QBManager

Rotational Motion

150170 A disc of moment of inertia $2 kg - m^{2}$ revolving with $8 rad / s$ is placed on another side of moment of inertia $4 kg - m^{2}$ revolving $4 rad / s$ . What is the angular frequency of composite disc?

1

4 rad / s

2

\frac{3}{16} rad / s

3

\frac{16}{3} rad / s

4

\frac{16}{5} rad / s

Explanation:

C Given,
Disc-1 Disc-2
$I_{1} = 2 {kgm}^{2}$ $I_{2} = 4 {kgm}^{2}$
$ω_{1} = 8 rad / \sec$
$ω_{2} = 4 rad / \sec$
Applying law of conservation of angular momentum-
$I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω$
$2 \times 8 + 4 \times 4 = 6 \times ω$
$ω = \frac{32}{6}$
Hence, $ω = \frac{16}{3} rad / s$

JIPMER-2018

Rotational Motion

150171 Two blocks of masses $10 kg$ and $4 kg$ are connected by a spring of negligible mass and placed on a horizontal frictionless surface. An impulse gives a velocity of $14 m / s$ to the heavier block in the direction of the lighter block. What is the velocity of the centre of mass?

1

5 m / s

2

10 m / s

3

20 m / s

4

30 m / s

Explanation:

B Given, mass of the blocks,
$m_{1} = 10 kg, m_{2} = 4 kg$
and their velocities,
original image
Just after collision,
Velocity of centre of mass,
$v_{cm} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$= \frac{10 \times 14 + 4 \times 0}{10 + 4} = \frac{140}{14} = 10 m / s$

COMEDK 2018

Rotational Motion

150172 The ratio of the radii of gyration of a circular disc and a circular ring of the same radius and same mass about a tangential axis perpendicular to plane of disc or ring is

1

\sqrt{3} : 2

2

2 : 3

3

1 : 2

4

\sqrt{5} : \sqrt{6}

Explanation:

A original image
$∵ I = {mk}^{2}$ $k \to$ radius of gyration)
Moment of inertia of a circular disc about tangential axis perpendicular to plane of disc,
$I_{disc} = \frac{1}{2} {MR}^{2} + {MR}^{2} = \frac{3 {MR}^{2}}{2}$
${MK}_{disc}^{2} = \frac{3 {MR}^{2}}{2}$
$K_{disc} = \sqrt{\frac{3}{2}} R$
Moment of inertia of a ring for a similar axis,
$I_{ring} = {MR}^{2} + {MR}^{2} = 2 {MR}^{2}$
${MK}_{ring}^{2} = 2 {MR}^{2}$
$K_{ring} = \sqrt{2} R$
From eq. (i) $\div$ (ii),
$∴ \frac{K_{disc}}{K_{ring}} = \frac{\sqrt{\frac{3}{2}} R}{\sqrt{2} R}$
$\frac{K_{disc}}{K_{ring}} = \frac{\sqrt{3}}{2}$
$K_{disc} : K_{ring} = \sqrt{3} : 2$

COMEDK 2018

Rotational Motion

150174 The moment of inertia of a uniform horizontal solid cylinder of mass $M$ about an axis passing through its edge and perpendicular to the axis of the cylinder, when its length is 6 times its radius $R$ , is

1

\frac{49 {MR}^{2}}{4}

2

\frac{39 {MR}^{2}}{4}

3

\frac{36 {MR}^{2}}{11}

4

\frac{3 {MR}^{2}}{2}

Explanation:

A
At the moment of inertia of a solid cylinder about perpendicular axis passing through its center
$I_{CM} = \frac{{ML}^{2}}{12} + \frac{{MR}^{2}}{4}$
Moment of inertia about $AB$ , Parallel to axis passing through its center and perpendicular to its plane
$I = I_{CM} + M {(\frac{L}{2})}^{2} = \frac{{ML}^{2}}{12} + \frac{{MR}^{2}}{4} + \frac{{ML}^{2}}{4}$
$I = \frac{M (6 R)^{2}}{12} + \frac{M R^{2}}{4} + \frac{M (6 R)^{2}}{4} (∵ L = 6 R$
$= \frac{36 {MR}^{2}}{12} + \frac{{MR}^{2}}{4} + \frac{36 {MR}^{2}}{4}$
$I = \frac{147}{12} {MR}^{2} = \frac{49}{4} {MR}^{2}$
$I = \frac{49}{4} {MR}^{2}$

CG PET -2018

Rotational Motion

150170 A disc of moment of inertia $2 kg - m^{2}$ revolving with $8 rad / s$ is placed on another side of moment of inertia $4 kg - m^{2}$ revolving $4 rad / s$ . What is the angular frequency of composite disc?

