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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365968 A spool is pulled horizontally by two equal and opposite forces as shown in figure. Which of the following statements are correct?
supporting img

1 The centre of mass moves towards right

2 The centre of mass moves towards left

3 The net torque about the centre of mass of the spool is zero

4 The centre of mass remains stationary

Explanation:

The two forces produce torque in clockwise direction. In the absence of friction the body slips backward. So friction acts in forward direction. So $C . M$ moves towards right side

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365969 Two cylinders having radii $2 R$ and $R$ and moment of inertia $4 I$ and $I$ about their central axes are supported by axles perpendicular to their planes. The large cylinder is initially rotating clockwise with angular velocity $ω_{0}$ . The small cylinder is moved to the right until it touches the large cylinder and is caused to rotate by the frictional force between the two. Eventually slipping ceases and the two cylinders rotate at constant rates in opposite directions. During this

1 Both the angular momentum and kinetic energy are conserved

2 Kinetic energy is conserved

3 Angular momentum of system is conserved

4 Neither the angular momentum nor the kinetic energy is conserved

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365970 A particle of mass ' $m$ ' is rigidly attached at ' $A$ ' to a ring of mass ' $3 m$ ' and radius ' $r$ '. The system is released from rest and rolls without sliding. The angular acceleration of ring just after release is
supporting img

1

\frac{g}{6 r}

2

\frac{g}{4 r}

3

\frac{g}{8 r}

4

\frac{g}{2 r}

Explanation:

The distance of $CM$ from the ring centre O
$x = \frac{3 m (0) + m (r)}{3 m + m} = \frac{r}{4}$
We can apply torque equation about point of contact as the ring is in pure rolling.
supporting img
$τ_{P} = I_{P} α$
$\begin{aligned} 4 m g (\frac{r}{4}) = [(3 m r^{2} + 3 m r^{2}) + m (A P)^{2}] α \\ \Rightarrow m g r = [6 m r^{2} + m (\sqrt{2} r)^{2}] α \\ \Rightarrow m g r = 8 m r^{2} α \Rightarrow α = \frac{g}{8 r} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365971 A disc is rolling without slipping on a surface. The radius of the disc is $R$ . At $t = 0$ , the top most point on the disc is $A$ as shown in figure. When the disc completes half of its rotation, the displacement of point $A$ from its initial position is
supporting img

1

2 R \sqrt{(1 + 4 π^{2})}

2

R \sqrt{(π^{2} + 1)}

3

R \sqrt{(π^{2} + 4)}

4

2 R

Explanation:

Consider motion of the rolling disc as shown in the figure.
supporting img
For pure rolling motion, during half of the rotation the new locations of $O$ and $A$ will $O^{'}$ and $A^{'}$ as shown.
Thus, displacement of point $A$ is given by
$A A^{'} = \sqrt{{(O A^{'})}^{2} + {(O A)}^{2}}$
$= \sqrt{{(π R)}^{2} + {(2 R)}^{2}} = R \sqrt{(π^{2} + 4)}$

JEE - 2023

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365968 A spool is pulled horizontally by two equal and opposite forces as shown in figure. Which of the following statements are correct?
supporting img

1 The centre of mass moves towards right

2 The centre of mass moves towards left

3 The net torque about the centre of mass of the spool is zero

4 The centre of mass remains stationary

Explanation:

The two forces produce torque in clockwise direction. In the absence of friction the body slips backward. So friction acts in forward direction. So $C . M$ moves towards right side

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365969 Two cylinders having radii $2 R$ and $R$ and moment of inertia $4 I$ and $I$ about their central axes are supported by axles perpendicular to their planes. The large cylinder is initially rotating clockwise with angular velocity $ω_{0}$ . The small cylinder is moved to the right until it touches the large cylinder and is caused to rotate by the frictional force between the two. Eventually slipping ceases and the two cylinders rotate at constant rates in opposite directions. During this

