QBManager

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366259 A stick of length $\frac{10}{3}$ metre is held vertically with one end on a horizontal floor. It is then allowed to fall on the floor. Assuming that the end of the stick in contact with the floor and it does not slip, the speed of the other end of the rod when it hits the floor is
(Take $g = 10 m / s^{2}$ )

1

5 m / s

2

10 m / s

3

9 m / s

4

3 m / s

Explanation:

Moment of inertia of the rod about the lower end,
$I = I_{c} + M h^{2}$
$= \frac{M L^{2}}{12} + M {(\frac{L}{2})}^{2} = \frac{M L^{2}}{3}$
From conservation of energy,
$\frac{1}{2} I ω^{2} = M g \frac{L}{2}$
$\Rightarrow ω = \sqrt{\frac{M g L}{I}}$
$= \sqrt{\frac{3 g}{L}} (I = \frac{M L^{2}}{3})$
$∴ v = ω L = \sqrt{3 g L}$
$= \sqrt{3 \times 10 \times \frac{10}{3}}$
$= 10 m / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366260 A rope is wound around a hollow cylinder of mass $3 k g$ and radius $40 c m$ . What is the angular acceleration of the cylinder if the rope is pulled with a force of $30 N ?$

1

0.25 r a d s^{- 2}

2

25 r a d s^{- 2}

3

5 m s^{- 2}

4

25 m s^{- 2}

Explanation:

$m = 3 k g, r = 40 c m = 40 \times 10^{- 2} m, F = 30 N$
Moment of inertia of hollow cylinder about its axis
$= m r^{2} = 3 k g \times (0.4)^{2} m^{2} = 0.48 k g m^{2}$
The torque is given by, $t = I α$
where $I =$ moment of inertia, $α =$ angular acceleration
In the given case, $τ = r F$ , as the force is acting perpendicularly to the radial vector.
$∴ α = \frac{τ}{I} = \frac{F r}{m r^{2}} = \frac{F}{m r} = \frac{30}{3 \times 40 \times 10^{- 2}}$
$= \frac{30 \times 100}{3 \times 40}$
$α = 25 r a d s^{- 2}$

NEET - 2017

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366261 The given figure shows a force ${\vec{F}}_{1}$ acting on a particle of a rigid body rotating about a fixed axis. The particle describes a circular path with centre $C$ on the axis. The correct option is
supporting img

1 Small displacement

d s_{1} = r_{1} d θ

2 Small work done by

F_{1} d W_{1} = F_{1} r_{1} d θ \sin α_{1}

3 Torque produced by

F_{1} τ_{1} = r_{1} F_{1} \sin α_{1}

4 All the above

Explanation:

The displacement of the body is the small arc length given by $d s_{1} = r d θ$
The small work done by the force $F_{1}$ is
$\begin{aligned} d W_{1} = F_{1} d s_{1} (\cos ϕ_{1}) = F_{1} (r_{1} d θ) \cos (90 - α_{1}) \\ = F_{1} (r_{1} d θ) \sin α_{1} \end{aligned}$
The magnitude of the torque produced by the force $F_{1}$ is
$\begin{aligned} {\vec{τ}}_{1} = {\vec{r}}_{1} \times {\vec{F}}_{1} \\ τ_{1} = r_{1} F_{1} \sin (α_{1}) \end{aligned}$
So the correct option is (4)

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366262 When a 12000 joule of work is done on a fly wheel, its frequency of rotation increases from $10 H z t o 20 H z$ . The moment of inertia of flywheel about its axis of rotation is $(π^{2} = 10)$

1

1 k g m^{2}

2

2 k g m^{2}

3

1.688 k g m^{2}

4

1.5 k g m^{2}

Explanation:

