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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366074 A ceiling fan is rotating around a fixed axle as shown. The direction of angular velocity is along
supporting img

1

+ \hat{j}

2

- \hat{j}

3

+ \hat{k}

4

- \hat{k}

Explanation:

Using right hand thumb rule, direction of angular velocity is along $- v e z$ axis. Correct option is (4).

KCET - 2024

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366075 Two points $A$ and $B$ on a disc have velocities $v_{1}$ and $v_{2}$ respectively, at some moment. Their directions make angles $60^{\circ}$ and $30^{\circ}$ , respectively, with the line of separation as shown in the figure. The angular velocity of disc is $(A$ and $B$ are seperated by $^{'} d^{'}$ )
supporting img

1

\frac{v_{2}}{\sqrt{3} d}

2

\frac{v_{2}}{d}

3

\frac{v_{2} - v_{1}}{d}

4

\frac{\sqrt{3} v_{1}}{d}

Explanation:

For rigid body, separation between two points remain same
$v_{1} \cos 60^{\circ} = v_{2} \cos 30^{\circ}$
$\frac{v_{1}}{2} = \frac{\sqrt{3} v_{2}}{2} \Rightarrow v_{1} = \sqrt{3} v_{2}$
supporting img
$ω_{d i s c} | \frac{v_{2} \sin 30^{\circ} - v_{1} - \sin 60^{\circ}}{d} | = | \frac{\frac{v_{2}}{2} - \frac{\sqrt{3} v_{1}}{2}}{d} |$
$ω_{d i s c} | \frac{v_{2} - \sqrt{3} \times \sqrt{3} v_{2}}{2 d} | = \frac{2 v_{1}}{2 d} = \frac{v_{2}}{d}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366076 Two end points of a rod move with velocities $3 v$ and $v$ perpendicular to the rod and in the same direction, separated by a distance ' $r$ '. Then the angular velocity of the rod is

1

\frac{2 v}{r}

2

\frac{3 v}{r}

3

\frac{4 v}{r}

4

\frac{5 v}{r}

Explanation:

$ω_{r a d} = ω_{p o int} = (\frac{v_{r e l}}{r})$ .
$v_{rel}$ . being the velocity of one point w.r.t other.
' $r$ ' being the distance between them.
$= \frac{3 v - v}{r} = \frac{2 v}{r}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366077 Two particles $A$ and $B$ are situated at a distance $d = 2 m$ apart. Particle $A$ has a velocity of $10 m / s$ at an angle of $60^{\circ}$ and particle $B$ has velocity $v$ at an angle $30^{\circ}$ as shown in the figure. The distance $d$ between $A$ and $B$ is constant. The angular velocity of $B$ with respect to $A$ is:
supporting img

1

\frac{5}{\sqrt{3}} rad / s

2

\frac{10}{\sqrt{3}} rad / s

3

5 \sqrt{3} rad / s

4

10 \sqrt{3} rad / s

Explanation:

If distance between them remains constant, then
$v \cos 30^{\circ} = u \cos 60^{\circ}$
$\Rightarrow v \frac{\sqrt{3}}{2} = 10 \frac{1}{2} \Rightarrow v = \frac{10}{\sqrt{3}}$
$ω = (v \sin 30^{\circ} - u \sin 60^{\circ}) / d$
$= \frac{(\frac{10}{\sqrt{3}}) \times \frac{1}{2} - \frac{10 \times \sqrt{3}}{2}}{2} = - \frac{5}{\sqrt{3}} rad / s .$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366078 A solid cylinder of mass $2 k g$ and radius $4 c m$ is rotating about its axis at the rate of $3 r p m$ . The torque required to stop after $2 π$ revolutions is

1

2 \times 10^{- 6} N m

2

2 \times 10^{- 3} N m

3

12 \times 10^{- 4} N m

4

2 \times 10^{6} N m

Explanation:

From work energy theorem.
$W = \frac{1}{2} I (ω_{f}^{2} - ω_{i}^{2})$
$θ = 2 π \times 2 π = 4 π^{2} r a d$
$ω_{i} = 3 \times \frac{2 π}{60} r a d / s$
$\Rightarrow - τ θ = \frac{1}{2} \times \frac{1}{2} m r^{2} (0^{2} - ω_{i}^{2})$
$\Rightarrow τ = \frac{\frac{1}{2} \times \frac{1}{2} \times 2 \times {(4 \times 10^{- 2})}^{2} {(3 \times \frac{2 π}{60})}^{2}}{4 π^{2}}$
$\Rightarrow τ = 2 \times 10^{- 6} N m$

NEET - 2019

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366074 A ceiling fan is rotating around a fixed axle as shown. The direction of angular velocity is along
supporting img

1

+ \hat{j}

2

- \hat{j}

3

+ \hat{k}

4

- \hat{k}

Explanation:

Using right hand thumb rule, direction of angular velocity is along $- v e z$ axis. Correct option is (4).

