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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366276 If ' $I$ ' is the moment of inertia of a body and ' $ω$ ' is its angular velocity then its rotational kinetic energy is

1

\frac{1}{2} I ω^{2}

2

\frac{1}{2} I^{2} ω

3

\frac{1}{2} I ω

4

\frac{1}{2} I^{2} ω^{2}

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366277 Two uniform discs of equal mass but unequal radii are mounted on fixed horizontal axiles. Light strings are wrapped on each of the discs. The strings are pulled by constant equal forces $F$ for same amount of time as shown in the figure.
supporting img

1

L_{2} < L_{1}

2

L_{1} = L_{2}

3

K_{1} = K_{2}

4

K_{1} > K_{2}

Explanation:

$τ Δ t = Δ L$
$F r Δ t = L$
As $F & Δ t$ are same
$L_{2} > L_{1} (r_{2} > r_{1})$
$K = \frac{L^{2}}{2 I} = \frac{F^{2} r^{2} Δ t^{2}}{2 \frac{m r^{2}}{2}} = \frac{F^{2} Δ t^{2}}{m}$
$\Rightarrow K_{1} = K_{2}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366278 The moment of inertia of two freely rotating bodies $A$ and $B$ are $I_{A}$ and $I_{B}$ respectively. $I_{A} > I_{B}$ and their angular momenta are equal. If $K_{A}$ and $K_{B}$ are their kinetic energies, then

1

K_{A} = 2 K_{B}

2

K_{A} > K_{B}

3

K_{A} < K_{B}

4

K_{A} = K_{B}

Explanation:

Kinetic energy $K = \frac{L^{2}}{2 I}$
If angular momenta are equal then $K \propto \frac{1}{I}$
If $I_{A} > I_{B}$ and $K_{A} < K_{B}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366279 A cord is wound round the circumference of wheel of radius $r$ . The axis of the wheel is horizontal and moment of inertia about it is $I . A$ weight $m g$ is attached to the end of the cord and falls from the rest. After falling through a distance $h$ , the angular velocity of the wheel will be

1

\sqrt{2 g h}

2

\sqrt{\frac{2 g h}{I + m r}}

3

{[\frac{2 m g h}{I + m r^{2}}]}^{1 / 2}

4

{[\frac{2 m g h}{I + 2 m r^{2}}]}^{1 / 2}

Explanation:

From the conservation of energy
$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
As the cord does not slip $v = r ω$
$\Rightarrow ω = \sqrt{\frac{2 m g h}{m r^{2} + m k^{2}}}$
$= \sqrt{\frac{2 m g h}{m r^{2} + I}} = \sqrt{\frac{2 m g h}{I + m r^{2}}}$ .

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366280 A uniform rod of length $l$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$ . The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $θ$ , its angular velocity $ω$ is given by
supporting img

1

\sqrt{(\frac{6 g}{l})} \sin \frac{θ}{2}

2

\sqrt{(\frac{6 g}{l})} \cos \frac{θ}{2}

3

\sqrt{(\frac{6 g}{l})} \sin θ

4

\sqrt{(\frac{6 g}{l})} \cos θ

Explanation:

When the rod rotates through an angle $θ$ , the centre of gravity falls through a distance $h$ .
supporting img
From $Δ B C G^{'}$
$\cos θ = \frac{(l / 2) - h}{l / 2}$
or $h = \frac{l}{2} (1 - \cos θ)$
Decrease in $P E = m g \frac{l}{2} (1 - \cos θ) (1)$
The decrease in $P E$ is equal to the kinetic energy of rotation
$(K E)_{rotational} = \frac{1}{2} I ω^{2} = \frac{1}{2} (\frac{m l^{2}}{3}) ω^{2} (2)$
$[∵ I = \frac{m l^{2}}{3}]$
From Eqs. (1) and (2), we get
$\frac{1}{2} (\frac{m l^{2}}{3}) ω^{2} = m g \frac{l}{2} (1 - \cos θ)$
$\Rightarrow ω = \sqrt{\frac{6 g}{l}} \sin \frac{θ}{2}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366276 If ' $I$ ' is the moment of inertia of a body and ' $ω$ ' is its angular velocity then its rotational kinetic energy is

