Rotational Motion — Page 5 | NEET Physics

Q41

Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc D 1 has 2 kg mass and 0.2 m radius and initial angular velocity of 50 rad s

-

1 . Disc D 2 has 4 kg mass, 0.1 m radius and initial angular velocity of 200 rad s

-

1 . The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity (in rad s

-

1 ) of the system is

A 60

B 100

C 120

D 40

Correct Answer

Option B

Solution

Given:m 1 = 2 kg m 2 = 4 kg r 1 = 0.2 m r 2 = 0.1 m w 1 = 50 rad s –1 w 2 = 200 rad s –1 As, angular momentum, I 1 W 1 = I 2 W 2 = Constant

{W_f} = {{{I_1}{W_1} + {I_2}{W_2}} \over {{I_1} + {I_2}}} = {{{1 \over 2}{m_1}r_1^2{w_1} + {1 \over 2}{m_2}r_2^2{w_2}} \over {{1 \over 2}{m_1}r_1^2 + {1 \over 2}{m_2}r_2^2}}

By putting the value of m 1 , m 2 , r 1 , r 2 and solving we get = 100 rad s –1

Q42

The ratio of radii of gyration of a circular ring and a circular disc, of the same mass and radius, about an axis passing through their centres and perpendicular to their planes are

A

1:\sqrt 2

B 3 : 2

C 2 : 1

D

\sqrt 2 :1

Correct Answer

Option D

Solution

$\because$

I = M{K^2}

$\therefore$

K = \sqrt {{I \over M}}

I ring = MR 2 and I disc =

{1 \over 2}

MR 2

{{{K_1}} \over {{K_2}}} = \sqrt {{{{I_1}} \over {{I_2}}}} = \sqrt {{{M{R^2}} \over {\left( {{{M{R^2}} \over 2}} \right)}}} = \sqrt 2 :1

Q43

A small object of uniform density rolls up a curved surface with an initial velocity 'v'. It reaches upto a maximum height of

{{3{v^2}} \over {4g}}

with respect to the initial position. The object is

A hollow sphere

B disc

C ring

D solid sphere

Correct Answer

Option B

Solution

The kinetic energy of the rolling object is converted into potential energy at height

h = \left( { {{3{v^2}} \over {4g}}} \right)

So by the law of conservation of mechanical energy, we have

{1 \over 2}M{v^2} + {1 \over 2}I{\omega ^2} = Mgh

\left( \because{\omega = {v \over R}} \right)

{1 \over 2}M{v^2} + {1 \over 2}I{\left( {{v \over R}} \right)^2} = Mg\left( {{{3{v^2}} \over {4g}}} \right)

{1 \over 2}I{{{v^2}} \over {{R^2}}} = {3 \over 4}M{v^2} - {1 \over 2}M{v^2}

{1 \over 2}I{{{v^2}} \over {{R^2}}} = {1 \over 4}M{v^2}

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

Hence, the object is disc.

Q44

A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is

A

{{2g} \over L}

B

{{2g} \over {2L}}

C

{{3g} \over {2L}}

D

{g \over L}

Correct Answer

Option C

Solution

When the string is cut, the rod will rotate about P.

Let $\alpha$ be initial angular acceleration of the rod.Then Torque,

\tau = I\alpha = {{M{L^2}} \over 3}\alpha

...(i) (Moment of inertia of the rod about one end =

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

) Also,

\tau = Mg{L \over 2}

...(ii) Equating (i) and (ii), we get

Mg{L \over 2} = {{M{L^2}} \over 3}\alpha \Rightarrow \alpha = {{3g} \over {2L}}

Q45

A circular platform is mounted on a frictionless vertical axle. Its radius R = 2 m and its moment of inertia about the axle is 200 kg m 2 . It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 ms

-

1 relative to the ground. Time taken by the man to complete one revolution is

A

\pi

s

B

{{3\pi } \over 2}

s

C 2

\pi

s

D

{\pi \over 2}

s

Correct Answer

Option C

Solution

Using conservation L i = 0 (Initial moment) L f = mvR – I $\omega$ (Final moment) According to the conservation of momentum L i = L f $\Rightarrow$ mvR – I. $\omega$ = 0 mvR = I.

$\omega$

\omega = \left( {{1 \over 2}} \right)

\left( {v + \omega R} \right)t = 2\pi R

t\left( {1 + {1 \over 2} \times 2} \right) = 2\pi \times 2

t = 2 $\pi$ sec.

Q46

A car of mass m is moving on a level circular track of radius R. If

\mu

s represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by

A

\sqrt {{\mu _s}mRg}

B

\sqrt {{{Rg} \over {{\mu _s}}}}

C

\sqrt {{{mRg} \over {{\mu _s}}}}

D

\sqrt {{\mu _s}Rg}

Correct Answer

Option D

Solution

For smooth driving maximum speed of car v then

{{m{v^2}} \over R} = {\mu _s}mg

v = \sqrt {{\mu _s}Rg}

Q47

The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through

A B

B C

C D

D A

Correct Answer

Option A

Solution

According to the theorem of parallel axes, I = I CM + Ma 2 As a is maximum for point B. Therefore I is maximum about B.

Q48

ABC is an equilateral triangle with O as its centre.

{\overrightarrow F _1},{\overrightarrow F _2}

and

{\overrightarrow F _3}

represent three forces acting along the sides AB, BC and AC respectively. If the total torque about O is zero then magnitude of

{\overrightarrow F _3}

is

A F 1 + F 2

B F 1

-

F 2

C

{{{F_1} + {F_2}} \over 2}

D 2(F 1 + F 2 )

Correct Answer

Option A

Solution

Let x be the distance of centre O of equilateral triangle from each side.

Total torque about O = 0 $\Rightarrow$ F 1 x + F 2 x – F 3 x = 0 or F 3 = F 1 + F 2

Q49

A car of mass 1000 kg negotiates a banked curve of radius 90 m on a frictionless road. If the banking angle is 45 o , the speed of the car is

A 20 m s

-

1

B 30 m s

-

1

C 5 m s

-

1

D 10 m s

-

1

Correct Answer

Option B

Solution

Here, m = 1000 kg, R = 90 m, $\theta$ = 45° For banking,

\tan \theta = {{{v^2}} \over {Rg}}

\Rightarrow v = \sqrt {Rg\tan \theta } = \sqrt {90 \times 10 \times \tan {{45}^o}} = 30\,m{s^{ - 1}}

Q50

When a mass is rotating in a plane about a fixed point, its angular momentum is directed along

A a line perpendicular to the plane of rotation

B the line making an angle of 45 o to the plane of rotation

C the radius

D the tangent to the orbit

Correct Answer

Option A

Solution

When a mass is rotating in a plane about a fixed point its angular momentum is directed along a line perpendicular to the plane of rotation.