Rotational Motion — Page 3 | JEE Physics

Q21

The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal plane. : (i) a ring of radius R, (ii) a solid cylinder of radius R/2 and (iii) a solid sphere of radius R/4 . If in each case, the speed of the centre of mass at the bottom of the incline is same, the ratio of the maximum heights they climb is :

A 20 : 15 : 14

B 4 : 3 : 2

C 2 : 3 : 4

D 10 : 15 : 7

Correct Answer

Option A

Solution

Total kinetic energy of a rolling body is given as

E_{\text{total }}=\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right]

where, $K$ is the radius of gyration. Using conservation law of energy,

\begin{array}{rlrl} \frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right] =m g h \\\\ \text{ or } h =\frac{v^2}{2 g}\left[1+\frac{K^2}{R^2}\right] \end{array}

For ring, $\dfrac{K^2}{R^2}=1$

\Rightarrow h_1=\frac{v^2}{2 g}[1+1]=\frac{2 v^2}{2 g}=\frac{v^2}{g}

For solid cylinder, $\dfrac{K^2}{R^2}=\dfrac{(R / 2 \sqrt{2})^2}{(R / 2)^2}$

\begin{aligned} & =\frac{R^2}{8} \times \frac{4}{R^2}=\frac{1}{2} \\\\ \Rightarrow h_2 & =\frac{v^2}{2 g}\left[1+\frac{1}{2}\right]=\frac{3 v^2}{4 g} \end{aligned}

For solid sphere, $\dfrac{K^2}{R^2}=\dfrac{2}{5}$

\Rightarrow h_3=\frac{v^2}{2 g}\left[1+\frac{2}{5}\right]=\frac{7 v^2}{10 g}

So,the ratio of $h_1, h_2$ and $h_3$ is

\begin{aligned} h_1: h_2: h_3 & =\frac{v^2}{g}: \frac{3 v^2}{4 g}: \frac{7}{10} \frac{v^2}{g} \\\\ & =1: \frac{3}{4}: \frac{7}{10}=20: 15: 14 \end{aligned}

Q22

A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of

\theta

, where

\theta

is the angle by which it has rotated, is given as k

\theta

2. If its moment of inertia is I then the angular acceleration of the disc is :

A

{k \over {4I}}\theta

B

{k \over {I}}\theta

C

{k \over {2I}}\theta

D

{2k \over {I}}\theta

Correct Answer

Option D

Solution

Kinetic energy KE =

{1 \over 2}l{\omega ^2} = k{\theta ^2}

\Rightarrow {\omega ^2} = {{2k{\theta ^2}} \over l} \Rightarrow \omega = \sqrt {{{2k} \over l}} \theta

.... (A) Differentiate (A) wrt time $\to$

{{d\omega } \over {dt}} = \alpha = \sqrt {{{2k} \over l}} \left( {{{d\theta } \over {dt}}} \right)

\Rightarrow \alpha = \sqrt {{{2k} \over l}} .\sqrt {{{2k} \over l}} \theta \,\{ by\,(1)\}

\Rightarrow \alpha = {{2k} \over l}\theta \,

Q23

Moment of inertia of a body about a given axis is 1.5 kg m2. Initially the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular accleration of 20 rad/s2 must be applied about the axis for a duration of :-

A 2.5 s

B 3 s

C 5s

D 2 s

Correct Answer

Option D

Solution

KE =

{1 \over 2}I{\omega ^2} = 1200

(given)

\Rightarrow \omega = 40\,rad/s

\Rightarrow \omega = {\omega _0} + \alpha t

\Rightarrow 40 = 0 + (20)t

\Rightarrow t = 2\,\sec

Q24

A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of

{{7M} \over 8}

and is converted into a uniform disc of radius 2R. The second part is converted into a uniform solid sphere. Let I1 be the moment of inertia of the disc about its axis and I2 be the moment of inertia of the new sphere about its axis. The ratio I1/I2 is given by :