1

4 rad / s

2

\frac{3}{16} rad / s

3

\frac{16}{3} rad / s

4

\frac{16}{5} rad / s

Explanation:

C Given,
Disc-1 Disc-2
$I_{1} = 2 {kgm}^{2}$ $I_{2} = 4 {kgm}^{2}$
$ω_{1} = 8 rad / \sec$
$ω_{2} = 4 rad / \sec$
Applying law of conservation of angular momentum-
$I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω$
$2 \times 8 + 4 \times 4 = 6 \times ω$
$ω = \frac{32}{6}$
Hence, $ω = \frac{16}{3} rad / s$

JIPMER-2018

Rotational Motion

150171 Two blocks of masses $10 kg$ and $4 kg$ are connected by a spring of negligible mass and placed on a horizontal frictionless surface. An impulse gives a velocity of $14 m / s$ to the heavier block in the direction of the lighter block. What is the velocity of the centre of mass?

1

5 m / s

2

10 m / s

3

20 m / s

4

30 m / s

Explanation:

B Given, mass of the blocks,
$m_{1} = 10 kg, m_{2} = 4 kg$
and their velocities,
original image
Just after collision,
Velocity of centre of mass,
$v_{cm} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$= \frac{10 \times 14 + 4 \times 0}{10 + 4} = \frac{140}{14} = 10 m / s$

COMEDK 2018

Rotational Motion

150172 The ratio of the radii of gyration of a circular disc and a circular ring of the same radius and same mass about a tangential axis perpendicular to plane of disc or ring is

1

\sqrt{3} : 2

2

2 : 3

3

1 : 2

4

\sqrt{5} : \sqrt{6}

Explanation:

A original image
$∵ I = {mk}^{2}$ $k \to$ radius of gyration)
Moment of inertia of a circular disc about tangential axis perpendicular to plane of disc,
$I_{disc} = \frac{1}{2} {MR}^{2} + {MR}^{2} = \frac{3 {MR}^{2}}{2}$
${MK}_{disc}^{2} = \frac{3 {MR}^{2}}{2}$
$K_{disc} = \sqrt{\frac{3}{2}} R$
Moment of inertia of a ring for a similar axis,
$I_{ring} = {MR}^{2} + {MR}^{2} = 2 {MR}^{2}$
${MK}_{ring}^{2} = 2 {MR}^{2}$
$K_{ring} = \sqrt{2} R$
From eq. (i) $\div$ (ii),
$∴ \frac{K_{disc}}{K_{ring}} = \frac{\sqrt{\frac{3}{2}} R}{\sqrt{2} R}$
$\frac{K_{disc}}{K_{ring}} = \frac{\sqrt{3}}{2}$
$K_{disc} : K_{ring} = \sqrt{3} : 2$

COMEDK 2018

Rotational Motion

150174 The moment of inertia of a uniform horizontal solid cylinder of mass $M$ about an axis passing through its edge and perpendicular to the axis of the cylinder, when its length is 6 times its radius $R$ , is

1

\frac{49 {MR}^{2}}{4}

2

\frac{39 {MR}^{2}}{4}

3

\frac{36 {MR}^{2}}{11}

4

\frac{3 {MR}^{2}}{2}

Explanation:

A
At the moment of inertia of a solid cylinder about perpendicular axis passing through its center
$I_{CM} = \frac{{ML}^{2}}{12} + \frac{{MR}^{2}}{4}$
Moment of inertia about $AB$ , Parallel to axis passing through its center and perpendicular to its plane
$I = I_{CM} + M {(\frac{L}{2})}^{2} = \frac{{ML}^{2}}{12} + \frac{{MR}^{2}}{4} + \frac{{ML}^{2}}{4}$
$I = \frac{M (6 R)^{2}}{12} + \frac{M R^{2}}{4} + \frac{M (6 R)^{2}}{4} (∵ L = 6 R$
$= \frac{36 {MR}^{2}}{12} + \frac{{MR}^{2}}{4} + \frac{36 {MR}^{2}}{4}$
$I = \frac{147}{12} {MR}^{2} = \frac{49}{4} {MR}^{2}$
$I = \frac{49}{4} {MR}^{2}$

CG PET -2018

Rotational Motion

150170 A disc of moment of inertia $2 kg - m^{2}$ revolving with $8 rad / s$ is placed on another side of moment of inertia $4 kg - m^{2}$ revolving $4 rad / s$ . What is the angular frequency of composite disc?

