Rotational Motion — Page 2 | NEET Physics

Q11

The angular acceleration of a body, moving along the circumference of a circle, is :

A along the radius towards the centre

B along the tangent to its position

C along the axis of rotation

D along the radius, away from centre

Correct Answer

Option C

Solution

Along the axis of rotation.

Q12

An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is :

A 0.07 kg-m 2

B 0.7 kg-m 2

C 3.22 kg-m 2

D 30.8 kg-m 2

Correct Answer

Option B

Solution

From work - energy theorem

W = \Delta k

(change in Kinetic Energy) In rotation,

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

484 = {1 \over 2}I(\omega _f^2 - \omega _i^2)

\Rightarrow 484 = {1 \over 2}I\left[ {{{\left( {2\pi {{360} \over {60}}} \right)}^2} - {{\left( {2\pi \times {{60} \over {60}}} \right)}^2}} \right]

\Rightarrow 484 = {1 \over 2}I4{\pi ^2}(36 - 1)

\Rightarrow I \simeq 0.7

kg-m 2

Q13

The angular speed of a fly wheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in rad/s 2 is

A 2

\pi

B 4

\pi

C 12

\pi

D 104

\pi

Correct Answer

Option B

Solution

Angular acceleration

\alpha = {{{\omega _f} - {\omega _i}} \over t}

{\omega _f} = 3120 \times {{2\pi } \over {60}}

rad/s

{\omega _i} =1200 \times {{2\pi } \over {60}}

rad/s

\Rightarrow \alpha = {{(3120 - 1200)} \over {16}} \times {{2\pi } \over {60}} = 4\pi

Q14

The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is

A 2 : 1

B

\sqrt2

: 1

C 4 : 1

D 1 :

\sqrt2

Correct Answer

Option B

Solution

{l_1} = {{M{R^2}} \over 2}

{k_1} = \sqrt {{{{l_1}} \over M}}

= {R \over {\sqrt 2 }}

{l_2} = {{M{R^2}} \over 4}

{k_2} = \sqrt {{{{l_2}} \over M}}

= {R \over 2}

{{{k_1}} \over {{k_2}}} = {{{R \over {\sqrt 2 }}} \over {{R \over 2}}}

= \sqrt 2 :1

Q15

From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 90

^\circ

sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the center of the ring and perpendicular to the plane of the ring is 'K' times 'MR 2 '. Then the value of 'K' is :

A

{1 \over 8}

B

{3 \over 4}

C

{7 \over 8}

D

{1 \over 4}

Correct Answer

Option B

Solution

Given that, Mass of Ring = M; Radius of Ring = R Now 90° arc is removed from circular ring, then Mass removed =

{M \over 4}

$\therefore$ Mass of remaining portion =

{{3M} \over 4}

I = MR 2 I' =

{{3M} \over 4}

R 2 I' =

{3 \over 4}

I $\therefore$ K =

{3 \over 4}

Q16

A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass 'm' is suspended from the rod at 160 cm mark as shown in the figure. Find the value of 'm' such that the rod is in equilibrium. (g = 10 m/s 2 )

A

{1 \over {12}}

kg

B

{1 \over {2}}

kg

C

{1 \over {3}}

kg

D

{1 \over {6}}

kg

Correct Answer

Option A

Solution

By balancing torque 2g $\times$ 20 = 0.5g $\times$ 60 + mg $\times$ 120 $\Rightarrow$ m =

{{0.5} \over 6}

kg =

{1 \over {12}}

kg

Q17

Find the torque about the origin when a force of 3

\hat j

N acts on a particle whose position vector is 2

\hat k

m.

A 6

\hat j

N m

B

-

6

\hat i

N m

C 6

\hat k

N m

D 6

\hat i

N m

Correct Answer

Option B

Solution

Torque is the moment of Force applied.

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

\overrightarrow \tau = 2\hat k \times 3\hat j

\overrightarrow \tau = - 6\hat i

Nm

Q18

A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after 2

\pi

revolutions is :

A 2 × 10 –3 N m

B 2 × 10 –6 N m

C 2 × 10 6 N m

D 12 × 10 –4 N m

Correct Answer

Option B

Solution

Acccording to work energy theorem, W =

{1 \over 2}\left( {\omega _f^2 - \omega _i^2} \right)

Given that, $\theta$ = 2 $\pi$ revolution/minute $\theta$ = 2 $\pi$ $\times$ 2 $\pi$ = 4 $\pi$ 2 rad

{\omega _i} = 3 \times {{2\pi } \over {60}}

rad/s

{\omega _f} = 0

rad/s By putting the value of

{\omega _f}

and

{\omega _i}

, we get - $\tau$ $\theta$ =

{1 \over 2} \times {1 \over 2}m{r^2}\left( {0 - \omega _i^2} \right)

$\Rightarrow$ $\tau$ =

- {{{1 \over 2} \times {1 \over 2} \times 2 \times \left( {4 \times {{10}^{ - 2}}} \right) \times {{\left( { - 3 \times {{2\pi } \over {60}}} \right)}^2}} \over {4{\pi ^2}}}

= 2 $\times$ 10 -6 Nm

Q19

Two particles A and B are moving in uniform circular motion in concentric circles of radii r A and r B with speed v A and v B respecitively. Their time period of rotation is the same. The ratio of angular speed of A to that of B will be :

A v A : v B

B 1 : 1

C r B : r A

D r A : r B

Correct Answer

Option B

Solution

Since, the time period for both the particle in same So, T A = T B = T Angular velocity for A, (

{\omega _A}

) =

{{2\pi } \over {{T_A}}}

Angular velocity for B, (

{\omega _B}

) =

{{2\pi } \over {{T_B}}}

Now, the required ratio is

{{{\omega _A}} \over {{\omega _B}}} = {{{T_B}} \over {{T_A}}} = {T \over T} = 1

Q20

A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its centre of mass has speed of 20 cm/s. How much work is needed to stop it?

A 1 J

B 2 J

C 3 J

D 30 kJ

Correct Answer

Option C

Solution

\omega = {V \over R} = {{0.2} \over 2} =

0.1 rad/sec Apply the law of conservation of energy, Work required = change in kinetic energy Since, final KE = 0 And, initial KE =

{1 \over 2}m{v^2} + {1 \over 2}I{\omega ^2}

=

{1 \over 2}m{v^2} + {1 \over 2}{{M{l^2}} \over 2}{\omega ^2}

=

{1 \over 2}\left( {100} \right){\left( {0.2} \right)^2} + {1 \over 2}{{100 \times {{\left( 2 \right)}^2}} \over 2}{\left( {0.1} \right)^2}

= 1 +

{1 \over 2} \times 4

= 3 J