Given, work done on the flywheel $W =$ $12000 J$ ,
initial frequency $f_{1} = 10 H z$ ,
and final frequency $f_{2} = 20 H z$ .
Initial angular velocity,
$ω_{1} = 2 π f_{1} = 2 π \times 10 = 20 π r a d / s .$
Final angular velocity,
$ω_{2} = 2 π f_{2} = 2 π \times 20 = 40 π r a d / s$
If I is the moment of inertia of the flywheel, then using work - energy theorem,
work done on the flywheel $=$ change in its rotational kinetic energy
$\Rightarrow W = \frac{1}{2} I (ω_{2}^{2} - ω_{1}^{2})$
$\Rightarrow 12000 = \frac{1}{2} I (1600 π^{2} - 400 π^{2})$
$= \frac{1}{2} I (1200 \times 10)$
$\Rightarrow I = \frac{12000 \times 2}{12000} = 2 k g m^{2}$

MHTCET - 2019

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366263 Three objects, $A$ : (a solid sphere), $B$ : (a thin circular disc) and $C$ : (a circular ring), each have the same mass $M$ and radius $R$ . They all spin with the same angular speed $ω$ about their own symmetry axes. The amounts of work $(W)$ required to bring them to rest, would satisfy the relation

1

W_{B} > W_{A} > W_{C}

2

W_{A} > W_{C} > W_{B}

3

W_{A} > W_{B} > W_{C}

4

W_{C} > W_{B} > W_{A}

Explanation:

Rotational $K E K = \frac{1}{2} I ω^{2}$
$\begin{aligned} K_{A} = \frac{1}{2} \times \frac{2}{5} m r^{2} \times ω^{2} \\ K_{B} = \frac{1}{2} \times \frac{1}{2} m r^{2} \times ω^{2} \\ K_{C} = \frac{1}{2} \times m r^{2} \times ω^{2} \\ K_{A} < K_{B} < K_{C} \end{aligned}$
Work done to bring them to rest $= Δ K$ $W_{A} < W_{B} < W_{C}$

NEET - 2018

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366259 A stick of length $\frac{10}{3}$ metre is held vertically with one end on a horizontal floor. It is then allowed to fall on the floor. Assuming that the end of the stick in contact with the floor and it does not slip, the speed of the other end of the rod when it hits the floor is
(Take $g = 10 m / s^{2}$ )

1

5 m / s

2

10 m / s

3

9 m / s

4

3 m / s

Explanation:

Moment of inertia of the rod about the lower end,
$I = I_{c} + M h^{2}$
$= \frac{M L^{2}}{12} + M {(\frac{L}{2})}^{2} = \frac{M L^{2}}{3}$
From conservation of energy,
$\frac{1}{2} I ω^{2} = M g \frac{L}{2}$
$\Rightarrow ω = \sqrt{\frac{M g L}{I}}$
$= \sqrt{\frac{3 g}{L}} (I = \frac{M L^{2}}{3})$
$∴ v = ω L = \sqrt{3 g L}$
$= \sqrt{3 \times 10 \times \frac{10}{3}}$
$= 10 m / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366260 A rope is wound around a hollow cylinder of mass $3 k g$ and radius $40 c m$ . What is the angular acceleration of the cylinder if the rope is pulled with a force of $30 N ?$

1

0.25 r a d s^{- 2}

2

25 r a d s^{- 2}

3

5 m s^{- 2}

4

25 m s^{- 2}

Explanation:

$m = 3 k g, r = 40 c m = 40 \times 10^{- 2} m, F = 30 N$
Moment of inertia of hollow cylinder about its axis
$= m r^{2} = 3 k g \times (0.4)^{2} m^{2} = 0.48 k g m^{2}$
The torque is given by, $t = I α$
where $I =$ moment of inertia, $α =$ angular acceleration
In the given case, $τ = r F$ , as the force is acting perpendicularly to the radial vector.
$∴ α = \frac{τ}{I} = \frac{F r}{m r^{2}} = \frac{F}{m r} = \frac{30}{3 \times 40 \times 10^{- 2}}$
$= \frac{30 \times 100}{3 \times 40}$
$α = 25 r a d s^{- 2}$

NEET - 2017

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366261 The given figure shows a force ${\vec{F}}_{1}$ acting on a particle of a rigid body rotating about a fixed axis. The particle describes a circular path with centre $C$ on the axis. The correct option is
supporting img

