Section -A
(Very Short Answer Type Questions)
Note:- Answer all questions. As per the nature of the question you delimit your answer in one word, one sentence or maximum up to 30 words. Each question carries 1 (one) mark.
(i). Write the moment of Inertia of a circular disc of mass MM and radius ‘ aa ‘ about a diameter.
(ii). Define moving axes and fixed axes.
(iii). Write the equation of Locus of an Invariable line.
(iv). Define the generalized co-ordinates.
Section-B
(Short Answer Questions)
Note :- Answer any two questions. Each answer should be given in 200 words. Each question carries 4 marks.
State and prove D^(‘)D^{\prime} Alembert’s principle.
A uniform rod of mass mm is placed at right angles to a smooth plane of inclination alpha\alpha with one end in contact with it. The rod is then released. Show that when the inclination to the plane is phi\phi, the reaction of the plane will be {(3(1-sin phi)^(2)+1)/((1+3cos^(2)phi)^(2))}mg cos alpha\left\{\frac{3(1-\sin \phi)^2+1}{\left(1+3 \cos ^2 \phi\right)^2}\right\} m g \cos \alpha.
A dice in the form of a portion of parabola bounded by its latus rectum and its axis has its vertex A fixed and is stuck by a blow through the end of its latus rectum perpendicular to its plane. Show that the dice starts revolving about a line through A inclined at an angle tan^(-1)((14)/(25))\tan ^{-1}\left(\frac{14}{25}\right) to the axis.
Use Lagrange’s equations to find the equation of motion of a simple pendulum.
Section-C
(Long Answer Questions)
Note :- Answer any one question. Each answer should be given in 800 words. Each question carries 08 marks.
(a) Three equal uniform rods AB,BC,DC\mathrm{AB}, \mathrm{BC}, \mathrm{DC} are smoothly jointed at B and C and the ends A and D are fastened to smooth fixed points whose distance a part is equal to the length of either rod. The frame being at rest in the form of a square. A blow I is given perpendicular to AB at its middle point and in the plane of the square. Show that the energy set up is (3I^(2))/(40(m))\frac{3 I^2}{40 \mathrm{~m}}, where m is the mass of each rod. Find also the blows at the joints A and C .
(b) Deduce the Lagrange’s equations from Hamilton’s Principle.
(a) A mass of fluid is in motion such that the lines of motion lie on the surface of coaxial cylinders, show that the equation of continuity is (del P)/(del t)+(1)/(r)(del(Pu))/(del theta)+(del(Pv))/(del z)=0\frac{\partial P}{\partial t}+\frac{1}{r} \frac{\partial(P u)}{\partial \theta}+\frac{\partial(P v)}{\partial z}=0. where u,v\mathrm{u}, \mathrm{v} are the velocity perpendicular and parallel to z .
(b) Find the Cauchy-Riemann equations in polar coordinates.
Answer:
Question:-01(a)
Write the moment of Inertia of a circular disc of mass MM and radius ‘ aa ‘ about a diameter.
Answer:
The moment of inertia of a circular disc of mass MM and radius aa about a diameter is:
I_(“diameter”)=(1)/(4)Ma^(2).I_{\text{diameter}} = \frac{1}{4} M a^2.
Derivation:
Moment of Inertia About the Axis Perpendicular to the Plane:
The moment of inertia of a circular disc about an axis perpendicular to its plane and passing through its center is:
I_(“perpendicular”)=(1)/(2)Ma^(2).I_{\text{perpendicular}} = \frac{1}{2} M a^2.
Using the Perpendicular Axis Theorem:
The perpendicular axis theorem states:
I_(z)=I_(x)+I_(y),I_z = I_x + I_y,
where I_(z)I_z is the moment of inertia about the axis perpendicular to the plane (through the center), and I_(x),I_(y)I_x, I_y are the moments of inertia about two perpendicular axes in the plane of the disc (e.g., diameters).
(1)/(2)Ma^(2)=I_(“diameter”)+I_(“diameter”).\frac{1}{2} M a^2 = I_{\text{diameter}} + I_{\text{diameter}}.
Solve for I_(“diameter”)I_{\text{diameter}}:
2I_(“diameter”)=(1)/(2)Ma^(2),2 I_{\text{diameter}} = \frac{1}{2} M a^2,
I_(“diameter”)=(1)/(4)Ma^(2).I_{\text{diameter}} = \frac{1}{4} M a^2.
