Rotational Motion — Page 2 | JEE Physics

Q11

A wheel is rotating freely with an angular speed

\omega

on a shaft. The moment of inertia of the wheel is I and the moment of inertia of the shaft is negligible. Another wheel of moment of inertia 3I initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is :

A 0

B

{5 \over 6}

C

{1 \over 4}

D

{3 \over 4}

Correct Answer

Option D

Solution

Applying Angular Momentum conservation

I\omega = (I + 3I)\omega '

$\Rightarrow$

\omega ' = {{I\omega } \over {4I}} = {\omega \over 4}

{k_i} = {1 \over 2}I{\omega ^2}

{k_f} = {1 \over 2}(4I){(\omega ')^2}

= 2I{\left( {{\omega \over 4}} \right)^2} = {1 \over 8}I{\omega ^2}

Fractional loss

= {{{K_i} - {K_f}} \over {{K_i}}} = {{{1 \over 2}I{\omega ^2} - {1 \over 8}I{\omega ^2}} \over {{1 \over 2}I{\omega ^2}}}

=

{{{3 \over 8}I{\omega ^2}} \over {{1 \over 2}I{\omega ^2}}} = {3 \over 4}

Q12

Two coaxial discs, having moments of inertia I1 and I1/2, are rotating with respective angular velocities

\omega

1 and

\omega

1/2 , about their common axis. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If Ef and Ei are the final and initial total energies, then (Ef - Ei) is:

A

{{{I_1}\omega _1^2} \over {24}}

B

{{{I_1}\omega _1^2} \over {12}}

C

{3 \over 8}{I_1}\omega _1^2

D

{{{I_1}\omega _1^2} \over {6}}

Correct Answer

Option A

Solution

{E_i} = {1 \over 2}{I_I} \times \omega _1^2 + {1 \over 2}{I \over 2} \times {{\omega _1^2} \over 4}

= {{{I_1}\omega _1^2} \over 2}\left( {{9 \over 8}} \right) = {9 \over {16}}{I_1}\omega _1^2

{I_1}{\omega _1} + {{{I_1}{\omega _1}} \over 4} = {{3{I_1}} \over 2}\omega ;{5 \over 4}{I_1}{\omega _1} = {{3{I_1}} \over 2}\omega

\omega = {5 \over 6}{\omega _1};{E_f} = {1 \over 2} \times {{3{I_1}} \over 2} \times {{25} \over {36}}\omega _1^2

= {{25} \over {48}}{I_1}\omega _1^2

\Rightarrow {E_f} - {E_i} = {I_1}\omega _1^2{{25} \over {49}} - {{ - 2} \over {48}}{I_2}\omega _1^2

= {{25} \over {48}}{I_1}\omega _1^2

\Rightarrow {E_f} - {E_i} = {I_1}\omega _1^2\left( {{{25} \over {48}} - {9 \over {16}}} \right) = {{ - 2} \over {48}}{I_1}\omega _1^2

= {{ - {I_1}\omega _1^2} \over {24}}

Q13

Let

\overrightarrow F

be the force acting on a particle having position vector

\overrightarrow r ,

and

\overrightarrow \tau

be the torque of this force about the origin. Then

A

\overrightarrow {r.} \overrightarrow \tau = 0\,\,

and

\overrightarrow {F.} \overrightarrow \tau \ne 0\,\,

B

\overrightarrow {r.} \vec \tau \ne 0{\mkern 1mu} {\mkern 1mu}

and

\overrightarrow {F.} \overrightarrow \tau = 0\,\,

C

\overrightarrow {r.} \vec \tau \ne 0{\mkern 1mu}

and

\overrightarrow {F.} \overrightarrow \tau \ne 0

D

\overrightarrow {r.} \vec \tau = 0{\mkern 1mu}

and

\overrightarrow {F.} \overrightarrow \tau = 0\,\,

Correct Answer

Option D

Solution

As we know

\overrightarrow \tau = \overrightarrow r \times \overrightarrow F

So the angle between

\overrightarrow \tau

and

\overrightarrow r

is

{90^ \circ }

and the angle between

\overrightarrow t

and

\overrightarrow F

is also

{90^ \circ }.