1 Small displacement

d s_{1} = r_{1} d θ

2 Small work done by

F_{1} d W_{1} = F_{1} r_{1} d θ \sin α_{1}

3 Torque produced by

F_{1} τ_{1} = r_{1} F_{1} \sin α_{1}

4 All the above

Explanation:

The displacement of the body is the small arc length given by $d s_{1} = r d θ$
The small work done by the force $F_{1}$ is
$\begin{aligned} d W_{1} = F_{1} d s_{1} (\cos ϕ_{1}) = F_{1} (r_{1} d θ) \cos (90 - α_{1}) \\ = F_{1} (r_{1} d θ) \sin α_{1} \end{aligned}$
The magnitude of the torque produced by the force $F_{1}$ is
$\begin{aligned} {\vec{τ}}_{1} = {\vec{r}}_{1} \times {\vec{F}}_{1} \\ τ_{1} = r_{1} F_{1} \sin (α_{1}) \end{aligned}$
So the correct option is (4)

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366262 When a 12000 joule of work is done on a fly wheel, its frequency of rotation increases from $10 H z t o 20 H z$ . The moment of inertia of flywheel about its axis of rotation is $(π^{2} = 10)$

1

1 k g m^{2}

2

2 k g m^{2}

3

1.688 k g m^{2}

4

1.5 k g m^{2}

Explanation:

Given, work done on the flywheel $W =$ $12000 J$ ,
initial frequency $f_{1} = 10 H z$ ,
and final frequency $f_{2} = 20 H z$ .
Initial angular velocity,
$ω_{1} = 2 π f_{1} = 2 π \times 10 = 20 π r a d / s .$
Final angular velocity,
$ω_{2} = 2 π f_{2} = 2 π \times 20 = 40 π r a d / s$
If I is the moment of inertia of the flywheel, then using work - energy theorem,
work done on the flywheel $=$ change in its rotational kinetic energy
$\Rightarrow W = \frac{1}{2} I (ω_{2}^{2} - ω_{1}^{2})$
$\Rightarrow 12000 = \frac{1}{2} I (1600 π^{2} - 400 π^{2})$
$= \frac{1}{2} I (1200 \times 10)$
$\Rightarrow I = \frac{12000 \times 2}{12000} = 2 k g m^{2}$

MHTCET - 2019

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366263 Three objects, $A$ : (a solid sphere), $B$ : (a thin circular disc) and $C$ : (a circular ring), each have the same mass $M$ and radius $R$ . They all spin with the same angular speed $ω$ about their own symmetry axes. The amounts of work $(W)$ required to bring them to rest, would satisfy the relation

1

W_{B} > W_{A} > W_{C}

2

W_{A} > W_{C} > W_{B}

3

W_{A} > W_{B} > W_{C}

4

W_{C} > W_{B} > W_{A}

Explanation:

Rotational $K E K = \frac{1}{2} I ω^{2}$
$\begin{aligned} K_{A} = \frac{1}{2} \times \frac{2}{5} m r^{2} \times ω^{2} \\ K_{B} = \frac{1}{2} \times \frac{1}{2} m r^{2} \times ω^{2} \\ K_{C} = \frac{1}{2} \times m r^{2} \times ω^{2} \\ K_{A} < K_{B} < K_{C} \end{aligned}$
Work done to bring them to rest $= Δ K$ $W_{A} < W_{B} < W_{C}$

NEET - 2018

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366259 A stick of length $\frac{10}{3}$ metre is held vertically with one end on a horizontal floor. It is then allowed to fall on the floor. Assuming that the end of the stick in contact with the floor and it does not slip, the speed of the other end of the rod when it hits the floor is
(Take $g = 10 m / s^{2}$ )

1

5 m / s

2

10 m / s

3

9 m / s

4

3 m / s

Explanation:

Moment of inertia of the rod about the lower end,
$I = I_{c} + M h^{2}$
$= \frac{M L^{2}}{12} + M {(\frac{L}{2})}^{2} = \frac{M L^{2}}{3}$
From conservation of energy,
$\frac{1}{2} I ω^{2} = M g \frac{L}{2}$
$\Rightarrow ω = \sqrt{\frac{M g L}{I}}$
$= \sqrt{\frac{3 g}{L}} (I = \frac{M L^{2}}{3})$
$∴ v = ω L = \sqrt{3 g L}$
$= \sqrt{3 \times 10 \times \frac{10}{3}}$
$= 10 m / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366260 A rope is wound around a hollow cylinder of mass $3 k g$ and radius $40 c m$ . What is the angular acceleration of the cylinder if the rope is pulled with a force of $30 N ?$

