Question 1 |

The area moment of inertia about the y-axis of a linearly tapered section shown in
the figure is ______ m^4.
(Answer in integer)

1254 | |

3625 | |

3024 | |

2542 |

Question 1 Explanation:

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}

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}

Question 2 |

A spherical ball weighing 2 kg is dropped from a height of 4.9 m onto an
immovable rigid block as shown in the figure. If the collision is perfectly elastic,
what is the momentum vector of the ball (in kg m/s) just after impact?

Take the acceleration due to gravity to be g=9.8 m/s^2. Options have been rounded off to one decimal place.

Take the acceleration due to gravity to be g=9.8 m/s^2. Options have been rounded off to one decimal place.

19.6\hat{i} | |

19.6\hat{j} | |

17.0\hat{i}+9.8\hat{j} | |

9.8\hat{i}+17.0\hat{j} |

Question 2 Explanation:

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}

Question 3 |

The options show frames consisting of rigid bars connected by pin joints. Which
one of the frames is non-rigid?

A | |

B | |

C | |

D |

Question 3 Explanation:

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.

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

Question 4 |

The lengths of members BC and CE in the frame
shown in the figure are equal. All the members are
rigid and lightweight, and the friction at the joints
is negligible. Two forces of magnitude Q \gt 0 are
applied as shown, each at the mid-length of the
respective member on which it acts.

Which one or more of the following members do not carry any load (force)?

Which one or more of the following members do not carry any load (force)?

AB | |

CD | |

EF | |

GH |

Question 4 Explanation:

If at a point 3 member are meeting and two are colinear then in 3rd member force will be zero.

F_{GH}=0

Question 5 |

A rigid homogeneous uniform block of mass 1 kg,
height h= 0.4 m and width b= 0.3 m is pinned at one
corner and placed upright in a uniform gravitational
field (g = 9.81 m/s^2), supported by a roller in the
configuration shown in the figure. A short duration
(impulsive) force F, producing an impulse I_F, is
applied at a height of d = 0.3 m from the bottom
as shown. Assume all joints to be frictionless. The
minimum value of I_F
required to topple the block is

0.953 Ns | |

1.403 Ns | |

0.814 Ns | |

1.172 ns |

Question 5 Explanation:

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

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

There are 5 questions to complete.

Excuse me sir where is Engineering economic 2021 set1 and set 2 questions?

Sorry engineering that is mechanics!