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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365667 A girl is sitting on a weightless platform. Initially the girl is rotating about the vertical axis as shown in the figure. If frictional force is neglected and girl bends her hand, then
supporting img

1

I_{g i r l}

will reduce

2

ω_{g i r l}

will reduce

3

I_{g i r l}

will increase

4 None of the above

Explanation:

As there is no external torque, so if the girl bends her hands, her moment of inertia about the rotational axis will decrease. By conservation of angualr momentum,
$L = I ω =$ constant. So in order to keep $L$ constant, if $I$ is decreasing, then $ω$ will increase.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365668 The angular speed of a body changes from $ω_{1}$ and $ω_{2}$ without allowing a torque but due to change in its moment of inertia. The ratio of radii of gyration in the two cases is

1

ω_{1} : ω_{2}

2

ω_{2} : ω_{1}

3

\sqrt{ω_{2}} : \sqrt{ω_{1}}

4

\sqrt{ω_{1}} : \sqrt{ω_{2}}

Explanation:

$L = I ω = M K^{2} ω =$ constant $∴ \frac{K_{1}}{K_{2}} = \sqrt{\frac{ω_{2}}{ω_{1}}}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365669 A disc of mass $2 k g$ and radius $0.2 m$ is rotating with angular velocity $30 r a d s^{- 1} .$ What is angular velocity, if a mass of $0.25 k g$ is put on periphery of the disc?

1

24 r a d s^{- 1}

2

36 r a d s^{- 1}

3

15 r a d s^{- 1}

4

26 r a d s^{- 1}

Explanation:

If no external torque acts on a system of particles, then angular momentum of the system remains constant, i.e. $τ = 0$
$\Rightarrow \frac{d L}{d t} = 0 \Rightarrow L = I ω =$ constant
$\Rightarrow I_{1} ω_{1} = I_{2} ω_{2}$
$\Rightarrow \frac{1}{2} M r^{2} ω_{1} = \frac{1}{2} (M + 2 m) r^{2} ω_{2} (1)$
Given, $M = 2 k g, m = 0.25 k g,$
$r = 0.2 m^{2}$ and $ω_{1} = 30 r a d s^{- 1} .$
Hence, after putting the given values in Eq. (1), we get
$\frac{1}{2} \times 2 \times (0.2)^{2} \times 30$
$= \frac{1}{2} \times (2 + 2 \times 0.25) (0.2)^{2} \times ω_{2}$
or $1.2 = 0.05 ω_{2}$ or $ω_{2} = 24 r a d s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365670 An ice skater starts a spin with her arms stretched out to the sides. She balances on the tip of one skate to turn without friction. She then pulls her arms in so that her moment of inertia decreases by a factor of 2 . In the process of her doing so, her kinetic energy becomes $n$ times the initial kinetic energy. Find the value of $n$ is

1 2

2 4

3 8

4 10

Explanation:

Her angular momentum stays constant as $I$ is cut in half and $ω$ doubles. Then $\frac{1}{2} I ω^{2}$ doubles. So $n = 2$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365667 A girl is sitting on a weightless platform. Initially the girl is rotating about the vertical axis as shown in the figure. If frictional force is neglected and girl bends her hand, then
supporting img

1

I_{g i r l}

will reduce

2

ω_{g i r l}

will reduce

3

I_{g i r l}

will increase

4 None of the above

Explanation:

As there is no external torque, so if the girl bends her hands, her moment of inertia about the rotational axis will decrease. By conservation of angualr momentum,
$L = I ω =$ constant. So in order to keep $L$ constant, if $I$ is decreasing, then $ω$ will increase.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365668 The angular speed of a body changes from $ω_{1}$ and $ω_{2}$ without allowing a torque but due to change in its moment of inertia. The ratio of radii of gyration in the two cases is

1

ω_{1} : ω_{2}

2

ω_{2} : ω_{1}

3

\sqrt{ω_{2}} : \sqrt{ω_{1}}

4

\sqrt{ω_{1}} : \sqrt{ω_{2}}

Explanation:

$L = I ω = M K^{2} ω =$ constant $∴ \frac{K_{1}}{K_{2}} = \sqrt{\frac{ω_{2}}{ω_{1}}}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365669 A disc of mass $2 k g$ and radius $0.2 m$ is rotating with angular velocity $30 r a d s^{- 1} .$ What is angular velocity, if a mass of $0.25 k g$ is put on periphery of the disc?