1

0.25 r a d s^{- 2}

2

25 r a d s^{- 2}

3

5 m s^{- 2}

4

25 m s^{- 2}

Explanation:

$m = 3 k g, r = 40 c m = 40 \times 10^{- 2} m, F = 30 N$
Moment of inertia of hollow cylinder about its axis
$= m r^{2} = 3 k g \times (0.4)^{2} m^{2} = 0.48 k g m^{2}$
The torque is given by, $t = I α$
where $I =$ moment of inertia, $α =$ angular acceleration
In the given case, $τ = r F$ , as the force is acting perpendicularly to the radial vector.
$∴ α = \frac{τ}{I} = \frac{F r}{m r^{2}} = \frac{F}{m r} = \frac{30}{3 \times 40 \times 10^{- 2}}$
$= \frac{30 \times 100}{3 \times 40}$
$α = 25 r a d s^{- 2}$

NEET - 2017

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366261 The given figure shows a force ${\vec{F}}_{1}$ acting on a particle of a rigid body rotating about a fixed axis. The particle describes a circular path with centre $C$ on the axis. The correct option is
supporting img

1 Small displacement

d s_{1} = r_{1} d θ

2 Small work done by

F_{1} d W_{1} = F_{1} r_{1} d θ \sin α_{1}

3 Torque produced by

F_{1} τ_{1} = r_{1} F_{1} \sin α_{1}

4 All the above

Explanation:

The displacement of the body is the small arc length given by $d s_{1} = r d θ$
The small work done by the force $F_{1}$ is
$\begin{aligned} d W_{1} = F_{1} d s_{1} (\cos ϕ_{1}) = F_{1} (r_{1} d θ) \cos (90 - α_{1}) \\ = F_{1} (r_{1} d θ) \sin α_{1} \end{aligned}$
The magnitude of the torque produced by the force $F_{1}$ is
$\begin{aligned} {\vec{τ}}_{1} = {\vec{r}}_{1} \times {\vec{F}}_{1} \\ τ_{1} = r_{1} F_{1} \sin (α_{1}) \end{aligned}$
So the correct option is (4)

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366262 When a 12000 joule of work is done on a fly wheel, its frequency of rotation increases from $10 H z t o 20 H z$ . The moment of inertia of flywheel about its axis of rotation is $(π^{2} = 10)$

1

1 k g m^{2}

2

2 k g m^{2}

3

1.688 k g m^{2}

4

1.5 k g m^{2}

Explanation:

Given, work done on the flywheel $W =$ $12000 J$ ,
initial frequency $f_{1} = 10 H z$ ,
and final frequency $f_{2} = 20 H z$ .
Initial angular velocity,
$ω_{1} = 2 π f_{1} = 2 π \times 10 = 20 π r a d / s .$
Final angular velocity,
$ω_{2} = 2 π f_{2} = 2 π \times 20 = 40 π r a d / s$
If I is the moment of inertia of the flywheel, then using work - energy theorem,
work done on the flywheel $=$ change in its rotational kinetic energy
$\Rightarrow W = \frac{1}{2} I (ω_{2}^{2} - ω_{1}^{2})$
$\Rightarrow 12000 = \frac{1}{2} I (1600 π^{2} - 400 π^{2})$
$= \frac{1}{2} I (1200 \times 10)$
$\Rightarrow I = \frac{12000 \times 2}{12000} = 2 k g m^{2}$

MHTCET - 2019

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366263 Three objects, $A$ : (a solid sphere), $B$ : (a thin circular disc) and $C$ : (a circular ring), each have the same mass $M$ and radius $R$ . They all spin with the same angular speed $ω$ about their own symmetry axes. The amounts of work $(W)$ required to bring them to rest, would satisfy the relation