1 Both the angular momentum and kinetic energy are conserved

2 Kinetic energy is conserved

3 Angular momentum of system is conserved

4 Neither the angular momentum nor the kinetic energy is conserved

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365970 A particle of mass ' $m$ ' is rigidly attached at ' $A$ ' to a ring of mass ' $3 m$ ' and radius ' $r$ '. The system is released from rest and rolls without sliding. The angular acceleration of ring just after release is
supporting img

1

\frac{g}{6 r}

2

\frac{g}{4 r}

3

\frac{g}{8 r}

4

\frac{g}{2 r}

Explanation:

The distance of $CM$ from the ring centre O
$x = \frac{3 m (0) + m (r)}{3 m + m} = \frac{r}{4}$
We can apply torque equation about point of contact as the ring is in pure rolling.
supporting img
$τ_{P} = I_{P} α$
$\begin{aligned} 4 m g (\frac{r}{4}) = [(3 m r^{2} + 3 m r^{2}) + m (A P)^{2}] α \\ \Rightarrow m g r = [6 m r^{2} + m (\sqrt{2} r)^{2}] α \\ \Rightarrow m g r = 8 m r^{2} α \Rightarrow α = \frac{g}{8 r} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365971 A disc is rolling without slipping on a surface. The radius of the disc is $R$ . At $t = 0$ , the top most point on the disc is $A$ as shown in figure. When the disc completes half of its rotation, the displacement of point $A$ from its initial position is
supporting img

1

2 R \sqrt{(1 + 4 π^{2})}

2

R \sqrt{(π^{2} + 1)}

3

R \sqrt{(π^{2} + 4)}

4

2 R

Explanation:

Consider motion of the rolling disc as shown in the figure.
supporting img
For pure rolling motion, during half of the rotation the new locations of $O$ and $A$ will $O^{'}$ and $A^{'}$ as shown.
Thus, displacement of point $A$ is given by
$A A^{'} = \sqrt{{(O A^{'})}^{2} + {(O A)}^{2}}$
$= \sqrt{{(π R)}^{2} + {(2 R)}^{2}} = R \sqrt{(π^{2} + 4)}$

JEE - 2023

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365968 A spool is pulled horizontally by two equal and opposite forces as shown in figure. Which of the following statements are correct?
supporting img

1 The centre of mass moves towards right

2 The centre of mass moves towards left

3 The net torque about the centre of mass of the spool is zero

4 The centre of mass remains stationary

Explanation:

The two forces produce torque in clockwise direction. In the absence of friction the body slips backward. So friction acts in forward direction. So $C . M$ moves towards right side

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365969 Two cylinders having radii $2 R$ and $R$ and moment of inertia $4 I$ and $I$ about their central axes are supported by axles perpendicular to their planes. The large cylinder is initially rotating clockwise with angular velocity $ω_{0}$ . The small cylinder is moved to the right until it touches the large cylinder and is caused to rotate by the frictional force between the two. Eventually slipping ceases and the two cylinders rotate at constant rates in opposite directions. During this

1 Both the angular momentum and kinetic energy are conserved

2 Kinetic energy is conserved

3 Angular momentum of system is conserved

4 Neither the angular momentum nor the kinetic energy is conserved

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365970 A particle of mass ' $m$ ' is rigidly attached at ' $A$ ' to a ring of mass ' $3 m$ ' and radius ' $r$ '. The system is released from rest and rolls without sliding. The angular acceleration of ring just after release is
supporting img

1

\frac{g}{6 r}

2

\frac{g}{4 r}

3

\frac{g}{8 r}

4

\frac{g}{2 r}

Explanation:

The distance of $CM$ from the ring centre O
$x = \frac{3 m (0) + m (r)}{3 m + m} = \frac{r}{4}$
We can apply torque equation about point of contact as the ring is in pure rolling.
supporting img
$τ_{P} = I_{P} α$
$\begin{aligned} 4 m g (\frac{r}{4}) = [(3 m r^{2} + 3 m r^{2}) + m (A P)^{2}] α \\ \Rightarrow m g r = [6 m r^{2} + m (\sqrt{2} r)^{2}] α \\ \Rightarrow m g r = 8 m r^{2} α \Rightarrow α = \frac{g}{8 r} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365971 A disc is rolling without slipping on a surface. The radius of the disc is $R$ . At $t = 0$ , the top most point on the disc is $A$ as shown in figure. When the disc completes half of its rotation, the displacement of point $A$ from its initial position is
supporting img

