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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365654 A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time a viscous fluid of mass $m$ is dropped at the centre and is allowed to spread out and finally falls from the disc. The angular velocity during this period:

1 Decreases continuously

2 Decreases initially and increases again

3 Remains unaltered

4 Increases continuously

Explanation:

From law of conservation of angular momentum if no external torque is acting upon body
$L = I ω = constant$
When liquid is dropped, mass increases hence $I$ increases $(I = m r^{2})$ . So, $ω$ decreases, but as soon as the liquid starts flowing out from the disc $ω$ increases again.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365655 A ring of mass $M$ and radius $R$ is rotating with angular speed $ω$ about a fixed vertical axis passing through its centre $O$ with two point masses each of mass $\frac{M}{8}$ at rest at $O$ . These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is $\frac{8}{9} ω$ and one of the masses is at a distance of $\frac{3}{5} R$ from O. At this instant the distance of the other mass from $O$ is
supporting img

1

\frac{3}{5} R

2

\frac{2}{3} R

3

\frac{4}{5} R

4

\frac{1}{3} R

Explanation:

By conservation of angular moment
supporting img
$M R^{2} ω = [M R^{2} + \frac{M}{8} \frac{9 R^{2}}{25} + \frac{M d^{2}}{8}] \frac{8 ω}{9}$
$R^{2} = (\frac{200 R^{2} + 9 R^{2} + 25 d^{2}}{8 \times 25}) \frac{8}{9}$
$225 R^{2} - 209 R^{2} = 25 d^{2}$
$\Rightarrow d = \frac{4 R}{5}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365656 A girl of mass $M$ stands on the rim of a frictionless merry-go-round of radius $R$ and rotational inertia $I$ that is not moving. She throws a rock of mass $m$ horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is $v$ , afterwards the linear speed of the girl is

1

\frac{m v R^{2}}{I + (M - m) R^{2}}

2

\frac{m v R^{2}}{I + (M + m) R^{2}}

3

\frac{m v R^{2}}{I + M R^{2}}

4

\frac{(m + M) v R^{2}}{I + M R^{2}}

Explanation:

The initial angular momentum of the system is zero. The final angular momentum of the girl-plus-merrygo-round is $(I + M R^{2}) ω$ , which we will take to be positive. The final angular momentum we associate with the throw rock is negative: $- m R v$ , where $v$ is the speed of the rock relative to the ground.
From angular momentum conservation
$0 = (I + M R^{2}) ω - m R v \Rightarrow ω = \frac{m R v}{I + M R^{2}}$
The girl's linear speed is $R ω = \frac{m v R^{2}}{I + M R^{2}}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365657 A man stands on a rotating platform with his arms stretched holding a $5 k g$ weight in each hand. The angular speed of the platform is 1.2 rev $s^{- 1}$ . The moment of inertia of the man together with the platform may be taken to be constant and equal to $6 k g m^{2}$ . If the man brings his arms close to his chest with the distance of each weight from the axis changing from $100 c m$ to $20 c m$ . The new angular speed of the platform is

1

2 r e v s^{- 1}

2

3 r e v s^{- 1}

3

5 r e v s^{- 1}

4

6 r e v s^{- 1}

Explanation:

Initial moment of inertia,
$I_{1} = 6 + [2 \times 5 \times (1)^{2}] = 16 kg m^{2}$
Initial angular velocity, $ω_{1} = 1.2 rev s^{- 1}$
Initial angular momentum, $L_{1} = I_{1} ω_{1}$
Final moment of inertia,
$I_{2} = 6 + [2 \times 5 \times (0.2)^{2}] = 6.4 kg m^{2}$
Final angular speed $= ω_{2}$
Final angular momentum, $L_{2} = I_{2} ω_{2}$
According to law of conservation of angular momentum,
$L_{1} = L_{2}$ or $I_{1} ω_{1} = I_{2} ω_{2}$
$ω_{2} = \frac{I_{1} ω_{1}}{I_{2}} = \frac{(16 k g m^{2}) (1.2 r e v s^{- 1})}{6.4 k g m^{2}} = 3 r e v s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365654 A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time a viscous fluid of mass $m$ is dropped at the centre and is allowed to spread out and finally falls from the disc. The angular velocity during this period:

1 Decreases continuously

2 Decreases initially and increases again

3 Remains unaltered

4 Increases continuously

Explanation:

From law of conservation of angular momentum if no external torque is acting upon body
$L = I ω = constant$
When liquid is dropped, mass increases hence $I$ increases $(I = m r^{2})$ . So, $ω$ decreases, but as soon as the liquid starts flowing out from the disc $ω$ increases again.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365655 A ring of mass $M$ and radius $R$ is rotating with angular speed $ω$ about a fixed vertical axis passing through its centre $O$ with two point masses each of mass $\frac{M}{8}$ at rest at $O$ . These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is $\frac{8}{9} ω$ and one of the masses is at a distance of $\frac{3}{5} R$ from O. At this instant the distance of the other mass from $O$ is
supporting img

1

\frac{3}{5} R

2

\frac{2}{3} R

3

\frac{4}{5} R

4

\frac{1}{3} R

Explanation:

By conservation of angular moment
supporting img
$M R^{2} ω = [M R^{2} + \frac{M}{8} \frac{9 R^{2}}{25} + \frac{M d^{2}}{8}] \frac{8 ω}{9}$
$R^{2} = (\frac{200 R^{2} + 9 R^{2} + 25 d^{2}}{8 \times 25}) \frac{8}{9}$
$225 R^{2} - 209 R^{2} = 25 d^{2}$
$\Rightarrow d = \frac{4 R}{5}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365656 A girl of mass $M$ stands on the rim of a frictionless merry-go-round of radius $R$ and rotational inertia $I$ that is not moving. She throws a rock of mass $m$ horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is $v$ , afterwards the linear speed of the girl is

1

\frac{m v R^{2}}{I + (M - m) R^{2}}

2

\frac{m v R^{2}}{I + (M + m) R^{2}}

3

\frac{m v R^{2}}{I + M R^{2}}

4

\frac{(m + M) v R^{2}}{I + M R^{2}}

Explanation:

The initial angular momentum of the system is zero. The final angular momentum of the girl-plus-merrygo-round is $(I + M R^{2}) ω$ , which we will take to be positive. The final angular momentum we associate with the throw rock is negative: $- m R v$ , where $v$ is the speed of the rock relative to the ground.
From angular momentum conservation
$0 = (I + M R^{2}) ω - m R v \Rightarrow ω = \frac{m R v}{I + M R^{2}}$
The girl's linear speed is $R ω = \frac{m v R^{2}}{I + M R^{2}}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365657 A man stands on a rotating platform with his arms stretched holding a $5 k g$ weight in each hand. The angular speed of the platform is 1.2 rev $s^{- 1}$ . The moment of inertia of the man together with the platform may be taken to be constant and equal to $6 k g m^{2}$ . If the man brings his arms close to his chest with the distance of each weight from the axis changing from $100 c m$ to $20 c m$ . The new angular speed of the platform is

1

2 r e v s^{- 1}

2

3 r e v s^{- 1}

3

5 r e v s^{- 1}

4

6 r e v s^{- 1}

Explanation:

Initial moment of inertia,
$I_{1} = 6 + [2 \times 5 \times (1)^{2}] = 16 kg m^{2}$
Initial angular velocity, $ω_{1} = 1.2 rev s^{- 1}$
Initial angular momentum, $L_{1} = I_{1} ω_{1}$
Final moment of inertia,
$I_{2} = 6 + [2 \times 5 \times (0.2)^{2}] = 6.4 kg m^{2}$
Final angular speed $= ω_{2}$
Final angular momentum, $L_{2} = I_{2} ω_{2}$
According to law of conservation of angular momentum,
$L_{1} = L_{2}$ or $I_{1} ω_{1} = I_{2} ω_{2}$
$ω_{2} = \frac{I_{1} ω_{1}}{I_{2}} = \frac{(16 k g m^{2}) (1.2 r e v s^{- 1})}{6.4 k g m^{2}} = 3 r e v s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365654 A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time a viscous fluid of mass $m$ is dropped at the centre and is allowed to spread out and finally falls from the disc. The angular velocity during this period:

