Rotational Motion Questions — JEE Physics

Q1

Mass per unit area of a circular disc of radius

a

depends on the distance r from its centre as

\sigma \left( r \right)

= A + Br . The moment of inertia of the disc about the axis, perpendicular to the plane and assing through its centre is:

A

2\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)

B

\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)

C

2\pi {a^4}\left( {{{aA} \over 4} + {B \over 5}} \right)

D

2\pi {a^4}\left( {{A \over 4} + {B \over 5}} \right)

Correct Answer

Option A

Solution

dI = dm(r2) and dm = $\sigma$ 2 $\pi$ rdr $\therefore$ dI = $\sigma$ 2 $\pi$ rdr(r2) = $\sigma$ 2 $\pi$ r3dr Given

\sigma \left( r \right)

= A + Br $\therefore$ dI = 2 $\pi$ (A + Br)r3dr

\int {dI = \int\limits_0^a {2\pi {r^3}\left( {A + Br} \right)dr} }

$\Rightarrow$ I =

2\pi \left[ {{{A{r^4}} \over 4} + {{B{r^5}} \over 5}} \right]

=

2\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)

Q2

A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which on of the following will not be affected ?

A Angular velocity

B Angular momentum

C Moment of inertia

D Rotational kinetic energy

Correct Answer

Option B

Solution

Solid sphere is rotating in free space that means no external torque is operating on the sphere.

Angular momentum will remain the same since external torque is zero.

Q3

A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is :

A 8.75 × 10–3 J

B 1.13 × 10–3 J

C 8.75 × 10–4 J

D 6.25 × 10–4 J

Correct Answer

Option C

Solution

K.E =

{1 \over 2}m{V^2} + {1 \over 2}{I_{cm}}{\omega ^2}

=

{1 \over 2}m{V^2} + {1 \over 2} \times {2 \over 5}m{R^2} \times {{{V^2}} \over {{R^2}}}

=

{1 \over 2}m{V^2} + {1 \over 5}m{V^2}

=

{7 \over {10}}m{V^2}

=

{7 \over {10}} \times 0.5 \times 25 \times {10^{ - 4}}

=

{{35} \over 4} \times {10^{ - 4}}

= 8.75 × 10–4 J

Q4

If the angular momentum of a planet of mass m, moving around the Sun in a circular orbit is L, about the center of the Sun, its areal velocity is :

A

{L \over m}

B

{4L \over m}

C

{L \over 2m}

D

{2L \over m}

Correct Answer

Option C

Solution

dA =

{1 \over 2}

r2d $\theta$ $\therefore$

{{dA} \over {dt}} = {1 \over 2}{r^2}{{d\theta } \over {dt}}

$\Rightarrow$

{{dA} \over {dt}} = {1 \over 2}{r^2}\omega

. . . . . (1) We know, angular momentum, L =

mvr

=

m\left( {\omega r} \right)r

= mr2 $\omega$ $\therefore$ $\omega$ =

{L \over {m{r^2}}}

. . . . . (2) Put value of $\omega$ in equation(1),

{{dA} \over {dt}}

=

{1 \over 2}{r^2}

(

{L \over {m{r^2}}}

) =

{L \over {2m}}

Q5

A uniform solid cylinder of mass ' m ' and radius ' r ' rolls along an inclined rough plane of inclination

45^{\circ}

. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder's axis will be

A

\sqrt{2} \mathrm{~g}

B

\dfrac{1}{\sqrt{2}} \mathrm{~g}

C

\dfrac{1}{3 \sqrt{2}} \mathrm{~g}

D

\dfrac{\sqrt{2} g}{3}

Correct Answer

Option D

Solution

The acceleration of the center of mass for a rolling object down an incline is given by:

a = \frac{g \sin\theta}{1 + \frac{I}{m r^2}}

For a uniform solid cylinder, the moment of inertia about its central axis is:

I = \frac{1}{2} m r^2

Substituting this into the expression for acceleration:

a = \frac{g \sin\theta}{1 + \frac{\frac{1}{2} m r^2}{m r^2}} = \frac{g \sin\theta}{1 + \frac{1}{2}} = \frac{g \sin\theta}{\frac{3}{2}} = \frac{2}{3} g \sin\theta

For an incline of

45^\circ

, we have:

\sin 45^\circ = \frac{\sqrt{2}}{2}

Thus, the acceleration becomes:

a = \frac{2}{3} g \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2} g}{3}

Therefore, the linear acceleration of the cylinder's axis is:

\frac{\sqrt{2} g}{3}

Q6

Initial angular velocity of a circular disc of mass

M

is

{\omega _1}.

Then two small spheres of mass

m

are attached gently to diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?