1

W_{B} > W_{A} > W_{C}

2

W_{A} > W_{C} > W_{B}

3

W_{A} > W_{B} > W_{C}

4

W_{C} > W_{B} > W_{A}

Explanation:

Rotational $K E K = \frac{1}{2} I ω^{2}$
$\begin{aligned} K_{A} = \frac{1}{2} \times \frac{2}{5} m r^{2} \times ω^{2} \\ K_{B} = \frac{1}{2} \times \frac{1}{2} m r^{2} \times ω^{2} \\ K_{C} = \frac{1}{2} \times m r^{2} \times ω^{2} \\ K_{A} < K_{B} < K_{C} \end{aligned}$
Work done to bring them to rest $= Δ K$ $W_{A} < W_{B} < W_{C}$

NEET - 2018

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366259 A stick of length $\frac{10}{3}$ metre is held vertically with one end on a horizontal floor. It is then allowed to fall on the floor. Assuming that the end of the stick in contact with the floor and it does not slip, the speed of the other end of the rod when it hits the floor is
(Take $g = 10 m / s^{2}$ )

1

5 m / s

2

10 m / s

3

9 m / s

4

3 m / s

Explanation:

Moment of inertia of the rod about the lower end,
$I = I_{c} + M h^{2}$
$= \frac{M L^{2}}{12} + M {(\frac{L}{2})}^{2} = \frac{M L^{2}}{3}$
From conservation of energy,
$\frac{1}{2} I ω^{2} = M g \frac{L}{2}$
$\Rightarrow ω = \sqrt{\frac{M g L}{I}}$
$= \sqrt{\frac{3 g}{L}} (I = \frac{M L^{2}}{3})$
$∴ v = ω L = \sqrt{3 g L}$
$= \sqrt{3 \times 10 \times \frac{10}{3}}$
$= 10 m / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366260 A rope is wound around a hollow cylinder of mass $3 k g$ and radius $40 c m$ . What is the angular acceleration of the cylinder if the rope is pulled with a force of $30 N ?$

1

0.25 r a d s^{- 2}

2

25 r a d s^{- 2}

3

5 m s^{- 2}

4

25 m s^{- 2}

Explanation:

$m = 3 k g, r = 40 c m = 40 \times 10^{- 2} m, F = 30 N$
Moment of inertia of hollow cylinder about its axis
$= m r^{2} = 3 k g \times (0.4)^{2} m^{2} = 0.48 k g m^{2}$
The torque is given by, $t = I α$
where $I =$ moment of inertia, $α =$ angular acceleration
In the given case, $τ = r F$ , as the force is acting perpendicularly to the radial vector.
$∴ α = \frac{τ}{I} = \frac{F r}{m r^{2}} = \frac{F}{m r} = \frac{30}{3 \times 40 \times 10^{- 2}}$
$= \frac{30 \times 100}{3 \times 40}$
$α = 25 r a d s^{- 2}$

NEET - 2017

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366261 The given figure shows a force ${\vec{F}}_{1}$ acting on a particle of a rigid body rotating about a fixed axis. The particle describes a circular path with centre $C$ on the axis. The correct option is
supporting img

1 Small displacement

d s_{1} = r_{1} d θ

2 Small work done by

F_{1} d W_{1} = F_{1} r_{1} d θ \sin α_{1}

3 Torque produced by

F_{1} τ_{1} = r_{1} F_{1} \sin α_{1}

4 All the above

Explanation:

The displacement of the body is the small arc length given by $d s_{1} = r d θ$
The small work done by the force $F_{1}$ is
$\begin{aligned} d W_{1} = F_{1} d s_{1} (\cos ϕ_{1}) = F_{1} (r_{1} d θ) \cos (90 - α_{1}) \\ = F_{1} (r_{1} d θ) \sin α_{1} \end{aligned}$
The magnitude of the torque produced by the force $F_{1}$ is
$\begin{aligned} {\vec{τ}}_{1} = {\vec{r}}_{1} \times {\vec{F}}_{1} \\ τ_{1} = r_{1} F_{1} \sin (α_{1}) \end{aligned}$
So the correct option is (4)