1

2 R \sqrt{(1 + 4 π^{2})}

2

R \sqrt{(π^{2} + 1)}

3

R \sqrt{(π^{2} + 4)}

4

2 R

Explanation:

Consider motion of the rolling disc as shown in the figure.
supporting img
For pure rolling motion, during half of the rotation the new locations of $O$ and $A$ will $O^{'}$ and $A^{'}$ as shown.
Thus, displacement of point $A$ is given by
$A A^{'} = \sqrt{{(O A^{'})}^{2} + {(O A)}^{2}}$
$= \sqrt{{(π R)}^{2} + {(2 R)}^{2}} = R \sqrt{(π^{2} + 4)}$

JEE - 2023

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365968 A spool is pulled horizontally by two equal and opposite forces as shown in figure. Which of the following statements are correct?
supporting img

1 The centre of mass moves towards right

2 The centre of mass moves towards left

3 The net torque about the centre of mass of the spool is zero

4 The centre of mass remains stationary

Explanation:

The two forces produce torque in clockwise direction. In the absence of friction the body slips backward. So friction acts in forward direction. So $C . M$ moves towards right side

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365969 Two cylinders having radii $2 R$ and $R$ and moment of inertia $4 I$ and $I$ about their central axes are supported by axles perpendicular to their planes. The large cylinder is initially rotating clockwise with angular velocity $ω_{0}$ . The small cylinder is moved to the right until it touches the large cylinder and is caused to rotate by the frictional force between the two. Eventually slipping ceases and the two cylinders rotate at constant rates in opposite directions. During this

1 Both the angular momentum and kinetic energy are conserved

2 Kinetic energy is conserved

3 Angular momentum of system is conserved

4 Neither the angular momentum nor the kinetic energy is conserved

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365970 A particle of mass ' $m$ ' is rigidly attached at ' $A$ ' to a ring of mass ' $3 m$ ' and radius ' $r$ '. The system is released from rest and rolls without sliding. The angular acceleration of ring just after release is
supporting img

1

\frac{g}{6 r}

2

\frac{g}{4 r}

3

\frac{g}{8 r}

4

\frac{g}{2 r}

Explanation:

The distance of $CM$ from the ring centre O
$x = \frac{3 m (0) + m (r)}{3 m + m} = \frac{r}{4}$
We can apply torque equation about point of contact as the ring is in pure rolling.
supporting img
$τ_{P} = I_{P} α$
$\begin{aligned} 4 m g (\frac{r}{4}) = [(3 m r^{2} + 3 m r^{2}) + m (A P)^{2}] α \\ \Rightarrow m g r = [6 m r^{2} + m (\sqrt{2} r)^{2}] α \\ \Rightarrow m g r = 8 m r^{2} α \Rightarrow α = \frac{g}{8 r} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365971 A disc is rolling without slipping on a surface. The radius of the disc is $R$ . At $t = 0$ , the top most point on the disc is $A$ as shown in figure. When the disc completes half of its rotation, the displacement of point $A$ from its initial position is
supporting img

1

2 R \sqrt{(1 + 4 π^{2})}

2

R \sqrt{(π^{2} + 1)}

3

R \sqrt{(π^{2} + 4)}

4

2 R

Explanation:

Consider motion of the rolling disc as shown in the figure.
supporting img
For pure rolling motion, during half of the rotation the new locations of $O$ and $A$ will $O^{'}$ and $A^{'}$ as shown.
Thus, displacement of point $A$ is given by
$A A^{'} = \sqrt{{(O A^{'})}^{2} + {(O A)}^{2}}$
$= \sqrt{{(π R)}^{2} + {(2 R)}^{2}} = R \sqrt{(π^{2} + 4)}$

JEE - 2023

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