KCET - 2024

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366075 Two points $A$ and $B$ on a disc have velocities $v_{1}$ and $v_{2}$ respectively, at some moment. Their directions make angles $60^{\circ}$ and $30^{\circ}$ , respectively, with the line of separation as shown in the figure. The angular velocity of disc is $(A$ and $B$ are seperated by $^{'} d^{'}$ )
supporting img

1

\frac{v_{2}}{\sqrt{3} d}

2

\frac{v_{2}}{d}

3

\frac{v_{2} - v_{1}}{d}

4

\frac{\sqrt{3} v_{1}}{d}

Explanation:

For rigid body, separation between two points remain same
$v_{1} \cos 60^{\circ} = v_{2} \cos 30^{\circ}$
$\frac{v_{1}}{2} = \frac{\sqrt{3} v_{2}}{2} \Rightarrow v_{1} = \sqrt{3} v_{2}$
supporting img
$ω_{d i s c} | \frac{v_{2} \sin 30^{\circ} - v_{1} - \sin 60^{\circ}}{d} | = | \frac{\frac{v_{2}}{2} - \frac{\sqrt{3} v_{1}}{2}}{d} |$
$ω_{d i s c} | \frac{v_{2} - \sqrt{3} \times \sqrt{3} v_{2}}{2 d} | = \frac{2 v_{1}}{2 d} = \frac{v_{2}}{d}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366076 Two end points of a rod move with velocities $3 v$ and $v$ perpendicular to the rod and in the same direction, separated by a distance ' $r$ '. Then the angular velocity of the rod is

1

\frac{2 v}{r}

2

\frac{3 v}{r}

3

\frac{4 v}{r}

4

\frac{5 v}{r}

Explanation:

$ω_{r a d} = ω_{p o int} = (\frac{v_{r e l}}{r})$ .
$v_{rel}$ . being the velocity of one point w.r.t other.
' $r$ ' being the distance between them.
$= \frac{3 v - v}{r} = \frac{2 v}{r}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366077 Two particles $A$ and $B$ are situated at a distance $d = 2 m$ apart. Particle $A$ has a velocity of $10 m / s$ at an angle of $60^{\circ}$ and particle $B$ has velocity $v$ at an angle $30^{\circ}$ as shown in the figure. The distance $d$ between $A$ and $B$ is constant. The angular velocity of $B$ with respect to $A$ is:
supporting img

1

\frac{5}{\sqrt{3}} rad / s

2

\frac{10}{\sqrt{3}} rad / s

3

5 \sqrt{3} rad / s

4

10 \sqrt{3} rad / s

Explanation:

If distance between them remains constant, then
$v \cos 30^{\circ} = u \cos 60^{\circ}$
$\Rightarrow v \frac{\sqrt{3}}{2} = 10 \frac{1}{2} \Rightarrow v = \frac{10}{\sqrt{3}}$
$ω = (v \sin 30^{\circ} - u \sin 60^{\circ}) / d$
$= \frac{(\frac{10}{\sqrt{3}}) \times \frac{1}{2} - \frac{10 \times \sqrt{3}}{2}}{2} = - \frac{5}{\sqrt{3}} rad / s .$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366078 A solid cylinder of mass $2 k g$ and radius $4 c m$ is rotating about its axis at the rate of $3 r p m$ . The torque required to stop after $2 π$ revolutions is

1

2 \times 10^{- 6} N m

2

2 \times 10^{- 3} N m

3

12 \times 10^{- 4} N m

4

2 \times 10^{6} N m

Explanation:

From work energy theorem.
$W = \frac{1}{2} I (ω_{f}^{2} - ω_{i}^{2})$
$θ = 2 π \times 2 π = 4 π^{2} r a d$
$ω_{i} = 3 \times \frac{2 π}{60} r a d / s$
$\Rightarrow - τ θ = \frac{1}{2} \times \frac{1}{2} m r^{2} (0^{2} - ω_{i}^{2})$
$\Rightarrow τ = \frac{\frac{1}{2} \times \frac{1}{2} \times 2 \times {(4 \times 10^{- 2})}^{2} {(3 \times \frac{2 π}{60})}^{2}}{4 π^{2}}$
$\Rightarrow τ = 2 \times 10^{- 6} N m$

NEET - 2019

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366074 A ceiling fan is rotating around a fixed axle as shown. The direction of angular velocity is along
supporting img

1

+ \hat{j}

2

- \hat{j}

3

+ \hat{k}

4

- \hat{k}

Explanation:

Using right hand thumb rule, direction of angular velocity is along $- v e z$ axis. Correct option is (4).