We also know that the dot product of two vectors which have an angle of

{90^ \circ }

between them is zero. $\therefore$

\overrightarrow {r.} \vec \tau = 0{\mkern 1mu}

and

\overrightarrow {F.} \overrightarrow \tau = 0\,\,

Therefore

(d)

is the correct option.

Q14

A circular disc

X

of radius

R

is made from an iron plate of thickness

t,

and another disc

Y

of radius

4

R

is made from an iron plate of thickness

{t \over 4}.

Then the relation between the moment of inertia

{I_X}

and

{I_Y}

is

A

{I_Y} = 32{I_X}

B

{I_Y} = 16{I_X}

C

{I_Y} = {I_X}

D

{I_Y} = 64{I_X}

Correct Answer

Option D

Solution

We know that density

\left( d \right) = {{mass\left( M \right)} \over {volume\left( V \right)}}

$\therefore$ Mass of disc

M = d \times V = d \times \left( {\pi {R^2} \times t} \right).

The moment of inertia of any disc is

I = {1 \over 2}M{R^2}

$\therefore$

I = {1 \over 2}\left( {d \times \pi {R^2} \times t} \right){R^2} = {{\pi d} \over 2}t \times {R^4}

$\therefore$

{{{I_X}} \over {{I_Y}}} = {{{t_X}R_X^4} \over {{t_Y}R_Y^4}}

= {{t \times {R^4}} \over {{t \over 4} \times {{\left( {4R} \right)}^4}}}

= {1 \over {64}}

Q15

The moment of inertia of a uniform semicircular disc of mass

M

and radius

r

about a line perpendicular to the plane of the disc through the center is

A

{2 \over 5}M{r^2}

B

{1 \over 4}Mr

C

{1 \over 2}M{r^2}

D

M{r^2}

Correct Answer

Option C

Solution

Let mass of the semi circular disc = M Now assume a disc which is combination of two semi circular parts. Let

I

be the moment of inertia of the uniform semicircular disc. So

2I

will be the moment of inertia of the full circular disc and 2M will be the mass.

\Rightarrow 2I = {{2M{r^2}} \over 2}

\Rightarrow I = {{M{r^2}} \over 2}

Q16

An annular ring with inner and outer radii

{R_1}

and

{R_2}

is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring,

{{{F_1}} \over {{F_2}}}\,

is

A

{\left( {{{{R_1}} \over {{R_2}}}} \right)^2}

B

{{{{R_2}} \over {{R_1}}}}

C

{{{{R_1}} \over {{R_2}}}}

D

1

Correct Answer

Option C

Solution

Let the mass of each particle is m. Then force experienced by each particle,

F = m{\omega ^2}R

$\therefore$

{{{F_1}} \over {{F_2}}} = {{m{\omega ^2}{R_1}} \over {m{\omega ^2}{R_2}}}

$\Rightarrow$

{{{F_1}} \over {{F_2}}} = {{{R_1}} \over {{R_2}}}

Q17

Four point masses, each of value

m,

are placed at the corners of a square

ABCD

of side

l

. The moment of inertia of this system about an axis passing through

A

and parallel to

BD

is

A

2m{l^2}

B

\sqrt 3 m{l^2}

C

3m{l^2}

D

m{l^2}

Correct Answer

Option C

Solution

Let

{I_{A}}

is the moment of inertia about an axis passing through A and parallel to BD.

{I_{A}} = M.I\,

due to the point mass at

B+

M.I

due to the point mass at

D+

M.I

due to the point mass at

C.

