Q: A round disc of moment of inertia I _ 2 about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia I _ 1 rotating with an angular velocit

Total angular momentum of the system initially $ L _ i = I _ 1 w + I _ 2 (0)= I _ 1 w $ Total angular momentum of the system finally $L_ f = (I_ 1 +I_ 2 ) w_ 2 $ According to conservation of angular momentum i.e. $ L _ i = L _ f $ $ aligned & I _ 1 = ( I _ 1 + I _ 2 ) _ 2 \\ & _ 2 = I _ 1 I _ 1 + I _ 2 aligned $

Question 1

The moment of inertia of a cube of mass m and side a about one of its edges is equal to

Accepted Answer

From theorem of perpendicular axes, we have aligned I & =I_C+m ( a 2 )^2 \ & = [ m a^2 12 + m a^2 12 ]+ m a^2 2 \ & = 2 3 m a^2 aligned

Question 2

The torque of a force 5 i +3 j -7 k about the origin is . If the force acts on a particle whose position vector is 2 i +2 j + k , then the value of will be :

Accepted Answer

Given that torque, =5 i +3 j -7 k Position vector, r =2 i ++2 j + k aligned & = r F = | array ccc i & j & k \ 2 & 2 & 1 \ 5 & 3 & -7 array | \ & = i (-14-3)- j (-14-5)+ k (6-10) \ & =-17 i +19 j -4 k aligned

Question 3

A disc is rolling without slipping on a surface. The radius of the disc is R. At t=0, the top most point on the disc is A as shown in figure. When the disc completes half of its rotation, the displace

Accepted Answer

(a) From figure, Displacement, BA = (2 R )^2+( R )^2 = R 4+ ^2

Question 4

Four holes of radius 5 ~cm are cut from a thin square plate of 20 ~cm and mass 1 ~kg . The moment of inertia of the remaining portion about Z-axis is

Accepted Answer

Area mass density, = M 16 R^2 ( Area =4 R 4 R=16 R^2 ) Mass of each hole m_1= R^2= M 16 R^2 R^2= M 16 Distance between centre of plate and centre of hole x= (2 R)^2+(2 R)^2 2 = 2 2 R 2 . Moment of inertia of one hole at about Z-axis I_1= 1 2 m_1 R^2+m_1 x^2= 5 32 M R^2 Moment of inertia of whole plate about Z-axis, I= M(4 R)^2 6 = 8 3 M R^2 Required moment of Inertia I_0=I-4 I_1,= [ 8 3 -4 ( 5 32 ) ] M R^2= [ 8 3 - 5 8 ] M R^2 Given, R=5 ~cm and M=1 ~kg So, I_0= [ 8 3 - 5 8 ] 1 25 10^ -4 =0.0017

Question 5

A solid sphere of 80 ~kg and radius 15 ~m moving in a space becomes a circular disc of radius 20 ~m in 1 ~h . The rate of change of moment of Inertia in this process is

Accepted Answer

Given, mass of solid sphere =80 ~kg radius of solid sphere, R_s=15 ~m radius of circular disc, R_c=20 ~m and time =1 hour =60 minute =60 60 sec Moment of inertia of solid sphere, I_s= 2 5 M R^2 aligned & = 2 5 80 (15)^2 \ & =7200 ~kg - m ^2 aligned Similarly, moment of inertia of the disc, aligned I_c & = 1 2 M R_c^2 \ & = 1 2 80 (20)^2 \ & =16000 ~kg - m ^2 aligned aligned Rate of change of moment of Inertia & = I_c-I_s t \ & = 16000-7200 60 60 \ & = 22 9 ~kg - m ^2 ~s ^ -1 aligned

Question 6

A round disc of moment of inertia I _ 2 about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia I _ 1 rotating with an angular velocit

Accepted Answer

Total angular momentum of the system initially $ L _ i = I _ 1 w + I _ 2 (0)= I _ 1 w $ Total angular momentum of the system finally $L_ f = (I_ 1 +I_ 2 ) w_ 2 $ According to conservation of angular momentum i.e. $ L _ i = L _ f $ $ aligned & I _ 1 = ( I _ 1 + I _ 2 ) _ 2 \ & _ 2 = I _ 1 I _ 1 + I _ 2 aligned $

Question 7

Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle A B C of side ~cm (as shown in the figure). The moment of inertia of the system about a line A X perpendic

Accepted Answer

I _ AX = m ( AB )^ 2 + m ( OC )^ 2 = m l^ 2 + m (l 60^ )^ 2 =m l^ 2 + m l^ 2 4 = 5 4 ~m l^ 2

Rotational Motion — Previous Year Questions & Solutions