Rotational Motion — Page 4 | JEE Physics

Q31

Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are 0.1 kg-m2 and 10 rad s–1 respectively while those for the second one are 0.2 kg-m2 and 5 rad s–1 respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The kinetic energy of the combined system is :

A

{{20} \over 3}J

B

{{5} \over 3}J

C

{{10} \over 3}J

D

{{2} \over 3}J

Correct Answer

Option A

Solution

Angular momentum conserved for the system I1

{\omega _1}

+ I2

{\omega _2}

= (I1 + I2)

{\omega _f}

$\Rightarrow$ 0.1 × 10 + 0.2 × 5 = (0.1 + 0.2) ×

{\omega _f}

$\Rightarrow$

{\omega _f}

=

{{20} \over 3}

Kinetic energy of combined disc system =

{1 \over 2}\left( {{I_1} + {I_2}} \right)\omega _f^2

=

{1 \over 2}\left( {0.1 + 0.2} \right){\left( {{{20} \over 3}} \right)^2}

=

{{20} \over 3}J

Q32

If force

\overrightarrow F = 3\widehat i + 4\widehat j - 2\widehat k

acts on a particle position vector

2\widehat i + \widehat j + 2\widehat k

then, the torque about the origin will be :

A

3\widehat i + 4\widehat j - 2\widehat k

B

- 10\widehat i + 10\widehat j + 5\widehat k

C

10\widehat i + 5\widehat j - 10\widehat k

D

10\widehat i + \widehat j - 5\widehat k

Correct Answer

Option B

Solution

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

= (2\widehat i + \widehat j + 2\widehat k) \times (3\widehat i + 4\widehat j - 2\widehat k)

= - 10\widehat i + 10\widehat j + 5\widehat k

Q33

Four identical solid spheres each of mass 'm' and radius 'a' are placed with their centres on the four corners of a square of side 'b'. The moment of inertia of the system about one side of square where the axis of rotation is parallel to the plane of the square is :

A

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

B

{8 \over 5}m{a^2} + m{b^2}

C

{4 \over 5}m{a^2} + 2m{b^2}

D

{8 \over 5}m{a^2} + 2m{b^2}

Correct Answer

Option D

Solution

I = {2 \over 5}m{a^2} + {2 \over 5}m{a^2} + \left[ {{2 \over 5}m{a^2} + m{b^2}} \right] + [{2 \over 5}m{a^2} + m{b^2}]

I = 4 \times {2 \over 5}m{a^2} + 2m{b^2}

= {8 \over 5}m{a^2} + 2m{b^2}

Q34

A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance 'h', the square of angular velocity of wheel will be :

A

{{2mgh} \over {I + 2m{r^2}}}

B

{{2mgh} \over {I + m{r^2}}}

C 2gh

D

{{2gh} \over {I + m{r^2}}}

Correct Answer

Option B

Solution

Using energy conservation between A and B point

mgh = {1 \over 2}m{(wR)^2} + {1 \over 2}I{\omega ^2}

2mgh = (M{R^2} + I){\omega ^2}

{\omega ^2} = {{2mgh} \over {I + M{R^2}}}

Q35

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed

\omega

. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become :

A

\omega {M \over {M + m}}

B

\omega {{M + 2m} \over M}

C

\omega {M \over {M + 2m}}

D

\omega {{M - 2m} \over {M + 2m}}

Correct Answer

Option C

Solution

$\tau$ net = 0, so angular momentum is conserved By angular momentum conservation Ii $\omega$ i = If $\omega$ f (MR2) $\omega$ = (MR2 + 2mR2) $\omega$ f $\omega$ f =

{{(M{R^2})\omega } \over {M{R^2} + 2m{R^2}}} = {{M\omega } \over {M + 2m}}

{\omega _f} = {{M\omega } \over {M + 2m}}

Q36

Consider a uniform wire of mass M and length L. It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the center is :

A

{1 \over 4}{{M{L^2}} \over {{\pi ^2}}}

B

{1 \over 2}{{M{L^2}} \over {{\pi ^2}}}

C

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

D

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

Correct Answer

Option D

Solution

$\therefore$ From figure, L = $\pi$ R $\Rightarrow$ R =

{L \over \pi }

Moment of inertia about center O, I = MR2 = M

{\left( {{L \over \pi }} \right)^2}

=

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

Q37

Consider a situation in which a ring, a solid cylinder and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and having identical diameter. The correct statement for this situation is

A All of them will have same velocity.