1

\frac{1}{2} I ω^{2}

2

\frac{1}{2} I^{2} ω

3

\frac{1}{2} I ω

4

\frac{1}{2} I^{2} ω^{2}

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366277 Two uniform discs of equal mass but unequal radii are mounted on fixed horizontal axiles. Light strings are wrapped on each of the discs. The strings are pulled by constant equal forces $F$ for same amount of time as shown in the figure.
supporting img

1

L_{2} < L_{1}

2

L_{1} = L_{2}

3

K_{1} = K_{2}

4

K_{1} > K_{2}

Explanation:

$τ Δ t = Δ L$
$F r Δ t = L$
As $F & Δ t$ are same
$L_{2} > L_{1} (r_{2} > r_{1})$
$K = \frac{L^{2}}{2 I} = \frac{F^{2} r^{2} Δ t^{2}}{2 \frac{m r^{2}}{2}} = \frac{F^{2} Δ t^{2}}{m}$
$\Rightarrow K_{1} = K_{2}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366278 The moment of inertia of two freely rotating bodies $A$ and $B$ are $I_{A}$ and $I_{B}$ respectively. $I_{A} > I_{B}$ and their angular momenta are equal. If $K_{A}$ and $K_{B}$ are their kinetic energies, then

1

K_{A} = 2 K_{B}

2

K_{A} > K_{B}

3

K_{A} < K_{B}

4

K_{A} = K_{B}

Explanation:

Kinetic energy $K = \frac{L^{2}}{2 I}$
If angular momenta are equal then $K \propto \frac{1}{I}$
If $I_{A} > I_{B}$ and $K_{A} < K_{B}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366279 A cord is wound round the circumference of wheel of radius $r$ . The axis of the wheel is horizontal and moment of inertia about it is $I . A$ weight $m g$ is attached to the end of the cord and falls from the rest. After falling through a distance $h$ , the angular velocity of the wheel will be

1

\sqrt{2 g h}

2

\sqrt{\frac{2 g h}{I + m r}}

3

{[\frac{2 m g h}{I + m r^{2}}]}^{1 / 2}

4

{[\frac{2 m g h}{I + 2 m r^{2}}]}^{1 / 2}

Explanation:

From the conservation of energy
$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
As the cord does not slip $v = r ω$
$\Rightarrow ω = \sqrt{\frac{2 m g h}{m r^{2} + m k^{2}}}$
$= \sqrt{\frac{2 m g h}{m r^{2} + I}} = \sqrt{\frac{2 m g h}{I + m r^{2}}}$ .

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366280 A uniform rod of length $l$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$ . The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $θ$ , its angular velocity $ω$ is given by
supporting img

1

\sqrt{(\frac{6 g}{l})} \sin \frac{θ}{2}

2

\sqrt{(\frac{6 g}{l})} \cos \frac{θ}{2}

3

\sqrt{(\frac{6 g}{l})} \sin θ

4

\sqrt{(\frac{6 g}{l})} \cos θ

Explanation:

When the rod rotates through an angle $θ$ , the centre of gravity falls through a distance $h$ .
supporting img
From $Δ B C G^{'}$
$\cos θ = \frac{(l / 2) - h}{l / 2}$
or $h = \frac{l}{2} (1 - \cos θ)$
Decrease in $P E = m g \frac{l}{2} (1 - \cos θ) (1)$
The decrease in $P E$ is equal to the kinetic energy of rotation
$(K E)_{rotational} = \frac{1}{2} I ω^{2} = \frac{1}{2} (\frac{m l^{2}}{3}) ω^{2} (2)$
$[∵ I = \frac{m l^{2}}{3}]$
From Eqs. (1) and (2), we get
$\frac{1}{2} (\frac{m l^{2}}{3}) ω^{2} = m g \frac{l}{2} (1 - \cos θ)$
$\Rightarrow ω = \sqrt{\frac{6 g}{l}} \sin \frac{θ}{2}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366276 If ' $I$ ' is the moment of inertia of a body and ' $ω$ ' is its angular velocity then its rotational kinetic energy is

