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Rotational Motion

149770 Three bricks each of length $L$ and mass $M$ are arranged as shown from the wall figure. The distance of the centre of mass of the system from the wall is

1

L / 4

2

L / 2

3

(3 / 2) L

4

(11 / 12) L

5

(5 / 6) L

Explanation:

D
Distance from wall $(x_{1}) = \frac{L}{2}$
Distance from wall $(x_{2}) = \frac{L}{2} + \frac{L}{2} = L$
Distance from wall $(x_{3}) = \frac{L}{2} + \frac{L}{2} + \frac{L}{4} = \frac{5}{4} L$
Now,
$X_{c m} = \frac{M_{1} x_{1} + M_{2} x_{2} + M_{3} x_{3}}{M_{1} + M_{2} + M_{3}}$
$X_{c m} = \frac{M \times \frac{L}{2} + M \times L + M \times \frac{5 L}{4}}{3 M}$
$X_{cm} = \frac{\frac{11}{4} ML}{3 M}$
$X_{cm} = \frac{11}{12} L$

Kerala CEE - 2010

Rotational Motion

149771 Two bodies of mass $4 kg$ and $5 kg$ are moving along east and north directions with velocities 5 $m / s$ and $3 m / s$ respectively. Magnitude of the velocity of centre of mass of the system is

1

\frac{25}{9} m / s

2

\frac{9}{25} m / s

3

\frac{41}{9} m / s

4

\frac{16}{9} m / s

Explanation:

A Given,
$m_{1} = 4 kg$
$m_{2} = 5 kg$
$v_{1} = 5 \hat{i} m / s (x direction)$
$v_{2} = 3 \hat{j} / s (y direction)$
$v_{cm} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$v_{cm} = \frac{4 \times 5 \hat{i} + 5 \times 3 \hat{j}}{9}$
$v_{cm} = \frac{20}{9} \hat{i} + \frac{15}{9} \hat{j}$
$v_{cm} = \sqrt{{(\frac{20}{9})}^{2} + {(\frac{15}{9})}^{2}}$
$v_{cm} = \sqrt{\frac{400 + 225}{81}}$
$v_{cm} = \frac{25}{9} m / s$

AP EAMCET -2013

Rotational Motion

149772 A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ such that the circumferences of the discs touch. The centre of mass of the new disc is at a distance $α R$ from the centre of the bigger disc. The value of $α$ is

1

\frac{1}{2}

2

\frac{1}{3}

3

\frac{1}{4}

4

\frac{1}{6}

Explanation:

B We know that,
$σ = \frac{mass}{area}$
Mass of the bigger disc-
$M_{1} = σ \times π \times (2 R)^{2}$
$M_{1} = 4 π R^{2} σ$
Mass of the smaller disc-
$M_{2} = σ \times π R^{2}$

Centre of the mass of cutting disc,
$X_{c m} = \frac{M_{1} x_{1} + M_{2} x_{2}}{M_{1} + M_{2}}$
$X_{c m} = \frac{[4 π R^{2} σ \times 0] + [(- σ \times π R^{2}) \times R]}{[4 π R^{2} σ + (- σ π R^{2})]} = - \frac{R}{3}$
$α R = - \frac{R}{3} (∵ x_{c m} = α R)$
$| α | = | - \frac{1}{3} |$
$α = \frac{1}{3}$

AP EAMCET -2011

Rotational Motion

269277 Centre of mass of the earth-moon system lies

1 on the surface of the earth

2 on the surface of the moon

3 with in the earth

4 at the midpoint of the line joining their centers

Rotational Motion

149770 Three bricks each of length $L$ and mass $M$ are arranged as shown from the wall figure. The distance of the centre of mass of the system from the wall is

1

L / 4

2

L / 2

3

(3 / 2) L

4

(11 / 12) L

5

(5 / 6) L

Explanation:

D
Distance from wall $(x_{1}) = \frac{L}{2}$
Distance from wall $(x_{2}) = \frac{L}{2} + \frac{L}{2} = L$
Distance from wall $(x_{3}) = \frac{L}{2} + \frac{L}{2} + \frac{L}{4} = \frac{5}{4} L$
Now,
$X_{c m} = \frac{M_{1} x_{1} + M_{2} x_{2} + M_{3} x_{3}}{M_{1} + M_{2} + M_{3}}$
$X_{c m} = \frac{M \times \frac{L}{2} + M \times L + M \times \frac{5 L}{4}}{3 M}$
$X_{cm} = \frac{\frac{11}{4} ML}{3 M}$
$X_{cm} = \frac{11}{12} L$

Kerala CEE - 2010

Rotational Motion

149771 Two bodies of mass $4 kg$ and $5 kg$ are moving along east and north directions with velocities 5 $m / s$ and $3 m / s$ respectively. Magnitude of the velocity of centre of mass of the system is

