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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365693 A simplified bola consists of two (instead of the usual three) small but heavy balls connected to a common point by identical lengths of string. To launch it, you hold one of the balls overhead and swing the other ball in a horizontal circle about your wrist. You then let go of the bola which very soon starts rotating around its centre of mass, as shown below. Which one of the following statements holds true when the rotation axis changes from (1) to (2)?
supporting img

1 Angular momentum and angular speed both increase.

2 Angular momentum and angular speed remain the same.

3 Angular momentum is the same, but angular speed decreases.

4 Angular momentum is the same, but angular speed increases.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365694 A solid sphere of mass $M$ and radius $r$ slips on a rough horizontal surface. At some instant, it has translational velocity $v_{0}$ and angular velocity $ω (= \frac{v_{0}}{2 r})$ about its centre as shown in figure. The translational velocity after the sphere starts pure rolling motion is
supporting img

1

\frac{2 v_{0}}{7}

2

\frac{5 v_{0}}{7}

3

\frac{v_{0}}{7}

4

\frac{6 v_{0}}{7}

Explanation:

Angular momentum about point of contact will be conserved
$L_{i} = L_{f}$
$\begin{aligned} (\frac{2}{5} M r^{2}) \frac{v_{0}}{r} + M r v_{0} = (\frac{2}{5} M r^{2}) \frac{v}{r} + M r v \\ \Rightarrow v = \frac{6}{7} v_{0} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365695 A hoop of radius $r$ and mass $m$ rotating with an angular velocity $ω_{0}$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

1

\frac{r ω_{0}}{4}

2

\frac{r ω_{0}}{3}

3

\frac{r ω_{0}}{2}

4

r ω_{0}

Explanation:

supporting img
By conserving angular momentum about $P$
$\begin{aligned} L_{i} = L_{f} \\ m r^{2} ω_{0} = m v r + m r^{2} \times \frac{v_{0}}{r} \\ \Rightarrow v = \frac{ω_{0} r}{2} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365696 A cubical block of side $30 cm$ is moving with velocity $2 {ms}^{- 1}$ on a smooth horizontal surface. The surface has a bump at a point $O$ as shown in figure. The angular velocity (in $rad / s$ ) of the block immediately after it hits the bump, is :
supporting img

1 9.4

2 6.7

3 5.0

4 13.3

Explanation:

Angular momentum of the block is conserved about point $O$ . The moment of inertia of the cube about the coner is
$\begin{aligned} I_{0} = [\frac{m a^{2}}{6} + m {(\frac{a}{\sqrt{2}})}^{2}] = \frac{2 m a^{2}}{3} \\ m v \frac{a}{2} = \frac{2}{3} m a^{2} ω \Rightarrow ω = \frac{3 v}{4 a} = 5 rad / s \end{aligned}$

JEE - 2016

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365693 A simplified bola consists of two (instead of the usual three) small but heavy balls connected to a common point by identical lengths of string. To launch it, you hold one of the balls overhead and swing the other ball in a horizontal circle about your wrist. You then let go of the bola which very soon starts rotating around its centre of mass, as shown below. Which one of the following statements holds true when the rotation axis changes from (1) to (2)?
supporting img

1 Angular momentum and angular speed both increase.

2 Angular momentum and angular speed remain the same.

3 Angular momentum is the same, but angular speed decreases.

4 Angular momentum is the same, but angular speed increases.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365694 A solid sphere of mass $M$ and radius $r$ slips on a rough horizontal surface. At some instant, it has translational velocity $v_{0}$ and angular velocity $ω (= \frac{v_{0}}{2 r})$ about its centre as shown in figure. The translational velocity after the sphere starts pure rolling motion is
supporting img

1

\frac{2 v_{0}}{7}

2

\frac{5 v_{0}}{7}

3

\frac{v_{0}}{7}

4

\frac{6 v_{0}}{7}

Explanation:

Angular momentum about point of contact will be conserved
$L_{i} = L_{f}$
$\begin{aligned} (\frac{2}{5} M r^{2}) \frac{v_{0}}{r} + M r v_{0} = (\frac{2}{5} M r^{2}) \frac{v}{r} + M r v \\ \Rightarrow v = \frac{6}{7} v_{0} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365695 A hoop of radius $r$ and mass $m$ rotating with an angular velocity $ω_{0}$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

1

\frac{r ω_{0}}{4}

2

\frac{r ω_{0}}{3}

3

\frac{r ω_{0}}{2}

4

r ω_{0}

Explanation:

supporting img
By conserving angular momentum about $P$
$\begin{aligned} L_{i} = L_{f} \\ m r^{2} ω_{0} = m v r + m r^{2} \times \frac{v_{0}}{r} \\ \Rightarrow v = \frac{ω_{0} r}{2} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365696 A cubical block of side $30 cm$ is moving with velocity $2 {ms}^{- 1}$ on a smooth horizontal surface. The surface has a bump at a point $O$ as shown in figure. The angular velocity (in $rad / s$ ) of the block immediately after it hits the bump, is :
supporting img

1 9.4

2 6.7

3 5.0

4 13.3

Explanation:

Angular momentum of the block is conserved about point $O$ . The moment of inertia of the cube about the coner is
$\begin{aligned} I_{0} = [\frac{m a^{2}}{6} + m {(\frac{a}{\sqrt{2}})}^{2}] = \frac{2 m a^{2}}{3} \\ m v \frac{a}{2} = \frac{2}{3} m a^{2} ω \Rightarrow ω = \frac{3 v}{4 a} = 5 rad / s \end{aligned}$