1

\frac{1}{2} I ω^{2}

2

\frac{1}{2} I^{2} ω

3

\frac{1}{2} I ω

4

\frac{1}{2} I^{2} ω^{2}

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366277 Two uniform discs of equal mass but unequal radii are mounted on fixed horizontal axiles. Light strings are wrapped on each of the discs. The strings are pulled by constant equal forces $F$ for same amount of time as shown in the figure.
supporting img

1

L_{2} < L_{1}

2

L_{1} = L_{2}

3

K_{1} = K_{2}

4

K_{1} > K_{2}

Explanation:

$τ Δ t = Δ L$
$F r Δ t = L$
As $F & Δ t$ are same
$L_{2} > L_{1} (r_{2} > r_{1})$
$K = \frac{L^{2}}{2 I} = \frac{F^{2} r^{2} Δ t^{2}}{2 \frac{m r^{2}}{2}} = \frac{F^{2} Δ t^{2}}{m}$
$\Rightarrow K_{1} = K_{2}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366278 The moment of inertia of two freely rotating bodies $A$ and $B$ are $I_{A}$ and $I_{B}$ respectively. $I_{A} > I_{B}$ and their angular momenta are equal. If $K_{A}$ and $K_{B}$ are their kinetic energies, then

1

K_{A} = 2 K_{B}

2

K_{A} > K_{B}

3

K_{A} < K_{B}

4

K_{A} = K_{B}

Explanation:

Kinetic energy $K = \frac{L^{2}}{2 I}$
If angular momenta are equal then $K \propto \frac{1}{I}$
If $I_{A} > I_{B}$ and $K_{A} < K_{B}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366279 A cord is wound round the circumference of wheel of radius $r$ . The axis of the wheel is horizontal and moment of inertia about it is $I . A$ weight $m g$ is attached to the end of the cord and falls from the rest. After falling through a distance $h$ , the angular velocity of the wheel will be

1

\sqrt{2 g h}

2

\sqrt{\frac{2 g h}{I + m r}}

3

{[\frac{2 m g h}{I + m r^{2}}]}^{1 / 2}

4

{[\frac{2 m g h}{I + 2 m r^{2}}]}^{1 / 2}

Explanation:

From the conservation of energy
$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
As the cord does not slip $v = r ω$
$\Rightarrow ω = \sqrt{\frac{2 m g h}{m r^{2} + m k^{2}}}$
$= \sqrt{\frac{2 m g h}{m r^{2} + I}} = \sqrt{\frac{2 m g h}{I + m r^{2}}}$ .

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366280 A uniform rod of length $l$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$ . The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $θ$ , its angular velocity $ω$ is given by
supporting img

1

\sqrt{(\frac{6 g}{l})} \sin \frac{θ}{2}

2

\sqrt{(\frac{6 g}{l})} \cos \frac{θ}{2}

3

\sqrt{(\frac{6 g}{l})} \sin θ

4

\sqrt{(\frac{6 g}{l})} \cos θ

Explanation:

When the rod rotates through an angle $θ$ , the centre of gravity falls through a distance $h$ .
supporting img
From $Δ B C G^{'}$
$\cos θ = \frac{(l / 2) - h}{l / 2}$
or $h = \frac{l}{2} (1 - \cos θ)$
Decrease in $P E = m g \frac{l}{2} (1 - \cos θ) (1)$
The decrease in $P E$ is equal to the kinetic energy of rotation
$(K E)_{rotational} = \frac{1}{2} I ω^{2} = \frac{1}{2} (\frac{m l^{2}}{3}) ω^{2} (2)$
$[∵ I = \frac{m l^{2}}{3}]$
From Eqs. (1) and (2), we get
$\frac{1}{2} (\frac{m l^{2}}{3}) ω^{2} = m g \frac{l}{2} (1 - \cos θ)$
$\Rightarrow ω = \sqrt{\frac{6 g}{l}} \sin \frac{θ}{2}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366276 If ' $I$ ' is the moment of inertia of a body and ' $ω$ ' is its angular velocity then its rotational kinetic energy is