1 Decreases continuously

2 Decreases initially and increases again

3 Remains unaltered

4 Increases continuously

Explanation:

From law of conservation of angular momentum if no external torque is acting upon body
$L = I ω = constant$
When liquid is dropped, mass increases hence $I$ increases $(I = m r^{2})$ . So, $ω$ decreases, but as soon as the liquid starts flowing out from the disc $ω$ increases again.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365655 A ring of mass $M$ and radius $R$ is rotating with angular speed $ω$ about a fixed vertical axis passing through its centre $O$ with two point masses each of mass $\frac{M}{8}$ at rest at $O$ . These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is $\frac{8}{9} ω$ and one of the masses is at a distance of $\frac{3}{5} R$ from O. At this instant the distance of the other mass from $O$ is
supporting img

1

\frac{3}{5} R

2

\frac{2}{3} R

3

\frac{4}{5} R

4

\frac{1}{3} R

Explanation:

By conservation of angular moment
supporting img
$M R^{2} ω = [M R^{2} + \frac{M}{8} \frac{9 R^{2}}{25} + \frac{M d^{2}}{8}] \frac{8 ω}{9}$
$R^{2} = (\frac{200 R^{2} + 9 R^{2} + 25 d^{2}}{8 \times 25}) \frac{8}{9}$
$225 R^{2} - 209 R^{2} = 25 d^{2}$
$\Rightarrow d = \frac{4 R}{5}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365656 A girl of mass $M$ stands on the rim of a frictionless merry-go-round of radius $R$ and rotational inertia $I$ that is not moving. She throws a rock of mass $m$ horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is $v$ , afterwards the linear speed of the girl is

1

\frac{m v R^{2}}{I + (M - m) R^{2}}

2

\frac{m v R^{2}}{I + (M + m) R^{2}}

3

\frac{m v R^{2}}{I + M R^{2}}

4

\frac{(m + M) v R^{2}}{I + M R^{2}}

Explanation:

The initial angular momentum of the system is zero. The final angular momentum of the girl-plus-merrygo-round is $(I + M R^{2}) ω$ , which we will take to be positive. The final angular momentum we associate with the throw rock is negative: $- m R v$ , where $v$ is the speed of the rock relative to the ground.
From angular momentum conservation
$0 = (I + M R^{2}) ω - m R v \Rightarrow ω = \frac{m R v}{I + M R^{2}}$
The girl's linear speed is $R ω = \frac{m v R^{2}}{I + M R^{2}}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365657 A man stands on a rotating platform with his arms stretched holding a $5 k g$ weight in each hand. The angular speed of the platform is 1.2 rev $s^{- 1}$ . The moment of inertia of the man together with the platform may be taken to be constant and equal to $6 k g m^{2}$ . If the man brings his arms close to his chest with the distance of each weight from the axis changing from $100 c m$ to $20 c m$ . The new angular speed of the platform is

1

2 r e v s^{- 1}

2

3 r e v s^{- 1}

3

5 r e v s^{- 1}

4

6 r e v s^{- 1}

Explanation:

Initial moment of inertia,
$I_{1} = 6 + [2 \times 5 \times (1)^{2}] = 16 kg m^{2}$
Initial angular velocity, $ω_{1} = 1.2 rev s^{- 1}$
Initial angular momentum, $L_{1} = I_{1} ω_{1}$
Final moment of inertia,
$I_{2} = 6 + [2 \times 5 \times (0.2)^{2}] = 6.4 kg m^{2}$
Final angular speed $= ω_{2}$
Final angular momentum, $L_{2} = I_{2} ω_{2}$
According to law of conservation of angular momentum,
$L_{1} = L_{2}$ or $I_{1} ω_{1} = I_{2} ω_{2}$
$ω_{2} = \frac{I_{1} ω_{1}}{I_{2}} = \frac{(16 k g m^{2}) (1.2 r e v s^{- 1})}{6.4 k g m^{2}} = 3 r e v s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365654 A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time a viscous fluid of mass $m$ is dropped at the centre and is allowed to spread out and finally falls from the disc. The angular velocity during this period:

1 Decreases continuously

2 Decreases initially and increases again

3 Remains unaltered

4 Increases continuously

Explanation:

From law of conservation of angular momentum if no external torque is acting upon body
$L = I ω = constant$
When liquid is dropped, mass increases hence $I$ increases $(I = m r^{2})$ . So, $ω$ decreases, but as soon as the liquid starts flowing out from the disc $ω$ increases again.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365655 A ring of mass $M$ and radius $R$ is rotating with angular speed $ω$ about a fixed vertical axis passing through its centre $O$ with two point masses each of mass $\frac{M}{8}$ at rest at $O$ . These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is $\frac{8}{9} ω$ and one of the masses is at a distance of $\frac{3}{5} R$ from O. At this instant the distance of the other mass from $O$ is
supporting img

1

\frac{3}{5} R

2

\frac{2}{3} R

3

\frac{4}{5} R

4

\frac{1}{3} R

Explanation:

By conservation of angular moment
supporting img
$M R^{2} ω = [M R^{2} + \frac{M}{8} \frac{9 R^{2}}{25} + \frac{M d^{2}}{8}] \frac{8 ω}{9}$
$R^{2} = (\frac{200 R^{2} + 9 R^{2} + 25 d^{2}}{8 \times 25}) \frac{8}{9}$
$225 R^{2} - 209 R^{2} = 25 d^{2}$
$\Rightarrow d = \frac{4 R}{5}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365656 A girl of mass $M$ stands on the rim of a frictionless merry-go-round of radius $R$ and rotational inertia $I$ that is not moving. She throws a rock of mass $m$ horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is $v$ , afterwards the linear speed of the girl is

1

\frac{m v R^{2}}{I + (M - m) R^{2}}

2

\frac{m v R^{2}}{I + (M + m) R^{2}}

3

\frac{m v R^{2}}{I + M R^{2}}

4

\frac{(m + M) v R^{2}}{I + M R^{2}}

Explanation:

The initial angular momentum of the system is zero. The final angular momentum of the girl-plus-merrygo-round is $(I + M R^{2}) ω$ , which we will take to be positive. The final angular momentum we associate with the throw rock is negative: $- m R v$ , where $v$ is the speed of the rock relative to the ground.
From angular momentum conservation
$0 = (I + M R^{2}) ω - m R v \Rightarrow ω = \frac{m R v}{I + M R^{2}}$
The girl's linear speed is $R ω = \frac{m v R^{2}}{I + M R^{2}}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365657 A man stands on a rotating platform with his arms stretched holding a $5 k g$ weight in each hand. The angular speed of the platform is 1.2 rev $s^{- 1}$ . The moment of inertia of the man together with the platform may be taken to be constant and equal to $6 k g m^{2}$ . If the man brings his arms close to his chest with the distance of each weight from the axis changing from $100 c m$ to $20 c m$ . The new angular speed of the platform is

1

2 r e v s^{- 1}

2

3 r e v s^{- 1}

3

5 r e v s^{- 1}

4

6 r e v s^{- 1}

Explanation:

Initial moment of inertia,
$I_{1} = 6 + [2 \times 5 \times (1)^{2}] = 16 kg m^{2}$
Initial angular velocity, $ω_{1} = 1.2 rev s^{- 1}$
Initial angular momentum, $L_{1} = I_{1} ω_{1}$
Final moment of inertia,
$I_{2} = 6 + [2 \times 5 \times (0.2)^{2}] = 6.4 kg m^{2}$
Final angular speed $= ω_{2}$
Final angular momentum, $L_{2} = I_{2} ω_{2}$
According to law of conservation of angular momentum,
$L_{1} = L_{2}$ or $I_{1} ω_{1} = I_{2} ω_{2}$
$ω_{2} = \frac{I_{1} ω_{1}}{I_{2}} = \frac{(16 k g m^{2}) (1.2 r e v s^{- 1})}{6.4 k g m^{2}} = 3 r e v s^{- 1}$

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