Rotational Motion — Page 7 | JEE Physics

Q61

Angular momentum of the particle rotating with a central force is constant due to

A constant torque

B constant force

C constant linear momentum

D zero torque

Correct Answer

Option D

Solution

We know that

\overrightarrow {{\tau _c}} = {{d\overrightarrow {{L_c}} } \over {dt}}

where

\overrightarrow {{\tau _c}}

torque about the center of mass of the body and

\overrightarrow {{L_c}} =

Angular momentum about the center of mass of the body. Given that

\overrightarrow {{L_c}} =

constant. $\therefore$

{{d\overrightarrow {{L_c}} } \over {dt}}

= 0 $\Rightarrow$

\overrightarrow {{\tau _c}} = 0

[as

\overrightarrow {{\tau _c}} = {{d\overrightarrow {{L_c}} } \over {dt}}

]

Q62

In a physical balance working on the principle of moments, when 5 mg weight is placed on the left pan, the beam becomes horizontal. Both the empty pans of the balance are of equal mass. Which of the following statements is correct ?

A Left arm is longer than the right arm

B Both the arms are of same length

C Left arm is shorter than the right arm

D Every object that is weighed using this balance appears lighter than its actual weight.

Correct Answer

Option C

Solution

From principle of moment we know, the anticlockwise moment is equal to clockwise moment when a system is stable or balance.

\therefore\,\,\,

load $\times$ load arm = effect $\times$ effect arm When 5 mg weight is placed on the left pan, load arm shift to left side, hence left arm become shorter than right arm.

Q63

Moment of inertia of a circular wire of mass

M

and radius

R

about its diameter is

A

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

B

M{R^2}

C

2M{R^2}

D

{{M{R^2}} \over 4}

Correct Answer

Option A

Solution

Moment of Inertia of a circular wire about an axis

nn'

passing through the centre of the circle and perpendicular to the plane of the circle

= M{R^2}

As shown in the figure,

X

-axis and

Y

-axis lies in the plane of the ring. Then by perpendicular axis theorem

{I_X} + {I_Y} = {I_Z}

\Rightarrow 2{I_X} = M{R^2}\,

\left[ \, \right.

as

{I_X} - {I_Y}

(by symmetry) and

{I_Z} = M{R^2}

\left. \, \right]

$\therefore$

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

Q64

A particle performing uniform circular motion has angular frequency is doubled & its kinetic energy halved, then the new angular momentum is

A

{L \over 4}

B

2L

C

4L

D

{L \over 2}

Correct Answer

Option A

Solution

We know Rotational Kinetic Energy

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

Angular Momentum

L = I\omega \Rightarrow I = {L \over \omega }

$\therefore$ Initial

K.E. = {1 \over 2}{L \over \omega } \times {\omega ^2} = {1 \over 2}L\omega

Final

K.E'

=

{{K.E} \over 2}

=

{1 \over 2}{L'} \times 2\omega

$\therefore$

{{K.E} \over {K.E'}} = {{L \times \omega } \over {L' \times \omega '}}

\Rightarrow {{K.E} \over {{{K.E} \over 2}}} = {{L \times \omega } \over {L' \times 2\omega }}

$\therefore$

L' = {L \over 4}

Q65

Consider a uniform square plate of side

' a '

and mass

'm'

. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

A

{5 \over 6}m{a^2}

B

{1 \over 12}m{a^2}

C

{7 \over 12}m{a^2}

D

{2 \over 3}m{a^2}

Correct Answer

Option D

Solution

Moment of inertia for the square plate through O, perpendicular to the plate is

{I_{nn'}} = {1 \over {12}}m\left( {{a^2} + {a^2}} \right) = {{m{a^2}} \over 6}

Also,

DO = {{DB} \over 2} = {{\sqrt 2 a} \over 2} = {a \over {\sqrt 2 }}

According to parallel axis theorem

{{\mathop{\rm I}\nolimits} _{mm'}} = {I_{nn'}} + m{\left( {{a \over {\sqrt 2 }}} \right)^2}

= {{m{a^2}} \over 6} + {{m{a^2}} \over 2}

= {{m{a^2} + 3m{a^2}} \over 6}

= {2 \over 3}m{a^2}

Q66

A thin uniform rod of length

l

and mass

m

is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is

\omega

. Its center of mass rises to a maximum height of:

A

{1 \over 6}\,\,{{l\omega } \over g}

B

{1 \over 2}\,\,{{{l^2}{\omega ^2}} \over g}

C

{1 \over 6}\,\,{{{l^2}{\omega ^2}} \over g}

D

{1 \over 3}\,\,{{{l^2}{\omega ^2}} \over g}

Correct Answer

Option C

Solution

The moment of inertia of the rod about

O

is

{1 \over 2}m{\ell ^2}.