A 65

B 140

C 185

D 285

Correct Answer

Option B

Solution

{I_1} = {{\left( {{{7M} \over 8}} \right){{\left( {2R} \right)}^2}} \over 2} = {{7M \times 4{R^2}} \over {2 \times 8}} = {{7M{R^2}} \over 4}

{I_2} = {2 \over 5}{M \over 8}{\left( {{R \over 2}} \right)^2} = {{2M} \over {5 \times 8}}{{{R^2}} \over 4} = {{M{R^2}} \over {80}}

{{{I_1}} \over {{I_2}}} = {{7M{R^2} \times 80} \over {4M{R^2}}} = 140

Q25

The time dependence of the position of a particle of mass m = 2 is given by

\overrightarrow r \left( t \right) = 2t\widehat i - 3{t^2}\widehat j

. Its angular momentum, with respect to the origin, at time t = 2 is

A 36

\widehat k

B - 48

\widehat k

C

- 34\left( {\widehat k - \widehat i} \right)

D

48\left( {\widehat i + \widehat j} \right)

Correct Answer

Option B

Solution

\overrightarrow v = 2\widehat i - 6 + \widehat j

At t = 2

\overrightarrow v = 2\widehat i - 12\widehat j

\overrightarrow P = m\overrightarrow v = 4i - 24\widehat j

At t = 2

\overrightarrow r = 4\widehat i - 12\widehat j

\overrightarrow L = \overrightarrow r \times \overrightarrow P = \left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 4 & { - 12} & 0 \\ 4 & { - 24} & 0 \end{array} \right|

= \left\{ {4( - 2) + 4 \times 12} \right\}\widehat k

= \left( { - 96 + 48} \right)\widehat k

= \left( - \right)48\widehat k

Q26

Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is :

A 16 cm

B 12 cm

C 14 cm

D 18 cm

Correct Answer

Option A

Solution

Consider an element of radius x and thickness dx Mass of element, dm =

\sigma 2\pi x\left( {dx} \right)

Here, $\sigma$ = mass per unit area =

{m \over {\pi \left( {{R^2} - {r^2}} \right)}}

Moment of inertia of element, dI = (dm)x2 $\Rightarrow$ I =

\sigma 2\pi \int\limits_r^R {{x^3}dx}

=

\sigma 2\pi \left( {{{{R^4} - {r^4}} \over 4}} \right)

=

{m \over {\pi \left( {{R^2} - {r^2}} \right)}}{\pi \over 2}\left( {{R^4} - {r^4}} \right)

=

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

.....(i) Moment of inertia of thin cylinder of same mass, I = m

r_0^2

......(ii) $\Rightarrow$ m

r_0^2

=

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

$\Rightarrow$

r_0^2

= 250 $\Rightarrow$ r0

\simeq

16 cm

Q27

Consider two uniform discs of the same thickness and different radii R1 = R and R2 =

\alpha

R made of the same material. If the ratio of their moments of inertia I1 and I2 , respectively, about their axes is I1 : I2 = 1 : 16 then the value of

\alpha

is :

A

\sqrt 2

B 2

C

2\sqrt 2

D 4

Correct Answer

Option B

Solution

Moment of inertia of disc,

I = {{M{R^2}} \over 2} = {{\left[ {p\left( {\pi {R^2}} \right)t} \right]{R^2}} \over 2}

I = K{R^4}

{{{I_1}} \over {{I_2}}} = {\left( {{{{R_1}} \over {{R_2}}}} \right)^4}

{1 \over {16}} = {\left( {{R \over {\alpha R}}} \right)^4} \Rightarrow \alpha = {\left( {16} \right)^{{1 \over 4}}} = 2

Q28

The linear mass density of a thin rod AB of length L varies from A to B as

\lambda \left( x \right) = {\lambda _0}\left( {1 + {x \over L}} \right)

, where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is :

A

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

B

{5 \over {12}}M{L^2}

C

{7 \over {18}}M{L^2}

D

{3 \over 7}M{L^2}

Correct Answer

Option C

Solution

dm = $\lambda$ dx =

{\lambda _0}\left( {1 + {x \over L}} \right)

dx Integrating both side, we get

\int\limits_0^M {dm} = \int\limits_0^L {{\lambda _0}\left( {1 + {x \over L}} \right)} dx