1

\frac{1}{2} I ω^{2}

2

\frac{1}{2} I^{2} ω

3

\frac{1}{2} I ω

4

\frac{1}{2} I^{2} ω^{2}

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366277 Two uniform discs of equal mass but unequal radii are mounted on fixed horizontal axiles. Light strings are wrapped on each of the discs. The strings are pulled by constant equal forces $F$ for same amount of time as shown in the figure.
supporting img

1

L_{2} < L_{1}

2

L_{1} = L_{2}

3

K_{1} = K_{2}

4

K_{1} > K_{2}

Explanation:

$τ Δ t = Δ L$
$F r Δ t = L$
As $F & Δ t$ are same
$L_{2} > L_{1} (r_{2} > r_{1})$
$K = \frac{L^{2}}{2 I} = \frac{F^{2} r^{2} Δ t^{2}}{2 \frac{m r^{2}}{2}} = \frac{F^{2} Δ t^{2}}{m}$
$\Rightarrow K_{1} = K_{2}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366278 The moment of inertia of two freely rotating bodies $A$ and $B$ are $I_{A}$ and $I_{B}$ respectively. $I_{A} > I_{B}$ and their angular momenta are equal. If $K_{A}$ and $K_{B}$ are their kinetic energies, then

1

K_{A} = 2 K_{B}

2

K_{A} > K_{B}

3

K_{A} < K_{B}

4

K_{A} = K_{B}

Explanation:

Kinetic energy $K = \frac{L^{2}}{2 I}$
If angular momenta are equal then $K \propto \frac{1}{I}$
If $I_{A} > I_{B}$ and $K_{A} < K_{B}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366279 A cord is wound round the circumference of wheel of radius $r$ . The axis of the wheel is horizontal and moment of inertia about it is $I . A$ weight $m g$ is attached to the end of the cord and falls from the rest. After falling through a distance $h$ , the angular velocity of the wheel will be

1

\sqrt{2 g h}

2

\sqrt{\frac{2 g h}{I + m r}}

3

{[\frac{2 m g h}{I + m r^{2}}]}^{1 / 2}

4

{[\frac{2 m g h}{I + 2 m r^{2}}]}^{1 / 2}

Explanation:

From the conservation of energy
$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
As the cord does not slip $v = r ω$
$\Rightarrow ω = \sqrt{\frac{2 m g h}{m r^{2} + m k^{2}}}$
$= \sqrt{\frac{2 m g h}{m r^{2} + I}} = \sqrt{\frac{2 m g h}{I + m r^{2}}}$ .

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366280 A uniform rod of length $l$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$ . The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $θ$ , its angular velocity $ω$ is given by
supporting img

1

\sqrt{(\frac{6 g}{l})} \sin \frac{θ}{2}

2

\sqrt{(\frac{6 g}{l})} \cos \frac{θ}{2}

3

\sqrt{(\frac{6 g}{l})} \sin θ

4

\sqrt{(\frac{6 g}{l})} \cos θ

Explanation:

When the rod rotates through an angle $θ$ , the centre of gravity falls through a distance $h$ .
supporting img
From $Δ B C G^{'}$
$\cos θ = \frac{(l / 2) - h}{l / 2}$
or $h = \frac{l}{2} (1 - \cos θ)$
Decrease in $P E = m g \frac{l}{2} (1 - \cos θ) (1)$
The decrease in $P E$ is equal to the kinetic energy of rotation
$(K E)_{rotational} = \frac{1}{2} I ω^{2} = \frac{1}{2} (\frac{m l^{2}}{3}) ω^{2} (2)$
$[∵ I = \frac{m l^{2}}{3}]$
From Eqs. (1) and (2), we get
$\frac{1}{2} (\frac{m l^{2}}{3}) ω^{2} = m g \frac{l}{2} (1 - \cos θ)$
$\Rightarrow ω = \sqrt{\frac{6 g}{l}} \sin \frac{θ}{2}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366276 If ' $I$ ' is the moment of inertia of a body and ' $ω$ ' is its angular velocity then its rotational kinetic energy is