KCET - 2024

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366075 Two points $A$ and $B$ on a disc have velocities $v_{1}$ and $v_{2}$ respectively, at some moment. Their directions make angles $60^{\circ}$ and $30^{\circ}$ , respectively, with the line of separation as shown in the figure. The angular velocity of disc is $(A$ and $B$ are seperated by $^{'} d^{'}$ )
supporting img

1

\frac{v_{2}}{\sqrt{3} d}

2

\frac{v_{2}}{d}

3

\frac{v_{2} - v_{1}}{d}

4

\frac{\sqrt{3} v_{1}}{d}

Explanation:

For rigid body, separation between two points remain same
$v_{1} \cos 60^{\circ} = v_{2} \cos 30^{\circ}$
$\frac{v_{1}}{2} = \frac{\sqrt{3} v_{2}}{2} \Rightarrow v_{1} = \sqrt{3} v_{2}$
supporting img
$ω_{d i s c} | \frac{v_{2} \sin 30^{\circ} - v_{1} - \sin 60^{\circ}}{d} | = | \frac{\frac{v_{2}}{2} - \frac{\sqrt{3} v_{1}}{2}}{d} |$
$ω_{d i s c} | \frac{v_{2} - \sqrt{3} \times \sqrt{3} v_{2}}{2 d} | = \frac{2 v_{1}}{2 d} = \frac{v_{2}}{d}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366076 Two end points of a rod move with velocities $3 v$ and $v$ perpendicular to the rod and in the same direction, separated by a distance ' $r$ '. Then the angular velocity of the rod is

1

\frac{2 v}{r}

2

\frac{3 v}{r}

3

\frac{4 v}{r}

4

\frac{5 v}{r}

Explanation:

$ω_{r a d} = ω_{p o int} = (\frac{v_{r e l}}{r})$ .
$v_{rel}$ . being the velocity of one point w.r.t other.
' $r$ ' being the distance between them.
$= \frac{3 v - v}{r} = \frac{2 v}{r}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366077 Two particles $A$ and $B$ are situated at a distance $d = 2 m$ apart. Particle $A$ has a velocity of $10 m / s$ at an angle of $60^{\circ}$ and particle $B$ has velocity $v$ at an angle $30^{\circ}$ as shown in the figure. The distance $d$ between $A$ and $B$ is constant. The angular velocity of $B$ with respect to $A$ is:
supporting img

1

\frac{5}{\sqrt{3}} rad / s

2

\frac{10}{\sqrt{3}} rad / s

3

5 \sqrt{3} rad / s

4

10 \sqrt{3} rad / s

Explanation:

If distance between them remains constant, then
$v \cos 30^{\circ} = u \cos 60^{\circ}$
$\Rightarrow v \frac{\sqrt{3}}{2} = 10 \frac{1}{2} \Rightarrow v = \frac{10}{\sqrt{3}}$
$ω = (v \sin 30^{\circ} - u \sin 60^{\circ}) / d$
$= \frac{(\frac{10}{\sqrt{3}}) \times \frac{1}{2} - \frac{10 \times \sqrt{3}}{2}}{2} = - \frac{5}{\sqrt{3}} rad / s .$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366078 A solid cylinder of mass $2 k g$ and radius $4 c m$ is rotating about its axis at the rate of $3 r p m$ . The torque required to stop after $2 π$ revolutions is

1

2 \times 10^{- 6} N m

2

2 \times 10^{- 3} N m

3

12 \times 10^{- 4} N m

4

2 \times 10^{6} N m

Explanation:

From work energy theorem.
$W = \frac{1}{2} I (ω_{f}^{2} - ω_{i}^{2})$
$θ = 2 π \times 2 π = 4 π^{2} r a d$
$ω_{i} = 3 \times \frac{2 π}{60} r a d / s$
$\Rightarrow - τ θ = \frac{1}{2} \times \frac{1}{2} m r^{2} (0^{2} - ω_{i}^{2})$
$\Rightarrow τ = \frac{\frac{1}{2} \times \frac{1}{2} \times 2 \times {(4 \times 10^{- 2})}^{2} {(3 \times \frac{2 π}{60})}^{2}}{4 π^{2}}$
$\Rightarrow τ = 2 \times 10^{- 6} N m$

NEET - 2019

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366074 A ceiling fan is rotating around a fixed axle as shown. The direction of angular velocity is along
supporting img

1

+ \hat{j}

2

- \hat{j}

3

+ \hat{k}

4

- \hat{k}

Explanation:

Using right hand thumb rule, direction of angular velocity is along $- v e z$ axis. Correct option is (4).