1

\frac{25}{9} m / s

2

\frac{9}{25} m / s

3

\frac{41}{9} m / s

4

\frac{16}{9} m / s

Explanation:

A Given,
$m_{1} = 4 kg$
$m_{2} = 5 kg$
$v_{1} = 5 \hat{i} m / s (x direction)$
$v_{2} = 3 \hat{j} / s (y direction)$
$v_{cm} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$v_{cm} = \frac{4 \times 5 \hat{i} + 5 \times 3 \hat{j}}{9}$
$v_{cm} = \frac{20}{9} \hat{i} + \frac{15}{9} \hat{j}$
$v_{cm} = \sqrt{{(\frac{20}{9})}^{2} + {(\frac{15}{9})}^{2}}$
$v_{cm} = \sqrt{\frac{400 + 225}{81}}$
$v_{cm} = \frac{25}{9} m / s$

AP EAMCET -2013

Rotational Motion

149772 A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ such that the circumferences of the discs touch. The centre of mass of the new disc is at a distance $α R$ from the centre of the bigger disc. The value of $α$ is

1

\frac{1}{2}

2

\frac{1}{3}

3

\frac{1}{4}

4

\frac{1}{6}

Explanation:

B We know that,
$σ = \frac{mass}{area}$
Mass of the bigger disc-
$M_{1} = σ \times π \times (2 R)^{2}$
$M_{1} = 4 π R^{2} σ$
Mass of the smaller disc-
$M_{2} = σ \times π R^{2}$

Centre of the mass of cutting disc,
$X_{c m} = \frac{M_{1} x_{1} + M_{2} x_{2}}{M_{1} + M_{2}}$
$X_{c m} = \frac{[4 π R^{2} σ \times 0] + [(- σ \times π R^{2}) \times R]}{[4 π R^{2} σ + (- σ π R^{2})]} = - \frac{R}{3}$
$α R = - \frac{R}{3} (∵ x_{c m} = α R)$
$| α | = | - \frac{1}{3} |$
$α = \frac{1}{3}$

AP EAMCET -2011

Rotational Motion

269277 Centre of mass of the earth-moon system lies

1 on the surface of the earth

2 on the surface of the moon

3 with in the earth

4 at the midpoint of the line joining their centers

Rotational Motion

149770 Three bricks each of length $L$ and mass $M$ are arranged as shown from the wall figure. The distance of the centre of mass of the system from the wall is

1

L / 4

2

L / 2

3

(3 / 2) L

4

(11 / 12) L

5

(5 / 6) L

Explanation:

D
Distance from wall $(x_{1}) = \frac{L}{2}$
Distance from wall $(x_{2}) = \frac{L}{2} + \frac{L}{2} = L$
Distance from wall $(x_{3}) = \frac{L}{2} + \frac{L}{2} + \frac{L}{4} = \frac{5}{4} L$
Now,
$X_{c m} = \frac{M_{1} x_{1} + M_{2} x_{2} + M_{3} x_{3}}{M_{1} + M_{2} + M_{3}}$
$X_{c m} = \frac{M \times \frac{L}{2} + M \times L + M \times \frac{5 L}{4}}{3 M}$
$X_{cm} = \frac{\frac{11}{4} ML}{3 M}$
$X_{cm} = \frac{11}{12} L$

Kerala CEE - 2010

Rotational Motion

149771 Two bodies of mass $4 kg$ and $5 kg$ are moving along east and north directions with velocities 5 $m / s$ and $3 m / s$ respectively. Magnitude of the velocity of centre of mass of the system is

1

\frac{25}{9} m / s

2

\frac{9}{25} m / s

3

\frac{41}{9} m / s

4

\frac{16}{9} m / s

Explanation:

A Given,
$m_{1} = 4 kg$
$m_{2} = 5 kg$
$v_{1} = 5 \hat{i} m / s (x direction)$
$v_{2} = 3 \hat{j} / s (y direction)$
$v_{cm} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$v_{cm} = \frac{4 \times 5 \hat{i} + 5 \times 3 \hat{j}}{9}$
$v_{cm} = \frac{20}{9} \hat{i} + \frac{15}{9} \hat{j}$
$v_{cm} = \sqrt{{(\frac{20}{9})}^{2} + {(\frac{15}{9})}^{2}}$
$v_{cm} = \sqrt{\frac{400 + 225}{81}}$
$v_{cm} = \frac{25}{9} m / s$

AP EAMCET -2013

Rotational Motion

149772 A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ such that the circumferences of the discs touch. The centre of mass of the new disc is at a distance $α R$ from the centre of the bigger disc. The value of $α$ is

