Engineering Mechanics

Above figure is symmetrical about {x} axis so, moment of inertia of whole section will be two times of the one section above x-axis.

Moment of inertia of rectangular section OABC about y-axis i.e. OC.
=\frac{1}{3} \times \mathrm{OC} \times(\mathrm{AB})^{3}
=\frac{1}{3} \times d \times b^{3}
\mathrm{I}_{\mathrm{OC}}=\frac{1}{3} \times 3 \times(12)^{3}=1728

Moment of inertia of triangular section BCD about y-axis i.e. about DC
\mathrm{I}_{D C}=\frac{1}{12} \times(D C) \times(B C)^{3} =\frac{1}{12} \times 1.5 \times(12)^{3}=216

So, moment of inertia of tapered section OABD about y-axis:
\mathrm{I}_{\mathrm{y}^{\prime} \mathrm{y}^{\prime}}=\mathrm{I}_{\mathrm{OC}}-\mathrm{I}_{\mathrm{DC}}
=1728-216=1512
So, moment of inertia of whole tapered section
I_{y y}=2 l_{y^{\prime} y^{\prime}}=2 \times 1512=3024 \mathrm{~m}^{4}

Let '\theta' be the angle about line of impact through which ball will move often the impact.
Let 'u' be the vertical downward velocity of the ball before striking and 'v' be the velocity of ball after the impact which make an angle '\theta' with the line of impact. As ball fall freely under the gravity from height \mathrm{h}=4.9 \mathrm{m}, hence downward velocity '\mathrm{u}' at the instance of striking the rigid body
u=\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 4.9}
or, u=9.8 \mathrm{~m} / \mathrm{sec}
for perfectly elastic collision, e=1
As e=\frac{\text { relative velocity of seperation }}{\text { relative velocity of approach }}=1\;\;...(i)
As block is rigid, so block velocity =0.
So, along the line of impact,
relative velocity of approach =u \cos 30^{\circ}-0 =u \cos 30^{\circ}
relative velocity of separation along the line of impact =v \cos \theta-0=v \cos \theta
so, by equation (i)
u \cos 30^{\circ}=v \cos \theta \;\;\;...(ii)
In the direction normal to the impact, the component velocity is not affected so,
u \sin 30^{\circ}=v \sin \theta \;\;\;...(iii)
So, by equation (ii) and (iii)
V=\sqrt{u^{2} \cos ^{2} 30^{\circ}+u^{2} \sin ^{2} 30^{\circ}}
or, V=9.8 \mathrm{~m} / \mathrm{sec}
by equation (ii) and (iii)
\tan \theta=\tan 30^{\circ}, \theta=30^{\circ}
So, inclination to the plane for of the section
=90^{\circ}-30^{\circ}=60^{\circ}
So, momentum equation is given by during seperation:

\vec{P}_{s}=m v \cos 30^{\circ} \hat{i}+m v \cos 30^{\circ} \hat{j} =17 \hat{i}+9.8 \hat{j}

Nonrigid frames: Relative movement of joints is large then a frame said to be nonrigid.

Here, P force cause large deformation therefore. It is non rigid.

If at a point 3 member are meeting and two are colinear then in 3rd member force will be zero.
F_{GH}=0

Mass moment of inertia of block about hinge 'O' =\frac{M}{12}(h^2+b^2)+Mr^2
where, h=0.4m, b=0.3m
r=\sqrt{0.15^2+0.2^2}=0.25m


I_o=\frac{1}{2} \times (0.4^2+0.3^2)+1 \times 0.25^2=0.083 kg-m^2
For block to topple about 'O' it should just reach the position such that centre of mass reaches from point (G) to point (G')

[Beyond this the torque of gravity will rotate the block itself]
Now, y(increase in height of centre of mass)
= OG' - OP
= OG - OP
= 0.25 - 0.20 = 0.05

So the block should have initial angular velocity to just shift centre of mass at this position (i.e., G')
F.B.D of block is shown below.

\frac{d\vec{L}}{dt}=\vec{\tau _{ext}}\;(about\;\;O)
[ \vec{L} is angular momentum, \vec{\tau _{ext}} is external torque]
\tau _{ext} dt=d \vec{L}
F\Delta t \times d=L_f-L_i
(Because, Gravity force is small compared to impulsive force hence we neglect its torque)
I_f \times d =I_o \times \omega . . . (i)
By using conservation of mechanical energy
\frac{1}{2}I_o\omega ^2=mg\;y
\frac{1}{2}0.083 \times \omega ^2=1 \times 9.81 \times 0.05
\Rightarrow \omega =3.44 \; rad/s
Putting in eq.(i)
I_F \times 0.3 =0.083 \times 3.44
\Rightarrow I_F =0.951 N-s