Final Result:
The moment of inertia of a circular disc about a diameter is:
I_(“diameter”)=(1)/(4)Ma^(2).I_{\text{diameter}} = \frac{1}{4} M a^2.
Question:-01(b)
Define moving axes and fixed axes.
Answer:
Fixed Axes:
Definition: Fixed axes refer to a coordinate system or a set of reference axes that remain stationary in space, regardless of the motion of the object being studied.
Key Characteristics:
The axes do not move or rotate.
They provide an external frame of reference.
Typically used in problems involving motion relative to an inertial or non-inertial frame.
Applications:
Describing motion of objects relative to the ground.
Used in Newtonian mechanics for describing motion in an inertial frame of reference.
Example:
The xx, yy, and zz axes of a Cartesian coordinate system fixed relative to the Earth.
Moving Axes:
Definition: Moving axes refer to a coordinate system that moves or rotates along with the object being studied.
Key Characteristics:
The axes are attached to the moving body and follow its motion.
They are often used to simplify the description of motion relative to the body.
May involve translation, rotation, or both.
Applications:
Studying the motion of rigid bodies.
Used in problems involving rotating reference frames, such as gyroscopes or rotating machinery.
Essential in dynamics for analyzing relative motion.
Example:
Axes attached to a rolling wheel or a rotating spacecraft.
Question:-01(c)
Write the equation of Locus of an Invariable line.
Answer:
The locus of the invariable line can also be expressed in symmetric form:
Here, L_(x),L_(y),L_(z)L_x, L_y, L_z are constant because the angular momentum vector L\mathbf{L} remains fixed in the inertial frame.
Question:-01(d)
Define the generalized co-ordinates.
Answer:
Generalized Coordinates
Generalized coordinates are a set of parameters used to describe the configuration of a mechanical system in terms of its degrees of freedom (DOFs). They provide a minimal and efficient way to represent the system’s position, especially when the system is constrained.
Definition:
A set of generalized coordinates {q_(1),q_(2),dots,q_(n)}\left\{q_1, q_2, \ldots, q_n\right\} are independent variables that uniquely specify the configuration of a mechanical system with nn degrees of freedom.
Question:-02
State and prove D^(‘)D^{\prime} Alembert’s principle.
Answer:
D’Alembert’s Principle
Statement:
D’Alembert’s principle is a reformulation of Newton’s second law of motion that allows us to incorporate dynamics into the framework of statics. It states:
The difference between the applied forces and the inertial forces acting on a system is in equilibrium. Mathematically:
sumF_(“applied”)-sum ma=0,\sum \mathbf{F}_\text{applied} – \sum m \mathbf{a} = 0,
This equation shows that the total applied force F_(i)\mathbf{F}_i and the inertial force F_(“inertial,i”)\mathbf{F}_\text{inertial,i} are in equilibrium.
Step 3: Principle of Virtual Work
Consider a virtual displacement deltar_(i)\delta \mathbf{r}_i for the ii-th particle, which is consistent with the constraints of the system. The principle of virtual work states:
This is the mathematical statement of D’Alembert’s principle, which states that the dynamic system can be treated as a system in static equilibrium when inertial forces are included.
Applications of D’Alembert’s Principle
Rigid Body Dynamics: Simplifies the analysis of rotating and translating bodies.
Lagrangian Mechanics: Provides the foundation for deriving Lagrange’s equations of motion.
Constraint Systems: Easily handles systems with holonomic or non-holonomic constraints.
Conclusion
D’Alembert’s principle transforms a dynamic problem into a static equilibrium problem by introducing inertial forces, enabling a powerful method for analyzing mechanical systems. It is particularly useful in constrained motion and forms the basis of advanced formulations like Lagrangian and Hamiltonian mechanics.
Question:-03
A uniform rod of mass mm is placed at right angles to a smooth plane of inclination alpha\alpha with one end in contact with it. The rod is then released. Show that when the inclination to the plane is phi\phi, the reaction of the plane will be {(3(1-sin phi)^(2)+1)/((1+3cos^(2)phi)^(2))}mg cos alpha\left\{\frac{3(1-\sin \phi)^2+1}{\left(1+3 \cos ^2 \phi\right)^2}\right\} m g \cos \alpha.