1

4 rad / s

2

\frac{3}{16} rad / s

3

\frac{16}{3} rad / s

4

\frac{16}{5} rad / s

Explanation:

C Given,
Disc-1 Disc-2
$I_{1} = 2 {kgm}^{2}$ $I_{2} = 4 {kgm}^{2}$
$ω_{1} = 8 rad / \sec$
$ω_{2} = 4 rad / \sec$
Applying law of conservation of angular momentum-
$I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω$
$2 \times 8 + 4 \times 4 = 6 \times ω$
$ω = \frac{32}{6}$
Hence, $ω = \frac{16}{3} rad / s$

JIPMER-2018

Rotational Motion

150171 Two blocks of masses $10 kg$ and $4 kg$ are connected by a spring of negligible mass and placed on a horizontal frictionless surface. An impulse gives a velocity of $14 m / s$ to the heavier block in the direction of the lighter block. What is the velocity of the centre of mass?

1

5 m / s

2

10 m / s

3

20 m / s

4

30 m / s

Explanation:

B Given, mass of the blocks,
$m_{1} = 10 kg, m_{2} = 4 kg$
and their velocities,
original image
Just after collision,
Velocity of centre of mass,
$v_{cm} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$= \frac{10 \times 14 + 4 \times 0}{10 + 4} = \frac{140}{14} = 10 m / s$

COMEDK 2018

Rotational Motion

150172 The ratio of the radii of gyration of a circular disc and a circular ring of the same radius and same mass about a tangential axis perpendicular to plane of disc or ring is

1

\sqrt{3} : 2

2

2 : 3

3

1 : 2

4

\sqrt{5} : \sqrt{6}

Explanation:

A original image
$∵ I = {mk}^{2}$ $k \to$ radius of gyration)
Moment of inertia of a circular disc about tangential axis perpendicular to plane of disc,
$I_{disc} = \frac{1}{2} {MR}^{2} + {MR}^{2} = \frac{3 {MR}^{2}}{2}$
${MK}_{disc}^{2} = \frac{3 {MR}^{2}}{2}$
$K_{disc} = \sqrt{\frac{3}{2}} R$
Moment of inertia of a ring for a similar axis,
$I_{ring} = {MR}^{2} + {MR}^{2} = 2 {MR}^{2}$
${MK}_{ring}^{2} = 2 {MR}^{2}$
$K_{ring} = \sqrt{2} R$
From eq. (i) $\div$ (ii),
$∴ \frac{K_{disc}}{K_{ring}} = \frac{\sqrt{\frac{3}{2}} R}{\sqrt{2} R}$
$\frac{K_{disc}}{K_{ring}} = \frac{\sqrt{3}}{2}$
$K_{disc} : K_{ring} = \sqrt{3} : 2$

COMEDK 2018

Rotational Motion

150174 The moment of inertia of a uniform horizontal solid cylinder of mass $M$ about an axis passing through its edge and perpendicular to the axis of the cylinder, when its length is 6 times its radius $R$ , is

1

\frac{49 {MR}^{2}}{4}

2

\frac{39 {MR}^{2}}{4}

3

\frac{36 {MR}^{2}}{11}

4

\frac{3 {MR}^{2}}{2}

Explanation:

A
At the moment of inertia of a solid cylinder about perpendicular axis passing through its center
$I_{CM} = \frac{{ML}^{2}}{12} + \frac{{MR}^{2}}{4}$
Moment of inertia about $AB$ , Parallel to axis passing through its center and perpendicular to its plane
$I = I_{CM} + M {(\frac{L}{2})}^{2} = \frac{{ML}^{2}}{12} + \frac{{MR}^{2}}{4} + \frac{{ML}^{2}}{4}$
$I = \frac{M (6 R)^{2}}{12} + \frac{M R^{2}}{4} + \frac{M (6 R)^{2}}{4} (∵ L = 6 R$
$= \frac{36 {MR}^{2}}{12} + \frac{{MR}^{2}}{4} + \frac{36 {MR}^{2}}{4}$
$I = \frac{147}{12} {MR}^{2} = \frac{49}{4} {MR}^{2}$
$I = \frac{49}{4} {MR}^{2}$

CG PET -2018

Rotational Motion

150170 A disc of moment of inertia $2 kg - m^{2}$ revolving with $8 rad / s$ is placed on another side of moment of inertia $4 kg - m^{2}$ revolving $4 rad / s$ . What is the angular frequency of composite disc?