$\Rightarrow$ M =

{\lambda _0}L + {{{\lambda _0}{L^2}} \over {2L}}

=

{{3{\lambda _0}L} \over 2}

$\Rightarrow$

{\lambda _0} = {{2M} \over {3L}}

.....(1) Moment of inertia of small part dx is dI = dmx2 Integrating both side, we get

\int\limits_0^I {dI} = \int\limits_0^L {dm{x^2}}

$\Rightarrow$ I =

\int\limits_0^L {{\lambda _0}\left( {1 + {x \over L}} \right)} dx\,{x^2}

=

{\lambda _0}\int\limits_0^L {\left( {{x^2} + {{{x^3}} \over L}} \right)} dx

=

{\lambda _0}\left[ {{{{L^3}} \over 3} + {{{L^3}} \over 4}} \right]

=

{{7{L^3}{\lambda _0}} \over {12}}

Here by putting the value of $\lambda$ 0 form (1) I =

{{7{L^3}} \over {12}}\left( {{{2M} \over {3L}}} \right)

=

{7 \over {18}}M{L^2}

Q29

Moment of inertia of a cylinder of mass M, length L and radius R about an axis passing through its centre and perpendicular to the axis of the cylinder is I =

M\left( {{{{R^2}} \over 4} + {{{L^2}} \over {12}}} \right)

. If such a cylinder is to be made for a given mass of a material, the ratio

{L \over R}

for it to have minimum possible I is

A

{3 \over 2}

B

\sqrt {{3 \over 2}}

C

\sqrt {{2 \over 3}}

D

{{2 \over 3}}

Correct Answer

Option B

Solution

Given I =

M\left( {{{{R^2}} \over 4} + {{{L^2}} \over {12}}} \right)

M = $\rho$ .V = $\rho$ $\pi$ R2L $\Rightarrow$ R2 =

{M \over {\rho \pi L}}

$\therefore$ I =

M\left( {{M \over {4\rho \pi L}} + {{{L^2}} \over {12}}} \right)

$\Rightarrow$

{{dI} \over {dL}}

=

M\left( {{M \over {4\rho \pi }}\left( { - {1 \over {{L^2}}}} \right) - {{2L} \over {12}}} \right)

For minimum I,

{{dI} \over {dL}} = 0

$\therefore$

M\left( {{M \over {4\rho \pi }}\left( { - {1 \over {{L^2}}}} \right) - {{2L} \over {12}}} \right)

= 0 $\Rightarrow$

{{{M^2}} \over {4\rho \pi {L^2}}} = {{2LM} \over {12}}

$\Rightarrow$

{M \over {4\rho \pi {L^2}}} = {L \over 6}

$\Rightarrow$

{{\rho \pi {R^2}L} \over {4\rho \pi {L^2}}} = {L \over 6}

$\Rightarrow$

{{{R^2}} \over {{L^2}}} = {2 \over 3}

$\Rightarrow$

{R \over L} = \sqrt {{2 \over 3}}

$\Rightarrow$

{L \over R} = \sqrt {{3 \over 2}}

Q30

Consider a uniform rod of mass M = 4m and length

\ell

pivoted about its centre. A mass m moving with velocity v making angle

\theta = {\pi \over 4}

to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is :

A

{{3\sqrt 2 } \over 7}{v \over \ell }

B

{3 \over 7}{v \over \ell }

C

{3 \over {7\sqrt 2 }}{v \over \ell }

D

{4 \over 7}{v \over \ell }

Correct Answer

Option A

Solution

About hinge(O) net torque $\tau$ = 0 So angular momentum is conserved about hinge(O), Linitial = Lfinal m

\left( {{v \over {\sqrt 2 }}} \right)\left( {{l \over 2}} \right)

+ 0 =

\left[ {{{4m{l^2}} \over {12}} + m{{\left( {{l \over 2}} \right)}^2}} \right]\omega

$\Rightarrow$ $\omega$ =

{{3\sqrt 2 } \over 7}{v \over \ell }