1

\frac{1}{2} I ω^{2}

2

\frac{1}{2} I^{2} ω

3

\frac{1}{2} I ω

4

\frac{1}{2} I^{2} ω^{2}

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366277 Two uniform discs of equal mass but unequal radii are mounted on fixed horizontal axiles. Light strings are wrapped on each of the discs. The strings are pulled by constant equal forces $F$ for same amount of time as shown in the figure.
supporting img

1

L_{2} < L_{1}

2

L_{1} = L_{2}

3

K_{1} = K_{2}

4

K_{1} > K_{2}

Explanation:

$τ Δ t = Δ L$
$F r Δ t = L$
As $F & Δ t$ are same
$L_{2} > L_{1} (r_{2} > r_{1})$
$K = \frac{L^{2}}{2 I} = \frac{F^{2} r^{2} Δ t^{2}}{2 \frac{m r^{2}}{2}} = \frac{F^{2} Δ t^{2}}{m}$
$\Rightarrow K_{1} = K_{2}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366278 The moment of inertia of two freely rotating bodies $A$ and $B$ are $I_{A}$ and $I_{B}$ respectively. $I_{A} > I_{B}$ and their angular momenta are equal. If $K_{A}$ and $K_{B}$ are their kinetic energies, then

1

K_{A} = 2 K_{B}

2

K_{A} > K_{B}

3

K_{A} < K_{B}

4

K_{A} = K_{B}

Explanation:

Kinetic energy $K = \frac{L^{2}}{2 I}$
If angular momenta are equal then $K \propto \frac{1}{I}$
If $I_{A} > I_{B}$ and $K_{A} < K_{B}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366279 A cord is wound round the circumference of wheel of radius $r$ . The axis of the wheel is horizontal and moment of inertia about it is $I . A$ weight $m g$ is attached to the end of the cord and falls from the rest. After falling through a distance $h$ , the angular velocity of the wheel will be

1

\sqrt{2 g h}

2

\sqrt{\frac{2 g h}{I + m r}}

3

{[\frac{2 m g h}{I + m r^{2}}]}^{1 / 2}

4

{[\frac{2 m g h}{I + 2 m r^{2}}]}^{1 / 2}

Explanation:

From the conservation of energy
$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
As the cord does not slip $v = r ω$
$\Rightarrow ω = \sqrt{\frac{2 m g h}{m r^{2} + m k^{2}}}$
$= \sqrt{\frac{2 m g h}{m r^{2} + I}} = \sqrt{\frac{2 m g h}{I + m r^{2}}}$ .

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366280 A uniform rod of length $l$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$ . The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $θ$ , its angular velocity $ω$ is given by
supporting img

1

\sqrt{(\frac{6 g}{l})} \sin \frac{θ}{2}

2

\sqrt{(\frac{6 g}{l})} \cos \frac{θ}{2}

3

\sqrt{(\frac{6 g}{l})} \sin θ

4

\sqrt{(\frac{6 g}{l})} \cos θ

Explanation:

When the rod rotates through an angle $θ$ , the centre of gravity falls through a distance $h$ .
supporting img
From $Δ B C G^{'}$
$\cos θ = \frac{(l / 2) - h}{l / 2}$
or $h = \frac{l}{2} (1 - \cos θ)$
Decrease in $P E = m g \frac{l}{2} (1 - \cos θ) (1)$
The decrease in $P E$ is equal to the kinetic energy of rotation
$(K E)_{rotational} = \frac{1}{2} I ω^{2} = \frac{1}{2} (\frac{m l^{2}}{3}) ω^{2} (2)$
$[∵ I = \frac{m l^{2}}{3}]$
From Eqs. (1) and (2), we get
$\frac{1}{2} (\frac{m l^{2}}{3}) ω^{2} = m g \frac{l}{2} (1 - \cos θ)$
$\Rightarrow ω = \sqrt{\frac{6 g}{l}} \sin \frac{θ}{2}$