KCET - 2024

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366075 Two points $A$ and $B$ on a disc have velocities $v_{1}$ and $v_{2}$ respectively, at some moment. Their directions make angles $60^{\circ}$ and $30^{\circ}$ , respectively, with the line of separation as shown in the figure. The angular velocity of disc is $(A$ and $B$ are seperated by $^{'} d^{'}$ )
supporting img

1

\frac{v_{2}}{\sqrt{3} d}

2

\frac{v_{2}}{d}

3

\frac{v_{2} - v_{1}}{d}

4

\frac{\sqrt{3} v_{1}}{d}

Explanation:

For rigid body, separation between two points remain same
$v_{1} \cos 60^{\circ} = v_{2} \cos 30^{\circ}$
$\frac{v_{1}}{2} = \frac{\sqrt{3} v_{2}}{2} \Rightarrow v_{1} = \sqrt{3} v_{2}$
supporting img
$ω_{d i s c} | \frac{v_{2} \sin 30^{\circ} - v_{1} - \sin 60^{\circ}}{d} | = | \frac{\frac{v_{2}}{2} - \frac{\sqrt{3} v_{1}}{2}}{d} |$
$ω_{d i s c} | \frac{v_{2} - \sqrt{3} \times \sqrt{3} v_{2}}{2 d} | = \frac{2 v_{1}}{2 d} = \frac{v_{2}}{d}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366076 Two end points of a rod move with velocities $3 v$ and $v$ perpendicular to the rod and in the same direction, separated by a distance ' $r$ '. Then the angular velocity of the rod is

1

\frac{2 v}{r}

2

\frac{3 v}{r}

3

\frac{4 v}{r}

4

\frac{5 v}{r}

Explanation:

$ω_{r a d} = ω_{p o int} = (\frac{v_{r e l}}{r})$ .
$v_{rel}$ . being the velocity of one point w.r.t other.
' $r$ ' being the distance between them.
$= \frac{3 v - v}{r} = \frac{2 v}{r}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366077 Two particles $A$ and $B$ are situated at a distance $d = 2 m$ apart. Particle $A$ has a velocity of $10 m / s$ at an angle of $60^{\circ}$ and particle $B$ has velocity $v$ at an angle $30^{\circ}$ as shown in the figure. The distance $d$ between $A$ and $B$ is constant. The angular velocity of $B$ with respect to $A$ is:
supporting img

1

\frac{5}{\sqrt{3}} rad / s

2

\frac{10}{\sqrt{3}} rad / s

3

5 \sqrt{3} rad / s

4

10 \sqrt{3} rad / s

Explanation:

If distance between them remains constant, then
$v \cos 30^{\circ} = u \cos 60^{\circ}$
$\Rightarrow v \frac{\sqrt{3}}{2} = 10 \frac{1}{2} \Rightarrow v = \frac{10}{\sqrt{3}}$
$ω = (v \sin 30^{\circ} - u \sin 60^{\circ}) / d$
$= \frac{(\frac{10}{\sqrt{3}}) \times \frac{1}{2} - \frac{10 \times \sqrt{3}}{2}}{2} = - \frac{5}{\sqrt{3}} rad / s .$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366078 A solid cylinder of mass $2 k g$ and radius $4 c m$ is rotating about its axis at the rate of $3 r p m$ . The torque required to stop after $2 π$ revolutions is

1

2 \times 10^{- 6} N m

2

2 \times 10^{- 3} N m

3

12 \times 10^{- 4} N m

4

2 \times 10^{6} N m

Explanation:

From work energy theorem.
$W = \frac{1}{2} I (ω_{f}^{2} - ω_{i}^{2})$
$θ = 2 π \times 2 π = 4 π^{2} r a d$
$ω_{i} = 3 \times \frac{2 π}{60} r a d / s$
$\Rightarrow - τ θ = \frac{1}{2} \times \frac{1}{2} m r^{2} (0^{2} - ω_{i}^{2})$
$\Rightarrow τ = \frac{\frac{1}{2} \times \frac{1}{2} \times 2 \times {(4 \times 10^{- 2})}^{2} {(3 \times \frac{2 π}{60})}^{2}}{4 π^{2}}$
$\Rightarrow τ = 2 \times 10^{- 6} N m$

NEET - 2019

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366074 A ceiling fan is rotating around a fixed axle as shown. The direction of angular velocity is along
supporting img

1

+ \hat{j}

2

- \hat{j}

3

+ \hat{k}

4

- \hat{k}

Explanation:

Using right hand thumb rule, direction of angular velocity is along $- v e z$ axis. Correct option is (4).