A

\left( {{{M + m} \over M}} \right)\,\,{\omega _1}

B

\left( {{{M + m} \over m}} \right)\,\,{\omega _1}

C

\left( {{M \over {M + 4m}}} \right)\,\,{\omega _1}

D

\left( {{M \over {M + 2m}}} \right)\,\,{\omega _1}

Correct Answer

Option C

Solution

When two small spheres of mass

m

are attached gently, the external torque, about the axis of rotation, is zero. So,

{{d\overrightarrow L } \over {dt}} = \overline z

= 0

\overrightarrow L

= conserved So the angular momentum about the axis of rotation is conserved. $\therefore$

{I_1}{\omega _1} = {I_2}{\omega _2}

\Rightarrow {\omega _2} = {{{I_1}} \over {{I_2}}}{\omega _1}

Here Moment of inertia of Disc

{I_1} = {1 \over 2}M{R^2}

and After adding two sphere Moment of Inertia of disc and two sphere,

{I_2} = {1 \over 2}M{R^2} +

2\left( {{1 \over 2}m{R^2} + {1 \over 2}m{R^2}} \right)

$\therefore$

{\omega _2} = {{{1 \over 2}M{R^2}} \over {{1 \over 2}MR + 2m{R^2}}} \times {\omega _1} = {M \over {M + 4m}}{\omega _1}

Q7

Moment of inertia (M. I.) of four bodies, having same mass and radius, are reported as; I1 = M.I. of thin circular ring about its diameter, I2 = M.I. of circular disc about an axis perpendicular to disc and going through the centre, I3 = M.I. of solid cylinder about its axis and I4 = M.I. of solid sphere about its diameter. Then :

A I1 = I2 = I3 > I4

B I1 + I3 < I2 + I4

C I1 = I2 = I3 < I4

D I1 + I2 = I3 +

{5 \over 2}

I4

Correct Answer

Option A

Solution

Let M and R be the mass and radius of four bodies. Then, as per question, their moment of inertia are I1 =

{{M{R^2}} \over 2}

, I2 =

{{M{R^2}} \over 2}

, I3 =

{{M{R^2}} \over 2}

, I4 =

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

$\therefore$ I1 = I2 = I3 > I4

Q8

The torque due to the force

(2 \hat{i}+\hat{j}+2 \hat{k})

about the origin, acting on a particle whose position vector is

(\hat{i}+\hat{j}+\hat{k})

, would be

A

\hat{j}+\hat{k}

B

\hat{i}-\hat{k}

C

\hat{i}-\hat{j}+\hat{k}

D

\hat{i}+\hat{k}

Correct Answer

Option B

Solution

First, torque ( $\vec{\tau}$ ) is found by taking the cross product of the position vector ( $\vec{r}$ ) and the force vector ( $\vec{F}$ ): $\vec{\tau} = \vec{r} \times \vec{F}$ Here, $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{F} = 2\hat{i} + \hat{j} + 2\hat{k}$ .

We use a determinant to find the cross product:

\vec{\tau} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & 2 \\ \end{array} \right|

Expanding the determinant: $\vec{\tau} = \hat{i}[(1)(2)-(1)(1)] -\hat{j}[(1)(2)-(1)(2)] + \hat{k}[(1)(1)-(1)(2)]$ This gives: $\vec{\tau} = \hat{i}(2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 - 2)$ So, $\vec{\tau} = \hat{i} - 0\hat{j} - \hat{k}$

Q9

A solid sphere, a hollow sphere and a ring are released from top of an inclined plane (frictionless) so that they slide down the plane. Then maximum acceleration down the plane is for (no rolling)

A solid sphere

B hollow sphere

C ring

D all same

Correct Answer

Option D

Solution

Each bodies is sliding along the frictionless inclined plane and there is no rolling, therefore the acceleration of all the bodies is same

\left( {g\,\sin \,\theta } \right).

Q10

A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with a time period T1 and, (ii) back and forth in a direction perpendicular to its plane, with a period T2. The ratio

{{{T_1}} \over {{T_2}}}

will be :

A

{{\sqrt 2 } \over 3}

B

{2 \over {\sqrt 3 }}

C

{2 \over 3}

D

{3 \over {\sqrt 2 }}

Correct Answer

Option B

Solution

Moment of inertia in case (i) is I1 Moment of inertia in case (ii) is I2 I1 = 2MR2 I2 =

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

+ MR2 =

{3 \over 2}M{R^2}

T1 =

2\pi \sqrt {{{{I_1}} \over {Mgd}}}

and T2 =

2\pi \sqrt {{{{I_2}} \over {Mgd}}}

$\Rightarrow$

{{{T_1}} \over {{T_2}}} = \sqrt {{{{I_1}} \over {{I_2}}}}

=

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

=

{2 \over {\sqrt 3 }}