1

24 r a d s^{- 1}

2

36 r a d s^{- 1}

3

15 r a d s^{- 1}

4

26 r a d s^{- 1}

Explanation:

If no external torque acts on a system of particles, then angular momentum of the system remains constant, i.e. $τ = 0$
$\Rightarrow \frac{d L}{d t} = 0 \Rightarrow L = I ω =$ constant
$\Rightarrow I_{1} ω_{1} = I_{2} ω_{2}$
$\Rightarrow \frac{1}{2} M r^{2} ω_{1} = \frac{1}{2} (M + 2 m) r^{2} ω_{2} (1)$
Given, $M = 2 k g, m = 0.25 k g,$
$r = 0.2 m^{2}$ and $ω_{1} = 30 r a d s^{- 1} .$
Hence, after putting the given values in Eq. (1), we get
$\frac{1}{2} \times 2 \times (0.2)^{2} \times 30$
$= \frac{1}{2} \times (2 + 2 \times 0.25) (0.2)^{2} \times ω_{2}$
or $1.2 = 0.05 ω_{2}$ or $ω_{2} = 24 r a d s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365670 An ice skater starts a spin with her arms stretched out to the sides. She balances on the tip of one skate to turn without friction. She then pulls her arms in so that her moment of inertia decreases by a factor of 2 . In the process of her doing so, her kinetic energy becomes $n$ times the initial kinetic energy. Find the value of $n$ is

1 2

2 4

3 8

4 10

Explanation:

Her angular momentum stays constant as $I$ is cut in half and $ω$ doubles. Then $\frac{1}{2} I ω^{2}$ doubles. So $n = 2$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365667 A girl is sitting on a weightless platform. Initially the girl is rotating about the vertical axis as shown in the figure. If frictional force is neglected and girl bends her hand, then
supporting img

1

I_{g i r l}

will reduce

2

ω_{g i r l}

will reduce

3

I_{g i r l}

will increase

4 None of the above

Explanation:

As there is no external torque, so if the girl bends her hands, her moment of inertia about the rotational axis will decrease. By conservation of angualr momentum,
$L = I ω =$ constant. So in order to keep $L$ constant, if $I$ is decreasing, then $ω$ will increase.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365668 The angular speed of a body changes from $ω_{1}$ and $ω_{2}$ without allowing a torque but due to change in its moment of inertia. The ratio of radii of gyration in the two cases is

1

ω_{1} : ω_{2}

2

ω_{2} : ω_{1}

3

\sqrt{ω_{2}} : \sqrt{ω_{1}}

4

\sqrt{ω_{1}} : \sqrt{ω_{2}}

Explanation:

$L = I ω = M K^{2} ω =$ constant $∴ \frac{K_{1}}{K_{2}} = \sqrt{\frac{ω_{2}}{ω_{1}}}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365669 A disc of mass $2 k g$ and radius $0.2 m$ is rotating with angular velocity $30 r a d s^{- 1} .$ What is angular velocity, if a mass of $0.25 k g$ is put on periphery of the disc?

1

24 r a d s^{- 1}

2

36 r a d s^{- 1}

3

15 r a d s^{- 1}

4

26 r a d s^{- 1}

Explanation:

If no external torque acts on a system of particles, then angular momentum of the system remains constant, i.e. $τ = 0$
$\Rightarrow \frac{d L}{d t} = 0 \Rightarrow L = I ω =$ constant
$\Rightarrow I_{1} ω_{1} = I_{2} ω_{2}$
$\Rightarrow \frac{1}{2} M r^{2} ω_{1} = \frac{1}{2} (M + 2 m) r^{2} ω_{2} (1)$
Given, $M = 2 k g, m = 0.25 k g,$
$r = 0.2 m^{2}$ and $ω_{1} = 30 r a d s^{- 1} .$
Hence, after putting the given values in Eq. (1), we get
$\frac{1}{2} \times 2 \times (0.2)^{2} \times 30$
$= \frac{1}{2} \times (2 + 2 \times 0.25) (0.2)^{2} \times ω_{2}$
or $1.2 = 0.05 ω_{2}$ or $ω_{2} = 24 r a d s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365670 An ice skater starts a spin with her arms stretched out to the sides. She balances on the tip of one skate to turn without friction. She then pulls her arms in so that her moment of inertia decreases by a factor of 2 . In the process of her doing so, her kinetic energy becomes $n$ times the initial kinetic energy. Find the value of $n$ is

1 2

2 4

3 8

4 10

Explanation:

Her angular momentum stays constant as $I$ is cut in half and $ω$ doubles. Then $\frac{1}{2} I ω^{2}$ doubles. So $n = 2$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365667 A girl is sitting on a weightless platform. Initially the girl is rotating about the vertical axis as shown in the figure. If frictional force is neglected and girl bends her hand, then
supporting img

1

I_{g i r l}

will reduce

2

ω_{g i r l}

will reduce

3

I_{g i r l}

will increase

4 None of the above

Explanation:

As there is no external torque, so if the girl bends her hands, her moment of inertia about the rotational axis will decrease. By conservation of angualr momentum,
$L = I ω =$ constant. So in order to keep $L$ constant, if $I$ is decreasing, then $ω$ will increase.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365668 The angular speed of a body changes from $ω_{1}$ and $ω_{2}$ without allowing a torque but due to change in its moment of inertia. The ratio of radii of gyration in the two cases is

1

ω_{1} : ω_{2}

2

ω_{2} : ω_{1}

3

\sqrt{ω_{2}} : \sqrt{ω_{1}}

4

\sqrt{ω_{1}} : \sqrt{ω_{2}}

Explanation:

$L = I ω = M K^{2} ω =$ constant $∴ \frac{K_{1}}{K_{2}} = \sqrt{\frac{ω_{2}}{ω_{1}}}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365669 A disc of mass $2 k g$ and radius $0.2 m$ is rotating with angular velocity $30 r a d s^{- 1} .$ What is angular velocity, if a mass of $0.25 k g$ is put on periphery of the disc?

1

24 r a d s^{- 1}

2

36 r a d s^{- 1}

3

15 r a d s^{- 1}

4

26 r a d s^{- 1}

Explanation:

If no external torque acts on a system of particles, then angular momentum of the system remains constant, i.e. $τ = 0$
$\Rightarrow \frac{d L}{d t} = 0 \Rightarrow L = I ω =$ constant
$\Rightarrow I_{1} ω_{1} = I_{2} ω_{2}$
$\Rightarrow \frac{1}{2} M r^{2} ω_{1} = \frac{1}{2} (M + 2 m) r^{2} ω_{2} (1)$
Given, $M = 2 k g, m = 0.25 k g,$
$r = 0.2 m^{2}$ and $ω_{1} = 30 r a d s^{- 1} .$
Hence, after putting the given values in Eq. (1), we get
$\frac{1}{2} \times 2 \times (0.2)^{2} \times 30$
$= \frac{1}{2} \times (2 + 2 \times 0.25) (0.2)^{2} \times ω_{2}$
or $1.2 = 0.05 ω_{2}$ or $ω_{2} = 24 r a d s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365670 An ice skater starts a spin with her arms stretched out to the sides. She balances on the tip of one skate to turn without friction. She then pulls her arms in so that her moment of inertia decreases by a factor of 2 . In the process of her doing so, her kinetic energy becomes $n$ times the initial kinetic energy. Find the value of $n$ is

1 2

2 4

3 8

4 10

Explanation:

Her angular momentum stays constant as $I$ is cut in half and $ω$ doubles. Then $\frac{1}{2} I ω^{2}$ doubles. So $n = 2$