1

4 rad / s

2

\frac{3}{16} rad / s

3

\frac{16}{3} rad / s

4

\frac{16}{5} rad / s

Explanation:

C Given,
Disc-1 Disc-2
$I_{1} = 2 {kgm}^{2}$ $I_{2} = 4 {kgm}^{2}$
$ω_{1} = 8 rad / \sec$
$ω_{2} = 4 rad / \sec$
Applying law of conservation of angular momentum-
$I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω$
$2 \times 8 + 4 \times 4 = 6 \times ω$
$ω = \frac{32}{6}$
Hence, $ω = \frac{16}{3} rad / s$

JIPMER-2018

Rotational Motion

150171 Two blocks of masses $10 kg$ and $4 kg$ are connected by a spring of negligible mass and placed on a horizontal frictionless surface. An impulse gives a velocity of $14 m / s$ to the heavier block in the direction of the lighter block. What is the velocity of the centre of mass?

1

5 m / s

2

10 m / s

3

20 m / s

4

30 m / s

Explanation:

B Given, mass of the blocks,
$m_{1} = 10 kg, m_{2} = 4 kg$
and their velocities,
original image
Just after collision,
Velocity of centre of mass,
$v_{cm} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$= \frac{10 \times 14 + 4 \times 0}{10 + 4} = \frac{140}{14} = 10 m / s$

COMEDK 2018

Rotational Motion

150172 The ratio of the radii of gyration of a circular disc and a circular ring of the same radius and same mass about a tangential axis perpendicular to plane of disc or ring is

1

\sqrt{3} : 2

2

2 : 3

3

1 : 2

4

\sqrt{5} : \sqrt{6}

Explanation:

A original image
$∵ I = {mk}^{2}$ $k \to$ radius of gyration)
Moment of inertia of a circular disc about tangential axis perpendicular to plane of disc,
$I_{disc} = \frac{1}{2} {MR}^{2} + {MR}^{2} = \frac{3 {MR}^{2}}{2}$
${MK}_{disc}^{2} = \frac{3 {MR}^{2}}{2}$
$K_{disc} = \sqrt{\frac{3}{2}} R$
Moment of inertia of a ring for a similar axis,
$I_{ring} = {MR}^{2} + {MR}^{2} = 2 {MR}^{2}$
${MK}_{ring}^{2} = 2 {MR}^{2}$
$K_{ring} = \sqrt{2} R$
From eq. (i) $\div$ (ii),
$∴ \frac{K_{disc}}{K_{ring}} = \frac{\sqrt{\frac{3}{2}} R}{\sqrt{2} R}$
$\frac{K_{disc}}{K_{ring}} = \frac{\sqrt{3}}{2}$
$K_{disc} : K_{ring} = \sqrt{3} : 2$

COMEDK 2018

Rotational Motion

150174 The moment of inertia of a uniform horizontal solid cylinder of mass $M$ about an axis passing through its edge and perpendicular to the axis of the cylinder, when its length is 6 times its radius $R$ , is

1

\frac{49 {MR}^{2}}{4}

2

\frac{39 {MR}^{2}}{4}

3

\frac{36 {MR}^{2}}{11}

4

\frac{3 {MR}^{2}}{2}

Explanation:

A
At the moment of inertia of a solid cylinder about perpendicular axis passing through its center
$I_{CM} = \frac{{ML}^{2}}{12} + \frac{{MR}^{2}}{4}$
Moment of inertia about $AB$ , Parallel to axis passing through its center and perpendicular to its plane
$I = I_{CM} + M {(\frac{L}{2})}^{2} = \frac{{ML}^{2}}{12} + \frac{{MR}^{2}}{4} + \frac{{ML}^{2}}{4}$
$I = \frac{M (6 R)^{2}}{12} + \frac{M R^{2}}{4} + \frac{M (6 R)^{2}}{4} (∵ L = 6 R$
$= \frac{36 {MR}^{2}}{12} + \frac{{MR}^{2}}{4} + \frac{36 {MR}^{2}}{4}$
$I = \frac{147}{12} {MR}^{2} = \frac{49}{4} {MR}^{2}$
$I = \frac{49}{4} {MR}^{2}$

CG PET -2018

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