The maximum angular speed of the rod is when the rod is instantaneously vertical.

The energy of the rod in this conditions is

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

where

I

is the moment of inertia of the rod about

O.

when the rod is in its extreme portion, its angular velocity is zero momentarily.

In this case, the center of mass is raised through

h

, so the increase in potential energy is

mgh

. This is equal to kinetic energy

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

. $\therefore$

mgh = {1 \over 2}I{\omega ^2} = {1 \over 2}\left( {{1 \over 3}m{l^2}} \right){\omega ^2}

.

\Rightarrow h = {{{\ell ^2}{\omega ^2}} \over {6g}}

Q67

A pulley of radius

2

m

is rotated about its axis by a force

F = \left( {20t - 5{t^2}} \right)

newton (where

t

is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is

10kg

-

{m^2}

the number of rotation made by the pulley before its direction of motion is reversed, is:

A more than

3

but less than

6

B more than

6

but less than

9

C more than

9

D less than

3

Correct Answer

Option A

Solution

Given

F = 20t - 5{t^2}

, R = 2 m and

I

= 10 kg m2 Torque applied on pulley

\tau = FR

$\therefore$

\alpha = {{FR} \over I}

[ as

\tau = I\alpha

] $\Rightarrow$

\alpha = {{\left( {20t - 5{t^2}} \right) \times 2} \over {10}}

$\Rightarrow$

\alpha = 4t - {t^2}

\Rightarrow {{d\omega } \over {dt}} = 4t - {t^2}

\Rightarrow \int\limits_0^\omega {d\omega } = \int\limits_0^t {\left( {4t - {t^2}} \right)} dt

\Rightarrow \omega = 2{t^2} - {{{r^3}} \over 3}

( At

t = 0,6 \,s

\omega = 0

)

\omega = {{d\theta } \over {dt}} = 2{t^2} - {{{t^3}} \over 3}

\int\limits_0^\theta {d\theta } = \int\limits_0^6 {\left( {2{t^2} - {{{r^3}} \over 3}} \right)} dt

\Rightarrow \theta = 36rad\,\, \Rightarrow n = {{36} \over {2\pi }} < 6

Q68

A mass

m

hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass

m

and radius

R.

Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass

m,

if the string does not slip on the pulley, is:

A

g

B

{2 \over 3}g

C

{g \over 3}

D

{3 \over 2}g

Correct Answer

Option B

Solution

This is the free body diagram of pulley and mass For translation motion of the block,

mg - T = ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right)

For rotational motion of the pulley,

T\times R = I\alpha = I{a \over R}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)

where

\alpha =

angular acceleration of disc =

{a \over R}

and

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

(For circular disc) Solving

(1)

&

(2),

a = {{mg} \over {\left( {m + {I \over {{R^2}}}} \right)}} = {{mg} \over {m + {{m{R^2}} \over {2{R^2}}}}}

= {{2mg} \over {3m}} = {{2g} \over 3}

Q69

A thin horizontal circular disc is rotating about a vertical axis passing through its center. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc.

A continuously decreases

B continuously increases

C first increases and then decreases

D remains unchanged

Correct Answer

Option C

Solution

Here no external force is applied on the disc so Torque ( $\tau$ ) = 0. So angular momentum is conserved. That means

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

$\Rightarrow$

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

$\therefore$ Angular speed is inversely proportional to Moment of inertia. For disc

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

$\therefore$ Moment of Inertia is proportional to Mass.

As insect moves along a diameter, the effective mass of disc first decreases then increases and hence the moment of inertia first decreases then increases so from principle of conservation of angular momentum, angular speed, first increases then decreases.

Q70

A hoop of radius

r

and mass

m

rotating with an angular velocity

{\omega _0}

is placed on a rough horizontal surface. The initial velocity of the center of the hoop is zero. What will be the velocity of the center of the hoop when it cases to slip?

A

{{r{\omega _0}} \over 4}

B

{{r{\omega _0}} \over 3}

C

{{r{\omega _0}} \over 2}

D

{r{\omega _0}}

Correct Answer

Option C

Solution

From conservation of angular momentum at point of contact,

m{r^2}{\omega _0} = mvr + m{r^2}\omega

m{r^2}{\omega _0} = mvr + m{r^2}\left( {{v \over r}} \right)

[ as

v = r\omega

]

m{r^2}{\omega _0} = mvr + mvr

m{r^2}{\omega _0} = 2mvr

v = {{{\omega _0}r} \over 2}