JEE - 2016

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365693 A simplified bola consists of two (instead of the usual three) small but heavy balls connected to a common point by identical lengths of string. To launch it, you hold one of the balls overhead and swing the other ball in a horizontal circle about your wrist. You then let go of the bola which very soon starts rotating around its centre of mass, as shown below. Which one of the following statements holds true when the rotation axis changes from (1) to (2)?
supporting img

1 Angular momentum and angular speed both increase.

2 Angular momentum and angular speed remain the same.

3 Angular momentum is the same, but angular speed decreases.

4 Angular momentum is the same, but angular speed increases.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365694 A solid sphere of mass $M$ and radius $r$ slips on a rough horizontal surface. At some instant, it has translational velocity $v_{0}$ and angular velocity $ω (= \frac{v_{0}}{2 r})$ about its centre as shown in figure. The translational velocity after the sphere starts pure rolling motion is
supporting img

1

\frac{2 v_{0}}{7}

2

\frac{5 v_{0}}{7}

3

\frac{v_{0}}{7}

4

\frac{6 v_{0}}{7}

Explanation:

Angular momentum about point of contact will be conserved
$L_{i} = L_{f}$
$\begin{aligned} (\frac{2}{5} M r^{2}) \frac{v_{0}}{r} + M r v_{0} = (\frac{2}{5} M r^{2}) \frac{v}{r} + M r v \\ \Rightarrow v = \frac{6}{7} v_{0} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365695 A hoop of radius $r$ and mass $m$ rotating with an angular velocity $ω_{0}$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

1

\frac{r ω_{0}}{4}

2

\frac{r ω_{0}}{3}

3

\frac{r ω_{0}}{2}

4

r ω_{0}

Explanation:

supporting img
By conserving angular momentum about $P$
$\begin{aligned} L_{i} = L_{f} \\ m r^{2} ω_{0} = m v r + m r^{2} \times \frac{v_{0}}{r} \\ \Rightarrow v = \frac{ω_{0} r}{2} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365696 A cubical block of side $30 cm$ is moving with velocity $2 {ms}^{- 1}$ on a smooth horizontal surface. The surface has a bump at a point $O$ as shown in figure. The angular velocity (in $rad / s$ ) of the block immediately after it hits the bump, is :
supporting img

1 9.4

2 6.7

3 5.0

4 13.3

Explanation:

Angular momentum of the block is conserved about point $O$ . The moment of inertia of the cube about the coner is
$\begin{aligned} I_{0} = [\frac{m a^{2}}{6} + m {(\frac{a}{\sqrt{2}})}^{2}] = \frac{2 m a^{2}}{3} \\ m v \frac{a}{2} = \frac{2}{3} m a^{2} ω \Rightarrow ω = \frac{3 v}{4 a} = 5 rad / s \end{aligned}$

JEE - 2016

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365693 A simplified bola consists of two (instead of the usual three) small but heavy balls connected to a common point by identical lengths of string. To launch it, you hold one of the balls overhead and swing the other ball in a horizontal circle about your wrist. You then let go of the bola which very soon starts rotating around its centre of mass, as shown below. Which one of the following statements holds true when the rotation axis changes from (1) to (2)?
supporting img

1 Angular momentum and angular speed both increase.

2 Angular momentum and angular speed remain the same.

3 Angular momentum is the same, but angular speed decreases.

4 Angular momentum is the same, but angular speed increases.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365694 A solid sphere of mass $M$ and radius $r$ slips on a rough horizontal surface. At some instant, it has translational velocity $v_{0}$ and angular velocity $ω (= \frac{v_{0}}{2 r})$ about its centre as shown in figure. The translational velocity after the sphere starts pure rolling motion is
supporting img

1

\frac{2 v_{0}}{7}

2

\frac{5 v_{0}}{7}

3

\frac{v_{0}}{7}

4

\frac{6 v_{0}}{7}

Explanation:

Angular momentum about point of contact will be conserved
$L_{i} = L_{f}$
$\begin{aligned} (\frac{2}{5} M r^{2}) \frac{v_{0}}{r} + M r v_{0} = (\frac{2}{5} M r^{2}) \frac{v}{r} + M r v \\ \Rightarrow v = \frac{6}{7} v_{0} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365695 A hoop of radius $r$ and mass $m$ rotating with an angular velocity $ω_{0}$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

1

\frac{r ω_{0}}{4}

2

\frac{r ω_{0}}{3}

3

\frac{r ω_{0}}{2}

4

r ω_{0}

Explanation:

supporting img
By conserving angular momentum about $P$
$\begin{aligned} L_{i} = L_{f} \\ m r^{2} ω_{0} = m v r + m r^{2} \times \frac{v_{0}}{r} \\ \Rightarrow v = \frac{ω_{0} r}{2} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365696 A cubical block of side $30 cm$ is moving with velocity $2 {ms}^{- 1}$ on a smooth horizontal surface. The surface has a bump at a point $O$ as shown in figure. The angular velocity (in $rad / s$ ) of the block immediately after it hits the bump, is :
supporting img

1 9.4

2 6.7

3 5.0

4 13.3

Explanation:

Angular momentum of the block is conserved about point $O$ . The moment of inertia of the cube about the coner is
$\begin{aligned} I_{0} = [\frac{m a^{2}}{6} + m {(\frac{a}{\sqrt{2}})}^{2}] = \frac{2 m a^{2}}{3} \\ m v \frac{a}{2} = \frac{2}{3} m a^{2} ω \Rightarrow ω = \frac{3 v}{4 a} = 5 rad / s \end{aligned}$

JEE - 2016

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