B The ring has greatest and the cylinder has the least velocity of the centre of mass at the bottom of the inclined plane.

C The sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.

D The cylinder has the greatest and the sphere has the least velocity of the centre of mass at the bottom of the inclined plane.

Correct Answer

Option C

Solution

{{{K_T}} \over {{K_R}}} = {{M{R^2}} \over {{I_{CM}}}}

ICM is maximum for ring. $\Rightarrow$ v is least for ring.

Q38

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Moment of inertia of a circular disc of mass 'M' and radius 'R' about X, Y axes (passing through its plane) and Z-axis which is perpendicular to its plane were found to be Ix, Iy and Iz respectively. The respectively radii of gyration about all the three axes will be the same. Reason R : A rigid body making rotational motion has fixed mass and shape. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both A and R are correct but R is NOT the correct explanation of A.

B A is not correct but R is correct.

C A is correct but R is not correct.

D Both A and R are correct and R is the correct explanation of A.

Correct Answer

Option B

Solution

Iz = Ix + Iy (using perpendicular axis theorem) & I = mk2 (K : radius of gyration) so, mKz2 = mKx2 + mKy2 Kz2 = Kx2 + Ky2 so radius of gyration about axes x, y & z won't be same hence assertion A is not correct reason R is correct statement (property of a rigid body)

Q39

Two discs have moments of inertia I1 and I2 about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds,

\omega

1 and

\omega

2 respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by :

A

{{{I_1}{I_2}} \over {({I_1} + {I_2})}}{({\omega _1} - {\omega _2})^2}

B

{{{{({I_1} - {I_2})}^2}{\omega _1}{\omega _2}} \over {2({I_1} + {I_2})}}

C

{{{I_1}{I_2}} \over {2({I_1} + {I_2})}}{({\omega _1} - {\omega _2})^2}

D

{{{{({\omega _1} - {\omega _2})}^2}} \over {2({I_1} + {I_2})}}

Correct Answer

Option C

Solution

From conservation of angular momentum we get

{I_1}{\omega _1} + {I_2}{\omega _2} = ({I_1} + {I_2})\omega

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

{k_i} = {1 \over 2}{I_1}\omega _1^2 + {1 \over 2}{I_2}\omega _2^2

{k_f} = {1 \over 2}({I_1} + {I_2}){\omega ^2}

{k_i} - {k_f} = {1 \over 2}\left[ {{I_1}\omega _1^2 + {I_2}\omega _2^2 - {{{{({I_1}{\omega _1} + {I_2}{\omega _2})}^2}} \over {{I_1} + {I_2}}}} \right]

Solving above we get

{k_i} - {k_f} = {1 \over 2}\left( {{{{I_1}{I_2}} \over {{I_1} + {I_2}}}} \right){({\omega _1} - {\omega _2})^2}

Q40

A system consists of two identical spheres each of mass 1.5 kg and radius 50 cm at the end of light rod. The distance between the centres of the two spheres is 5 m. What will be the moment of inertia of the system about an axis perpendicular to the rod passing through its midpoint?

A 18.75 kgm2

B 1.905

\times

105 kgm2

C 19.05 kgm2

D 1.875

\times

105 kgm2

Correct Answer

Option C

Solution

M = 1.5 kg, r = 0.5 m, d =

{5 \over 2}

m

I = 2\left( {{2 \over 5}M{r^2} + M{d^2}} \right)

= 19.05 kgm2