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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365684 A man sits on a freely rotating stool holding two dumbbells, each of mass $2.0 k g$ . When his arms are extended horizontally [Fig. (a)], the dumbbells are $1.0 m$ from the axis of rotation and the man rotates with an angular speed of $0.50 r a d / s$ . The moment of inertia of the man plus stool is $5.0 k g \cdot m^{2}$ and is assumed to be constant. The man pulls the dumbbells inward horizontally to a position $0.50 m$ from the rotation axis [Fig. (b)]. Find the new angular speed of the man
supporting img

1

0.90 r a d / s

2

0.67 r a d / s

3

0.45 r a d / s

4

0.15 r a d / s

Explanation:

A man is seated on a stool that can rotate freely about a vertical axis with his arms stretched horizontally with equal weights in each hand. As no net external torque acts on the system that consists of the 'student, stool and dumbbells'. It means, the angular momentum of that system about the rotation axis must remain constant. The total angular momentum of the system of the student, the stool is conserved then,
${\vec{L}}_{i} = {\vec{L}}_{f} or I_{i} ω_{i} = I_{f} ω_{f}$
Moment of inertia of the system,
$I_{total} = I_{student} + I_{weights} = 5.0 + 2 ({m r}^{2})$
Initial moment of inertia, $r_{initial} = 1.0 m$
Thus, $I_{i} = 5.0 + 2 \times (2.0) (1.0)^{2} = 9.0 k g \cdot m^{2}$
Final moment of inertia, $r_{final} = 0.50 m$
Thus, $I_{f} = 5.0 + 2 (2.0) (0.50)^{2} = 5.0 k g \cdot m^{2}$
We now using conservation of angular momentum $I_{f} ω_{f} = I_{i} ω_{i}$
or $ω_{f} = (\frac{I_{i}}{I_{f}}) ω_{i} = (\frac{9.0}{5.0}) (0.50) = 0.90 r a d / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365685 When acrobat bends his body (assume no external torque)
supporting img

1

I_{acrobat}

decrease

2

I_{acrobat}

will increase

3

ω_{acrobat}

increase

4 Both (1) and (3)

Explanation:

By the conservation of angular momentum, we know that $I ω =$ constant. As no external torque is acting when the acrobat bends his body, then moment of inertia $I$ will decrease and hence $ω$ of acrobat will increase.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365686 Statement A :
If the total force on the body is zero, then the total linear momentum of the body does not change with time.
Statement B :
If the total torque on the rigid body is zero, the total angular momentum of the body does not change with time.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

$F = \frac{d P}{d T} = 0 \Rightarrow P =$ constant
$τ = \frac{d L}{d t} = 0 \Rightarrow L =$ constant

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365687 A thin circular ring of mass $M$ is rotating about its axis with a constant angular velocity $ω$ . Two objects, each of mass $m$ , are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity:

1

\frac{ω (M - 2 m)}{M - 2 m}

2

\frac{ω M}{M - m}

3

\frac{ω (M + 2 m)}{M}

4

\frac{M ω}{M + 2 m}

Explanation:

$I_{1} ω_{1} = I_{2} ω_{2}$
$\begin{aligned} M R^{2} ω = (M R^{2} + 2 m R^{2}) ω_{2} \\ ω_{2} = \frac{M ω}{M + 2 m} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365684 A man sits on a freely rotating stool holding two dumbbells, each of mass $2.0 k g$ . When his arms are extended horizontally [Fig. (a)], the dumbbells are $1.0 m$ from the axis of rotation and the man rotates with an angular speed of $0.50 r a d / s$ . The moment of inertia of the man plus stool is $5.0 k g \cdot m^{2}$ and is assumed to be constant. The man pulls the dumbbells inward horizontally to a position $0.50 m$ from the rotation axis [Fig. (b)]. Find the new angular speed of the man
supporting img

1

0.90 r a d / s

2

0.67 r a d / s

3

0.45 r a d / s

4

0.15 r a d / s

Explanation:

A man is seated on a stool that can rotate freely about a vertical axis with his arms stretched horizontally with equal weights in each hand. As no net external torque acts on the system that consists of the 'student, stool and dumbbells'. It means, the angular momentum of that system about the rotation axis must remain constant. The total angular momentum of the system of the student, the stool is conserved then,
${\vec{L}}_{i} = {\vec{L}}_{f} or I_{i} ω_{i} = I_{f} ω_{f}$
Moment of inertia of the system,
$I_{total} = I_{student} + I_{weights} = 5.0 + 2 ({m r}^{2})$
Initial moment of inertia, $r_{initial} = 1.0 m$
Thus, $I_{i} = 5.0 + 2 \times (2.0) (1.0)^{2} = 9.0 k g \cdot m^{2}$
Final moment of inertia, $r_{final} = 0.50 m$
Thus, $I_{f} = 5.0 + 2 (2.0) (0.50)^{2} = 5.0 k g \cdot m^{2}$
We now using conservation of angular momentum $I_{f} ω_{f} = I_{i} ω_{i}$
or $ω_{f} = (\frac{I_{i}}{I_{f}}) ω_{i} = (\frac{9.0}{5.0}) (0.50) = 0.90 r a d / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365685 When acrobat bends his body (assume no external torque)
supporting img

1

I_{acrobat}

decrease

2

I_{acrobat}

will increase

3

ω_{acrobat}

increase

4 Both (1) and (3)

Explanation:

By the conservation of angular momentum, we know that $I ω =$ constant. As no external torque is acting when the acrobat bends his body, then moment of inertia $I$ will decrease and hence $ω$ of acrobat will increase.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365686 Statement A :
If the total force on the body is zero, then the total linear momentum of the body does not change with time.
Statement B :
If the total torque on the rigid body is zero, the total angular momentum of the body does not change with time.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

$F = \frac{d P}{d T} = 0 \Rightarrow P =$ constant
$τ = \frac{d L}{d t} = 0 \Rightarrow L =$ constant

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365687 A thin circular ring of mass $M$ is rotating about its axis with a constant angular velocity $ω$ . Two objects, each of mass $m$ , are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity:

1

\frac{ω (M - 2 m)}{M - 2 m}

2

\frac{ω M}{M - m}

3

\frac{ω (M + 2 m)}{M}

4

\frac{M ω}{M + 2 m}

Explanation:

$I_{1} ω_{1} = I_{2} ω_{2}$
$\begin{aligned} M R^{2} ω = (M R^{2} + 2 m R^{2}) ω_{2} \\ ω_{2} = \frac{M ω}{M + 2 m} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365684 A man sits on a freely rotating stool holding two dumbbells, each of mass $2.0 k g$ . When his arms are extended horizontally [Fig. (a)], the dumbbells are $1.0 m$ from the axis of rotation and the man rotates with an angular speed of $0.50 r a d / s$ . The moment of inertia of the man plus stool is $5.0 k g \cdot m^{2}$ and is assumed to be constant. The man pulls the dumbbells inward horizontally to a position $0.50 m$ from the rotation axis [Fig. (b)]. Find the new angular speed of the man
supporting img

1

0.90 r a d / s

2

0.67 r a d / s

3

0.45 r a d / s

4

0.15 r a d / s

Explanation:

A man is seated on a stool that can rotate freely about a vertical axis with his arms stretched horizontally with equal weights in each hand. As no net external torque acts on the system that consists of the 'student, stool and dumbbells'. It means, the angular momentum of that system about the rotation axis must remain constant. The total angular momentum of the system of the student, the stool is conserved then,
${\vec{L}}_{i} = {\vec{L}}_{f} or I_{i} ω_{i} = I_{f} ω_{f}$
Moment of inertia of the system,
$I_{total} = I_{student} + I_{weights} = 5.0 + 2 ({m r}^{2})$
Initial moment of inertia, $r_{initial} = 1.0 m$
Thus, $I_{i} = 5.0 + 2 \times (2.0) (1.0)^{2} = 9.0 k g \cdot m^{2}$
Final moment of inertia, $r_{final} = 0.50 m$
Thus, $I_{f} = 5.0 + 2 (2.0) (0.50)^{2} = 5.0 k g \cdot m^{2}$
We now using conservation of angular momentum $I_{f} ω_{f} = I_{i} ω_{i}$
or $ω_{f} = (\frac{I_{i}}{I_{f}}) ω_{i} = (\frac{9.0}{5.0}) (0.50) = 0.90 r a d / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365685 When acrobat bends his body (assume no external torque)
supporting img

1

I_{acrobat}

decrease

2

I_{acrobat}

will increase

3

ω_{acrobat}

increase

4 Both (1) and (3)

Explanation:

By the conservation of angular momentum, we know that $I ω =$ constant. As no external torque is acting when the acrobat bends his body, then moment of inertia $I$ will decrease and hence $ω$ of acrobat will increase.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365686 Statement A :
If the total force on the body is zero, then the total linear momentum of the body does not change with time.
Statement B :
If the total torque on the rigid body is zero, the total angular momentum of the body does not change with time.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

$F = \frac{d P}{d T} = 0 \Rightarrow P =$ constant
$τ = \frac{d L}{d t} = 0 \Rightarrow L =$ constant

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365687 A thin circular ring of mass $M$ is rotating about its axis with a constant angular velocity $ω$ . Two objects, each of mass $m$ , are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity:

1

\frac{ω (M - 2 m)}{M - 2 m}

2

\frac{ω M}{M - m}

3

\frac{ω (M + 2 m)}{M}

4

\frac{M ω}{M + 2 m}

Explanation:

$I_{1} ω_{1} = I_{2} ω_{2}$
$\begin{aligned} M R^{2} ω = (M R^{2} + 2 m R^{2}) ω_{2} \\ ω_{2} = \frac{M ω}{M + 2 m} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365684 A man sits on a freely rotating stool holding two dumbbells, each of mass $2.0 k g$ . When his arms are extended horizontally [Fig. (a)], the dumbbells are $1.0 m$ from the axis of rotation and the man rotates with an angular speed of $0.50 r a d / s$ . The moment of inertia of the man plus stool is $5.0 k g \cdot m^{2}$ and is assumed to be constant. The man pulls the dumbbells inward horizontally to a position $0.50 m$ from the rotation axis [Fig. (b)]. Find the new angular speed of the man
supporting img

1

0.90 r a d / s

2

0.67 r a d / s

3

0.45 r a d / s

4

0.15 r a d / s

Explanation:

A man is seated on a stool that can rotate freely about a vertical axis with his arms stretched horizontally with equal weights in each hand. As no net external torque acts on the system that consists of the 'student, stool and dumbbells'. It means, the angular momentum of that system about the rotation axis must remain constant. The total angular momentum of the system of the student, the stool is conserved then,
${\vec{L}}_{i} = {\vec{L}}_{f} or I_{i} ω_{i} = I_{f} ω_{f}$
Moment of inertia of the system,
$I_{total} = I_{student} + I_{weights} = 5.0 + 2 ({m r}^{2})$
Initial moment of inertia, $r_{initial} = 1.0 m$
Thus, $I_{i} = 5.0 + 2 \times (2.0) (1.0)^{2} = 9.0 k g \cdot m^{2}$
Final moment of inertia, $r_{final} = 0.50 m$
Thus, $I_{f} = 5.0 + 2 (2.0) (0.50)^{2} = 5.0 k g \cdot m^{2}$
We now using conservation of angular momentum $I_{f} ω_{f} = I_{i} ω_{i}$
or $ω_{f} = (\frac{I_{i}}{I_{f}}) ω_{i} = (\frac{9.0}{5.0}) (0.50) = 0.90 r a d / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365685 When acrobat bends his body (assume no external torque)
supporting img

1

I_{acrobat}

decrease

2

I_{acrobat}

will increase

3

ω_{acrobat}

increase

4 Both (1) and (3)

Explanation:

By the conservation of angular momentum, we know that $I ω =$ constant. As no external torque is acting when the acrobat bends his body, then moment of inertia $I$ will decrease and hence $ω$ of acrobat will increase.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365686 Statement A :
If the total force on the body is zero, then the total linear momentum of the body does not change with time.
Statement B :
If the total torque on the rigid body is zero, the total angular momentum of the body does not change with time.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

$F = \frac{d P}{d T} = 0 \Rightarrow P =$ constant
$τ = \frac{d L}{d t} = 0 \Rightarrow L =$ constant

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365687 A thin circular ring of mass $M$ is rotating about its axis with a constant angular velocity $ω$ . Two objects, each of mass $m$ , are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity:

1

\frac{ω (M - 2 m)}{M - 2 m}

2

\frac{ω M}{M - m}

3

\frac{ω (M + 2 m)}{M}

4

\frac{M ω}{M + 2 m}

Explanation:

$I_{1} ω_{1} = I_{2} ω_{2}$
$\begin{aligned} M R^{2} ω = (M R^{2} + 2 m R^{2}) ω_{2} \\ ω_{2} = \frac{M ω}{M + 2 m} \end{aligned}$