Answer:
We are analyzing the motion of a uniform rod BCBC of mass mm and length 2a2a, placed at right angles to a smooth inclined plane ABAB. The inclined plane is at an angle alpha\alpha with the horizontal. The rod is initially at rest, and its center of mass GG is located at a distance BG=aBG = a from point BB. The vertical height of the center of mass GG above the inclined plane is:
GH=a cos alphaquad(from (/_\BGH)”)”.GH = a \cos \alpha \quad \text{(from \(\triangle BGH\))}.
When the rod is released, it begins to rotate and slide. After a time tt, let the rod make an angle phi\phi with the inclined plane ABAB. We analyze the motion of the center of mass GG and the forces acting on the rod.
Coordinates of the Center of Mass:
We define the axes with BABA as the horizontal axis and the perpendicular to BABA as the vertical axis. The coordinates of the center of mass GG are:
x=BN=a cos phi,quad y=GN=a sin phiquad(from (/_\BGN)”)”.x = BN = a \cos \phi, \quad y = GN = a \sin \phi \quad \text{(from \(\triangle BGN\))}.
Equation of Motion Perpendicular to the Plane:
The motion of the center of mass GG is constrained to be perpendicular to the inclined plane ABAB. The equation of motion along the yy-direction is:
my^(¨)=m(a cos phi(phi^(¨))-a sin phiphi^(˙)^(2))=-R-mg cos alpham \ddot{y} = m \left(a \cos \phi \ddot{\phi} – a \sin \phi \dot{\phi}^2 \right) = -R – mg \cos \alpha
where:
RR is the normal reaction force exerted by the inclined plane on the rod,
mg cos alphamg \cos \alpha is the component of the gravitational force perpendicular to the plane.
Moment About the Center of Mass GG:
Taking moments about the center of mass GG, the equation of rotational motion is:
m(a^(2))/(3)phi^(¨)=-Ra cos phim \frac{a^2}{3} \ddot{\phi} = -R a \cos \phi
where (a^(2))/(3)\frac{a^2}{3} is the moment of inertia of the rod about its center.
Vertical Height of the Center of Mass:
Initially, the vertical height of the center of mass GG above the inclined plane was:
GH=a cos alphaGH = a \cos \alpha
At any time tt, the vertical height of GG above the inclined plane is:
GQ=GN cos alpha=(a sin phi)cos alphaGQ = GN \cos \alpha = (a \sin \phi) \cos \alpha
The distance fallen by GG in the downward direction is:
“Distance fallen”=GH-GQ=a cos alpha-a sin phi cos alpha=a cos alpha(1-sin phi)\text{Distance fallen} = GH – GQ = a \cos \alpha – a \sin \phi \cos \alpha = a \cos \alpha (1 – \sin \phi)
Energy Equation:
The total energy of the system is conserved. The work done by gravity is equal to the kinetic energy of the rod. The energy equation is:
(m)/(2)[v^(2)+k^(2)phi^(˙)^(2)]=”Work done by gravity”\frac{m}{2} \left[v^2 + k^2 \dot{\phi}^2 \right] = \text{Work done by gravity}
where:
v^(2)=y^(˙)^(2)=(a cos phiphi^(˙))^(2)v^2 = \dot{y}^2 = (a \cos \phi \dot{\phi})^2,
k^(2)phi^(˙)^(2)=(a^(2))/(3)phi^(˙)^(2)k^2 \dot{\phi}^2 = \frac{a^2}{3} \dot{\phi}^2 is the rotational kinetic energy.
Substituting these expressions, the energy equation becomes:
(m)/(2)[a^(2)cos^(2)phiphi^(˙)^(2)+(a^(2))/(3)phi^(˙)^(2)]=mga cos alpha(1-sin phi)\frac{m}{2} \left[a^2 \cos^2 \phi \dot{\phi}^2 + \frac{a^2}{3} \dot{\phi}^2 \right] = m g a \cos \alpha (1 – \sin \phi)
Simplifying:
(ma^(2))/(6)[3cos^(2)phi+1]phi^(˙)^(2)=mga cos alpha(1-sin phi)\frac{m a^2}{6} \left[3 \cos^2 \phi + 1 \right] \dot{\phi}^2 = m g a \cos \alpha (1 – \sin \phi)