KCET - 2024

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366075 Two points $A$ and $B$ on a disc have velocities $v_{1}$ and $v_{2}$ respectively, at some moment. Their directions make angles $60^{\circ}$ and $30^{\circ}$ , respectively, with the line of separation as shown in the figure. The angular velocity of disc is $(A$ and $B$ are seperated by $^{'} d^{'}$ )
supporting img

1

\frac{v_{2}}{\sqrt{3} d}

2

\frac{v_{2}}{d}

3

\frac{v_{2} - v_{1}}{d}

4

\frac{\sqrt{3} v_{1}}{d}

Explanation:

For rigid body, separation between two points remain same
$v_{1} \cos 60^{\circ} = v_{2} \cos 30^{\circ}$
$\frac{v_{1}}{2} = \frac{\sqrt{3} v_{2}}{2} \Rightarrow v_{1} = \sqrt{3} v_{2}$
supporting img
$ω_{d i s c} | \frac{v_{2} \sin 30^{\circ} - v_{1} - \sin 60^{\circ}}{d} | = | \frac{\frac{v_{2}}{2} - \frac{\sqrt{3} v_{1}}{2}}{d} |$
$ω_{d i s c} | \frac{v_{2} - \sqrt{3} \times \sqrt{3} v_{2}}{2 d} | = \frac{2 v_{1}}{2 d} = \frac{v_{2}}{d}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366076 Two end points of a rod move with velocities $3 v$ and $v$ perpendicular to the rod and in the same direction, separated by a distance ' $r$ '. Then the angular velocity of the rod is

1

\frac{2 v}{r}

2

\frac{3 v}{r}

3

\frac{4 v}{r}

4

\frac{5 v}{r}

Explanation:

$ω_{r a d} = ω_{p o int} = (\frac{v_{r e l}}{r})$ .
$v_{rel}$ . being the velocity of one point w.r.t other.
' $r$ ' being the distance between them.
$= \frac{3 v - v}{r} = \frac{2 v}{r}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366077 Two particles $A$ and $B$ are situated at a distance $d = 2 m$ apart. Particle $A$ has a velocity of $10 m / s$ at an angle of $60^{\circ}$ and particle $B$ has velocity $v$ at an angle $30^{\circ}$ as shown in the figure. The distance $d$ between $A$ and $B$ is constant. The angular velocity of $B$ with respect to $A$ is:
supporting img

1

\frac{5}{\sqrt{3}} rad / s

2

\frac{10}{\sqrt{3}} rad / s

3

5 \sqrt{3} rad / s

4

10 \sqrt{3} rad / s

Explanation:

If distance between them remains constant, then
$v \cos 30^{\circ} = u \cos 60^{\circ}$
$\Rightarrow v \frac{\sqrt{3}}{2} = 10 \frac{1}{2} \Rightarrow v = \frac{10}{\sqrt{3}}$
$ω = (v \sin 30^{\circ} - u \sin 60^{\circ}) / d$
$= \frac{(\frac{10}{\sqrt{3}}) \times \frac{1}{2} - \frac{10 \times \sqrt{3}}{2}}{2} = - \frac{5}{\sqrt{3}} rad / s .$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366078 A solid cylinder of mass $2 k g$ and radius $4 c m$ is rotating about its axis at the rate of $3 r p m$ . The torque required to stop after $2 π$ revolutions is

1

2 \times 10^{- 6} N m

2

2 \times 10^{- 3} N m

3

12 \times 10^{- 4} N m

4

2 \times 10^{6} N m

Explanation:

From work energy theorem.
$W = \frac{1}{2} I (ω_{f}^{2} - ω_{i}^{2})$
$θ = 2 π \times 2 π = 4 π^{2} r a d$
$ω_{i} = 3 \times \frac{2 π}{60} r a d / s$
$\Rightarrow - τ θ = \frac{1}{2} \times \frac{1}{2} m r^{2} (0^{2} - ω_{i}^{2})$
$\Rightarrow τ = \frac{\frac{1}{2} \times \frac{1}{2} \times 2 \times {(4 \times 10^{- 2})}^{2} {(3 \times \frac{2 π}{60})}^{2}}{4 π^{2}}$
$\Rightarrow τ = 2 \times 10^{- 6} N m$

NEET - 2019

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