{I_{A}} = 2 \times m{\left( {{\textstyle{\ell \over {\sqrt 2 }}}} \right)^2} + m{\left( {\sqrt 2 \ell } \right)^2}

= m{\ell ^2} + 2m{\ell ^2} = 3m{\ell ^2}

Q18

A thin circular ring of mass

m

and radius

R

is rotating about its axis with a constant angular velocity

\omega

. Two objects each of mass

M

are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity

\omega ' =

A

{{\omega \left( {m + 2M} \right)} \over m}

B

{{\omega \left( {m - 2M} \right)} \over {\left( {m + 2M} \right)}}

C

{{\omega m} \over {\left( {m + M} \right)}}

D

{{\omega m} \over {\left( {m + 2M} \right)}}

Correct Answer

Option D

Solution

Here angular momentum is conserved. Applying conservation of angular momentum

I'\omega ' = I\omega \,\,

\left( {m{R^2} + 2M{R^2}} \right)\omega \,' = m{R^2}\omega

\Rightarrow \omega \,' = \omega \left[ {{m \over {m + 2M}}} \right]

Q19

From a solid sphere of mass

M

and radius

R

a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its face is:

A

{{4M{R^2}} \over {9\sqrt {3\pi } }}

B

{{4M{R^2}} \over {3\sqrt {3\pi } }}

C

{{M{R^2}} \over {32\sqrt {2\pi } }}

D

{{M{R^2}} \over {16\sqrt {2\pi } }}

Correct Answer

Option A

Solution

1.

Determine the side length of the cube: The cube with the maximum possible volume that can be cut from the sphere will have its diagonal equal to the diameter of the sphere.

Let the side length of the cube be 'a'.

Using Pythagoras in 3D, we have:

a^2 + a^2 + a^2 = (2R)^2

3a^2 = 4R^2

a = \sqrt{\frac{4R^2}{3}} = \frac{2R}{\sqrt{3}}

2. Calculate the mass of the cube: The volume of the cube is

V = a^3 = \left(\frac{2R}{\sqrt{3}}\right)^3 = \frac{8R^3}{3\sqrt{3}}

The density of the sphere (and hence the cube) is

\rho = \frac{M}{\frac{4}{3}\pi R^3}

The mass of the cube is

m = \rho V = \frac{M}{\frac{4}{3}\pi R^3} \cdot \frac{8R^3}{3\sqrt{3}} = \frac{2M}{\sqrt{3}\pi}

3.

Find the moment of inertia of the cube: The moment of inertia of a cube about an axis passing through its center and perpendicular to one of its faces is given by:

I = \frac{1}{6}ma^2

Substituting the values we found:

I = \frac{1}{6} \cdot \frac{2M}{\sqrt{3}\pi} \cdot \left(\frac{2R}{\sqrt{3}}\right)^2

I = \frac{4MR^2}{9\sqrt{3}\pi}

Therefore, the correct answer is Option A:

{{4M{R^2}} \over {9\sqrt {3\pi } }}

Q20

The moment of inertia of a uniform cylinder of length

l

and radius R about its perpendicular bisector is

I

. What is the ratio

{l \over R}

such that the moment of inertia is minimum?

A

{3 \over {\sqrt 2 }}

B

\sqrt {{3 \over 2}}

C

{{\sqrt 3 } \over 2}

D 1

Correct Answer

Option B

Solution

The volume of the cylinder V =

\pi {R^2}l

$\therefore$

{R^2} = {V \over {\pi l}}

We know, moment of inertia of a uniform cylinder of length

l

and radius R about its perpendicular bisector is,

I = {{M{l^2}} \over {12}} + {{M{R^2}} \over 4}

[ Putting

{R^2} = {V \over {\pi l}}

in this equation] $\Rightarrow$

I = {{M{l^2}} \over {12}} + {{MV} \over {4\pi l}}

Here

I

is a function of

l

as M and V are constant.

I

will be maximum or minimum when

{{{dI} \over {dl}}}

= 0.

\Rightarrow {{Ml} \over 6} - {{MV} \over {4\pi {l^2}}} = 0

\Rightarrow {{Ml} \over 6} = {{MV} \over {4\pi {l^2}}}

\Rightarrow {l \over 6} = {{\pi {R^2}l} \over {4\pi {l^2}}}

[ as

{V = \pi {R^2}l}

]

\Rightarrow {{{R^2}} \over {{l^2}}} = {4 \over 6}

\Rightarrow {l \over R} = \sqrt {{3 \over 2}}