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366262 When a 12000 joule of work is done on a fly wheel, its frequency of rotation increases from $10 H z t o 20 H z$ . The moment of inertia of flywheel about its axis of rotation is $(π^{2} = 10)$

1

1 k g m^{2}

2

2 k g m^{2}

3

1.688 k g m^{2}

4

1.5 k g m^{2}

Explanation:

Given, work done on the flywheel $W =$ $12000 J$ ,
initial frequency $f_{1} = 10 H z$ ,
and final frequency $f_{2} = 20 H z$ .
Initial angular velocity,
$ω_{1} = 2 π f_{1} = 2 π \times 10 = 20 π r a d / s .$
Final angular velocity,
$ω_{2} = 2 π f_{2} = 2 π \times 20 = 40 π r a d / s$
If I is the moment of inertia of the flywheel, then using work - energy theorem,
work done on the flywheel $=$ change in its rotational kinetic energy
$\Rightarrow W = \frac{1}{2} I (ω_{2}^{2} - ω_{1}^{2})$
$\Rightarrow 12000 = \frac{1}{2} I (1600 π^{2} - 400 π^{2})$
$= \frac{1}{2} I (1200 \times 10)$
$\Rightarrow I = \frac{12000 \times 2}{12000} = 2 k g m^{2}$

MHTCET - 2019

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366263 Three objects, $A$ : (a solid sphere), $B$ : (a thin circular disc) and $C$ : (a circular ring), each have the same mass $M$ and radius $R$ . They all spin with the same angular speed $ω$ about their own symmetry axes. The amounts of work $(W)$ required to bring them to rest, would satisfy the relation

1

W_{B} > W_{A} > W_{C}

2

W_{A} > W_{C} > W_{B}

3

W_{A} > W_{B} > W_{C}

4

W_{C} > W_{B} > W_{A}

Explanation:

Rotational $K E K = \frac{1}{2} I ω^{2}$
$\begin{aligned} K_{A} = \frac{1}{2} \times \frac{2}{5} m r^{2} \times ω^{2} \\ K_{B} = \frac{1}{2} \times \frac{1}{2} m r^{2} \times ω^{2} \\ K_{C} = \frac{1}{2} \times m r^{2} \times ω^{2} \\ K_{A} < K_{B} < K_{C} \end{aligned}$
Work done to bring them to rest $= Δ K$ $W_{A} < W_{B} < W_{C}$

NEET - 2018

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366259 A stick of length $\frac{10}{3}$ metre is held vertically with one end on a horizontal floor. It is then allowed to fall on the floor. Assuming that the end of the stick in contact with the floor and it does not slip, the speed of the other end of the rod when it hits the floor is
(Take $g = 10 m / s^{2}$ )

1

5 m / s

2

10 m / s

3

9 m / s

4

3 m / s

Explanation:

Moment of inertia of the rod about the lower end,
$I = I_{c} + M h^{2}$
$= \frac{M L^{2}}{12} + M {(\frac{L}{2})}^{2} = \frac{M L^{2}}{3}$
From conservation of energy,
$\frac{1}{2} I ω^{2} = M g \frac{L}{2}$
$\Rightarrow ω = \sqrt{\frac{M g L}{I}}$
$= \sqrt{\frac{3 g}{L}} (I = \frac{M L^{2}}{3})$
$∴ v = ω L = \sqrt{3 g L}$
$= \sqrt{3 \times 10 \times \frac{10}{3}}$
$= 10 m / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366260 A rope is wound around a hollow cylinder of mass $3 k g$ and radius $40 c m$ . What is the angular acceleration of the cylinder if the rope is pulled with a force of $30 N ?$

1

0.25 r a d s^{- 2}

2

25 r a d s^{- 2}

3

5 m s^{- 2}

4

25 m s^{- 2}

Explanation:

$m = 3 k g, r = 40 c m = 40 \times 10^{- 2} m, F = 30 N$
Moment of inertia of hollow cylinder about its axis
$= m r^{2} = 3 k g \times (0.4)^{2} m^{2} = 0.48 k g m^{2}$
The torque is given by, $t = I α$
where $I =$ moment of inertia, $α =$ angular acceleration
In the given case, $τ = r F$ , as the force is acting perpendicularly to the radial vector.
$∴ α = \frac{τ}{I} = \frac{F r}{m r^{2}} = \frac{F}{m r} = \frac{30}{3 \times 40 \times 10^{- 2}}$
$= \frac{30 \times 100}{3 \times 40}$
$α = 25 r a d s^{- 2}$

NEET - 2017

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366261 The given figure shows a force ${\vec{F}}_{1}$ acting on a particle of a rigid body rotating about a fixed axis. The particle describes a circular path with centre $C$ on the axis. The correct option is
supporting img

1 Small displacement

d s_{1} = r_{1} d θ

2 Small work done by

F_{1} d W_{1} = F_{1} r_{1} d θ \sin α_{1}

3 Torque produced by

F_{1} τ_{1} = r_{1} F_{1} \sin α_{1}

4 All the above

Explanation:

The displacement of the body is the small arc length given by $d s_{1} = r d θ$
The small work done by the force $F_{1}$ is
$\begin{aligned} d W_{1} = F_{1} d s_{1} (\cos ϕ_{1}) = F_{1} (r_{1} d θ) \cos (90 - α_{1}) \\ = F_{1} (r_{1} d θ) \sin α_{1} \end{aligned}$
The magnitude of the torque produced by the force $F_{1}$ is
$\begin{aligned} {\vec{τ}}_{1} = {\vec{r}}_{1} \times {\vec{F}}_{1} \\ τ_{1} = r_{1} F_{1} \sin (α_{1}) \end{aligned}$
So the correct option is (4)

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366262 When a 12000 joule of work is done on a fly wheel, its frequency of rotation increases from $10 H z t o 20 H z$ . The moment of inertia of flywheel about its axis of rotation is $(π^{2} = 10)$

1

1 k g m^{2}

2

2 k g m^{2}

3

1.688 k g m^{2}

4

1.5 k g m^{2}

Explanation:

Given, work done on the flywheel $W =$ $12000 J$ ,
initial frequency $f_{1} = 10 H z$ ,
and final frequency $f_{2} = 20 H z$ .
Initial angular velocity,
$ω_{1} = 2 π f_{1} = 2 π \times 10 = 20 π r a d / s .$
Final angular velocity,
$ω_{2} = 2 π f_{2} = 2 π \times 20 = 40 π r a d / s$
If I is the moment of inertia of the flywheel, then using work - energy theorem,
work done on the flywheel $=$ change in its rotational kinetic energy
$\Rightarrow W = \frac{1}{2} I (ω_{2}^{2} - ω_{1}^{2})$
$\Rightarrow 12000 = \frac{1}{2} I (1600 π^{2} - 400 π^{2})$
$= \frac{1}{2} I (1200 \times 10)$
$\Rightarrow I = \frac{12000 \times 2}{12000} = 2 k g m^{2}$

MHTCET - 2019

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366263 Three objects, $A$ : (a solid sphere), $B$ : (a thin circular disc) and $C$ : (a circular ring), each have the same mass $M$ and radius $R$ . They all spin with the same angular speed $ω$ about their own symmetry axes. The amounts of work $(W)$ required to bring them to rest, would satisfy the relation

1

W_{B} > W_{A} > W_{C}

2

W_{A} > W_{C} > W_{B}

3

W_{A} > W_{B} > W_{C}

4

W_{C} > W_{B} > W_{A}

Explanation:

Rotational $K E K = \frac{1}{2} I ω^{2}$
$\begin{aligned} K_{A} = \frac{1}{2} \times \frac{2}{5} m r^{2} \times ω^{2} \\ K_{B} = \frac{1}{2} \times \frac{1}{2} m r^{2} \times ω^{2} \\ K_{C} = \frac{1}{2} \times m r^{2} \times ω^{2} \\ K_{A} < K_{B} < K_{C} \end{aligned}$
Work done to bring them to rest $= Δ K$ $W_{A} < W_{B} < W_{C}$

NEET - 2018