1

\frac{1}{2}

2

\frac{1}{3}

3

\frac{1}{4}

4

\frac{1}{6}

Explanation:

B We know that,
$σ = \frac{mass}{area}$
Mass of the bigger disc-
$M_{1} = σ \times π \times (2 R)^{2}$
$M_{1} = 4 π R^{2} σ$
Mass of the smaller disc-
$M_{2} = σ \times π R^{2}$

Centre of the mass of cutting disc,
$X_{c m} = \frac{M_{1} x_{1} + M_{2} x_{2}}{M_{1} + M_{2}}$
$X_{c m} = \frac{[4 π R^{2} σ \times 0] + [(- σ \times π R^{2}) \times R]}{[4 π R^{2} σ + (- σ π R^{2})]} = - \frac{R}{3}$
$α R = - \frac{R}{3} (∵ x_{c m} = α R)$
$| α | = | - \frac{1}{3} |$
$α = \frac{1}{3}$

AP EAMCET -2011

Rotational Motion

269277 Centre of mass of the earth-moon system lies

1 on the surface of the earth

2 on the surface of the moon

3 with in the earth

4 at the midpoint of the line joining their centers

Rotational Motion

149770 Three bricks each of length $L$ and mass $M$ are arranged as shown from the wall figure. The distance of the centre of mass of the system from the wall is

1

L / 4

2

L / 2

3

(3 / 2) L

4

(11 / 12) L

5

(5 / 6) L

Explanation:

D
Distance from wall $(x_{1}) = \frac{L}{2}$
Distance from wall $(x_{2}) = \frac{L}{2} + \frac{L}{2} = L$
Distance from wall $(x_{3}) = \frac{L}{2} + \frac{L}{2} + \frac{L}{4} = \frac{5}{4} L$
Now,
$X_{c m} = \frac{M_{1} x_{1} + M_{2} x_{2} + M_{3} x_{3}}{M_{1} + M_{2} + M_{3}}$
$X_{c m} = \frac{M \times \frac{L}{2} + M \times L + M \times \frac{5 L}{4}}{3 M}$
$X_{cm} = \frac{\frac{11}{4} ML}{3 M}$
$X_{cm} = \frac{11}{12} L$

Kerala CEE - 2010

Rotational Motion

149771 Two bodies of mass $4 kg$ and $5 kg$ are moving along east and north directions with velocities 5 $m / s$ and $3 m / s$ respectively. Magnitude of the velocity of centre of mass of the system is

1

\frac{25}{9} m / s

2

\frac{9}{25} m / s

3

\frac{41}{9} m / s

4

\frac{16}{9} m / s

Explanation:

A Given,
$m_{1} = 4 kg$
$m_{2} = 5 kg$
$v_{1} = 5 \hat{i} m / s (x direction)$
$v_{2} = 3 \hat{j} / s (y direction)$
$v_{cm} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$v_{cm} = \frac{4 \times 5 \hat{i} + 5 \times 3 \hat{j}}{9}$
$v_{cm} = \frac{20}{9} \hat{i} + \frac{15}{9} \hat{j}$
$v_{cm} = \sqrt{{(\frac{20}{9})}^{2} + {(\frac{15}{9})}^{2}}$
$v_{cm} = \sqrt{\frac{400 + 225}{81}}$
$v_{cm} = \frac{25}{9} m / s$

AP EAMCET -2013

Rotational Motion

149772 A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ such that the circumferences of the discs touch. The centre of mass of the new disc is at a distance $α R$ from the centre of the bigger disc. The value of $α$ is

1

\frac{1}{2}

2

\frac{1}{3}

3

\frac{1}{4}

4

\frac{1}{6}

Explanation:

B We know that,
$σ = \frac{mass}{area}$
Mass of the bigger disc-
$M_{1} = σ \times π \times (2 R)^{2}$
$M_{1} = 4 π R^{2} σ$
Mass of the smaller disc-
$M_{2} = σ \times π R^{2}$

Centre of the mass of cutting disc,
$X_{c m} = \frac{M_{1} x_{1} + M_{2} x_{2}}{M_{1} + M_{2}}$
$X_{c m} = \frac{[4 π R^{2} σ \times 0] + [(- σ \times π R^{2}) \times R]}{[4 π R^{2} σ + (- σ π R^{2})]} = - \frac{R}{3}$
$α R = - \frac{R}{3} (∵ x_{c m} = α R)$
$| α | = | - \frac{1}{3} |$
$α = \frac{1}{3}$

AP EAMCET -2011

Rotational Motion

269277 Centre of mass of the earth-moon system lies

1 on the surface of the earth

2 on the surface of the moon

3 with in the earth

4 at the midpoint of the line joining their centers

Explanation:

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