

UPSEE - 2015

Rotational Motion

149714 Three identical spheres of mass $m$ each are placed at the corners of an equilateral triangle of side $2 m$ . Taking one of the corner as the origin, the position vector of the centre of mass is

1

\sqrt{3} (\hat{i} - \hat{j})

2

\frac{\hat{i}}{\sqrt{3}} + \hat{j}

3

\frac{\hat{i} + \hat{j}}{3}

4

\hat{i} + \frac{\hat{j}}{\sqrt{3}}

Explanation:

D
From figure,
$\sin 60^{\circ} = d / 2$
$\Rightarrow d = 2 \sin 60^{\circ}$
$= 2 \times \frac{\sqrt{3}}{2}$
$= \sqrt{3}$
The coordinates of centre of mass are-
$\overset{―}{x} = \frac{\sum m_{i} x_{i}}{\sum m_{i}} = \frac{m \times 0 + m \times 1 + m \times 2}{m + m + m} = 1$
$\overset{―}{y} = \frac{\sum m_{i} y_{i}}{\sum m_{i}} = \frac{m \times 0 + m (\sqrt{3}) + m \times 0}{m + m + m}$
$\overset{―}{y} = \frac{m \sqrt{3}}{3 m} = \frac{1}{\sqrt{3}}$
Hence, position vector of centre of mass is $(\hat{i} + \frac{\hat{j}}{\sqrt{3}})$

UPSEE - 2014

Rotational Motion

149715 Two blocks of masses $6 kg$ and $4 kg$ are placed on a frictionless surface and connected by a spring. If the heavier mass is given a velocity of $14 {ms}^{- 1}$ in the direction of lighter one, then the velocity gained by the centre of mass will be

1

7.4 {ms}^{- 1}

2

14 {ms}^{- 1}

3

8.4 {ms}^{- 1}

4

10.0 {ms}^{- 1}

Explanation:

C Given:
$m_{1} = 6 kg$
$m_{2} = 4 kg$
$v_{1} = 14 m / \sec .$
If the spring is massless, then no force will be exerted on the smaller mass i.e. Velocity of smaller mass will be zero.
Now, the velocity of center of mass (CM) is
$v_{CM} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$= \frac{6 \times 14 + 4 \times 0}{6 + 4}$
$v_{CM} = \frac{84}{10} = 8.4 m / \sec$
Hence, the velocity of COM is $8.4 m / \sec$

UPSEE - 2009

Rotational Motion

149717 The centre of mass of three particles of masses $1 kg, 2 kg$ and $3 kg$ is at $(3, 3, 3)$ with reference to a fixed coordinate system. Where should a fourth particle of mass $4 kg$ be placed, so that the centre of mass of the system of all particles shifts to a point $(1, 1, 1)$ ?

1

(- 1, - 1, - 1)

2

(- 2, - 2, - 2)

3

(2, 2, 2)

4

(1, 1, 1)

Explanation:

B Suppose particle of $4 kg$ mass is placed at point $(x, y, z)$ . Then the center of mass of $6 kg (1 kg +$ $2 kg + 3 kg$ ) mass and $4 kg$ mass must lie at the point (1, $1, 1)$ .
$x_{CM} = \frac{6 \times 3 + 4 \times x}{6 + 4} = 1$
$10 = 18 + 4 x$
$x = - 2 m$
Similarly,
$y_{C M} = \frac{6 \times 3 + 4 \times y}{6 + 4} = 1$
$10 = 18 + 4 y$
$y = - 2 m$
And for $z$ -coordinate,
$z_{C M} = \frac{6 \times 3 + 4 \times z}{6 + 4} = 1$
$10 = 18 + 4 z$
$z = - 2 m$
Hence, the fourth particle must be placed at the point $(- 2, - 2, - 2)$

JCECE-2010

Rotational Motion

149718 A small disc of radius $2 cm$ is cut from a disc of radius $6 cm$ . If the distance between their centers is $3.2 cm$ , what is the shift in the centre of mass of the disc?

1

0.4 cm

2

2.4 cm

3

1.8 cm

4

1.2 cm

Explanation:

A Given, $R = 6 cm, r = 2 cm$

Let the mass per unit area $= m$
$M =$ Mass of disc $= π R^{2} m = π (6)^{2} m = 36 π m$
$M_{1} =$ Mass of removed disc $= π (r)^{2} m = π (2)^{2} m = 4 π m$
$M_{2} =$ Mass of remaining portion $= M - M_{1}$
$= 36 π m - 4 π m = 32 π m$
Center of removed disc $r_{1} = 3.2 cm$
Center of remaining disc $= r_{2}$
Hence,
$M \times 0 = M_{1} r_{1} + M_{2} r_{2}$
$0 = M_{1} r_{1} + M_{2} r_{2}$
$r_{2} = [\frac{(- M_{1} r_{1})}{M_{2}}]$
$r_{2} = [\frac{(- 4 π m) \times 3.2}{32 π m}]$
$r_{2} = - 0.4 cm$
Negative sign indicate the shift from the centre.

JCECE-2007

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Question Bank Series

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