1. (i). Write the moment of Inertia of a circular disc of mass $M$$M$MM$M$ and radius ‘ $a$$a$aa$a$ ‘ 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.

2. State and prove $D^{\prime }$$D^{\prime }$D^(‘)D^{\prime}$D^{\prime }$ Alembert’s principle.
3. A uniform rod of mass $m$$m$mm$m$ is placed at right angles to a smooth plane of inclination $\alpha$$\alpha$alpha\alpha$\alpha$ with one end in contact with it. The rod is then released. Show that when the inclination to the plane is $\varphi$$\varphi$phi\phi$\varphi$, the reaction of the plane will be $\left\{\frac{3(1-\sin \varphi {)}^2+1}{{(1+3{\cos }^2\varphi )}^2}\right\}mg\cos \alpha$$\left\{\frac{3(1-\sin \varphi {)}^2+1}{{\left(1+3{\cos }^2\varphi \right)}^2}\right\}mg\cos \alpha${(3(1-sin phi)^(2)+1)/((1+3cos^(2)phi)^(2))}mg cos alpha\left\{\frac{3(1-\sin \phi)^2+1}{\left(1+3 \cos ^2 \phi\right)^2}\right\} m g \cos \alpha$\left\{\frac{3(1-\sin \varphi {)}^2+1}{{(1+3{\cos }^2\varphi )}^2}\right\}mg\cos \alpha$.
4. A dice in the form of a portion of parabola bounded by its latus rectum and its axis has its vertex A fixed and is stuck by a blow through the end of its latus rectum perpendicular to its plane. Show that the dice starts revolving about a line through A inclined at an angle ${\tan }^{-1}\left(\frac{14}{25}\right)$${\tan }^{-1}\left(\frac{14}{25}\right)$tan^(-1)((14)/(25))\tan ^{-1}\left(\frac{14}{25}\right)${\tan }^{-1}\left(\frac{14}{25}\right)$ to the axis.
5. Use Lagrange’s equations to find the equation of motion of a simple pendulum.

6. (a) Three equal uniform rods $\mathrm{AB},\mathrm{BC},\mathrm{DC}$$\mathrm{AB},\mathrm{BC},\mathrm{DC}$AB,BC,DC\mathrm{AB}, \mathrm{BC}, \mathrm{DC}$\mathrm{AB},\mathrm{BC},\mathrm{DC}$ are smoothly jointed at B and C and the ends A and D are fastened to smooth fixed points whose distance a part is equal to the length of either rod. The frame being at rest in the form of a square. A blow I is given perpendicular to AB at its middle point and in the plane of the square. Show that the energy set up is $\frac{3I^2}{40\mathrm{m}}$$\frac{3I^2}{40\mathrm{m}}$(3I^(2))/(40(m))\frac{3 I^2}{40 \mathrm{~m}}$\frac{3I^2}{40\mathrm{m}}$, where m is the mass of each rod. Find also the blows at the joints A and C .

7. (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 $\frac{\partial P}{\partial t}+\frac{1}{r}\frac{\partial (Pu)}{\partial \theta }+\frac{\partial (Pv)}{\partial z}=0$$\frac{\partial P}{\partial t}+\frac{1}{r}\frac{\partial (Pu)}{\partial \theta }+\frac{\partial (Pv)}{\partial z}=0$(del P)/(del t)+(1)/(r)(del(Pu))/(del theta)+(del(Pv))/(del z)=0\frac{\partial P}{\partial t}+\frac{1}{r} \frac{\partial(P u)}{\partial \theta}+\frac{\partial(P v)}{\partial z}=0$\frac{\partial P}{\partial t}+\frac{1}{r}\frac{\partial (Pu)}{\partial \theta }+\frac{\partial (Pv)}{\partial z}=0$. where $\mathrm{u},\mathrm{v}$$\mathrm{u},\mathrm{v}$u,v\mathrm{u}, \mathrm{v}$\mathrm{u},\mathrm{v}$ are the velocity perpendicular and parallel to z .

Answer:

Question:-01(a)

Write the moment of Inertia of a circular disc of mass $M$$M$MM$M$ and radius ‘ $a$$a$aa$a$ ‘ about a diameter.

Answer:

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Derivation:

1. 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:
   $$I_{\text{perpendicular}}=\frac{1}{2}M{a}^2.$$$$I_{\text{perpendicular}}=\frac{1}{2}M{a}^2.$$I_(“perpendicular”)=(1)/(2)Ma^(2).I_{\text{perpendicular}} = \frac{1}{2} M a^2.$$I_{\text{perpendicular}}=\frac{1}{2}M{a}^2.$$
2. Using the Perpendicular Axis Theorem:
   The perpendicular axis theorem states:
   $$I_z=I_x+I_y,$$$$I_z=I_x+I_y,$$I_(z)=I_(x)+I_(y),I_z = I_x + I_y,$$I_z=I_x+I_y,$$
   where $I_z$$I_z$I_(z)I_z$I_z$ is the moment of inertia about the axis perpendicular to the plane (through the center), and $I_x,I_y$$I_x,I_y$I_(x),I_(y)I_x, I_y$I_x,I_y$ are the moments of inertia about two perpendicular axes in the plane of the disc (e.g., diameters).
   For a circular disc, due to symmetry:
   $$I_x=I_y=I_{\text{diameter}}.$$$$I_x=I_y=I_{\text{diameter}}.$$I_(x)=I_(y)=I_(“diameter”).I_x = I_y = I_{\text{diameter}}.$$I_x=I_y=I_{\text{diameter}}.$$
   Substituting into the perpendicular axis theorem:
   $$\frac{1}{2}M{a}^2=I_{\text{diameter}}+I_{\text{diameter}}.$$$$\frac{1}{2}M{a}^2=I_{\text{diameter}}+I_{\text{diameter}}.$$(1)/(2)Ma^(2)=I_(“diameter”)+I_(“diameter”).\frac{1}{2} M a^2 = I_{\text{diameter}} + I_{\text{diameter}}.$$\frac{1}{2}M{a}^2=I_{\text{diameter}}+I_{\text{diameter}}.$$
3. Solve for $I_{\text{diameter}}$$I_{\text{diameter}}$I_(“diameter”)I_{\text{diameter}}$I_{\text{diameter}}$:
   $$2I_{\text{diameter}}=\frac{1}{2}M{a}^2,$$$$2I_{\text{diameter}}=\frac{1}{2}M{a}^2,$$2I_(“diameter”)=(1)/(2)Ma^(2),2 I_{\text{diameter}} = \frac{1}{2} M a^2,$$2I_{\text{diameter}}=\frac{1}{2}M{a}^2,$$
   $$I_{\text{diameter}}=\frac{1}{4}M{a}^2.$$$$I_{\text{diameter}}=\frac{1}{4}M{a}^2.$$I_(“diameter”)=(1)/(4)Ma^(2).I_{\text{diameter}} = \frac{1}{4} M a^2.$$I_{\text{diameter}}=\frac{1}{4}M{a}^2.$$

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Final Result:

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 $x$$x$xx$x$, $y$$y$yy$y$, and $z$$z$zz$z$ axes of a Cartesian coordinate system fixed relative to the Earth.

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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.

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Question:-01(c)

Write the equation of Locus of an Invariable line.

Answer:

Question:-01(d)

Define the generalized co-ordinates.

Answer:

Generalized Coordinates

Definition:

Question:-02

State and prove $D^{\prime }$$D^{\prime }$D^(‘)D^{\prime}$D^{\prime }$ Alembert’s principle.

Answer:

D’Alembert’s Principle

• ${\mathbf{F}}_{\text{applied}}$${\mathbf{F}}_{\text{applied}}$F_(“applied”)\mathbf{F}_\text{applied}${\mathbf{F}}_{\text{applied}}$ is the vector sum of all applied forces acting on the system,
• ${\mathbf{F}}_{\text{inertial}}=-m\mathbf{a}$${\mathbf{F}}_{\text{inertial}}=-m\mathbf{a}$F_(“inertial”)=-ma\mathbf{F}_\text{inertial} = -m \mathbf{a}${\mathbf{F}}_{\text{inertial}}=-m\mathbf{a}$ is the inertial force (negative of mass times acceleration),
• $m$$m$mm$m$ is the mass of the particle, and $\mathbf{a}$$\mathbf{a}$a\mathbf{a}$\mathbf{a}$ is its acceleration.

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Proof

Step 1: Newton’s Second Law

• ${\mathbf{F}}_i$${\mathbf{F}}_i$F_(i)\mathbf{F}_i${\mathbf{F}}_i$ is the total force acting on the $i$$i$ii$i$-th particle,
• $m_i$$m_i$m_(i)m_i$m_i$ is the mass of the $i$$i$ii$i$-th particle,
• ${\mathbf{a}}_i$${\mathbf{a}}_i$a_(i)\mathbf{a}_i${\mathbf{a}}_i$ is the acceleration of the $i$$i$ii$i$-th particle.

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Step 2: Include Inertial Forces

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Step 3: Principle of Virtual Work

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Step 4: Generalize for the System

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Applications of D’Alembert’s Principle

1. Rigid Body Dynamics: Simplifies the analysis of rotating and translating bodies.
2. Lagrangian Mechanics: Provides the foundation for deriving Lagrange’s equations of motion.
3. Constraint Systems: Easily handles systems with holonomic or non-holonomic constraints.

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Conclusion

Question:-03

A uniform rod of mass $m$$m$mm$m$ is placed at right angles to a smooth plane of inclination $\alpha$$\alpha$alpha\alpha$\alpha$ with one end in contact with it. The rod is then released. Show that when the inclination to the plane is $\varphi$$\varphi$phi\phi$\varphi$, the reaction of the plane will be $\left\{\frac{3(1-\sin \varphi {)}^2+1}{{(1+3{\cos }^2\varphi )}^2}\right\}mg\cos \alpha$$\left\{\frac{3(1-\sin \varphi {)}^2+1}{{\left(1+3{\cos }^2\varphi \right)}^2}\right\}mg\cos \alpha${(3(1-sin phi)^(2)+1)/((1+3cos^(2)phi)^(2))}mg cos alpha\left\{\frac{3(1-\sin \phi)^2+1}{\left(1+3 \cos ^2 \phi\right)^2}\right\} m g \cos \alpha$\left\{\frac{3(1-\sin \varphi {)}^2+1}{{(1+3{\cos }^2\varphi )}^2}\right\}mg\cos \alpha$.

Answer:

Coordinates of the Center of Mass:

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Equation of Motion Perpendicular to the Plane:

• $R$$R$RR$R$ is the normal reaction force exerted by the inclined plane on the rod,
• $mg\cos \alpha$$mg\cos \alpha$mg cos alphamg \cos \alpha$mg\cos \alpha$ is the component of the gravitational force perpendicular to the plane.

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Moment About the Center of Mass $G$$G$GG$G$:

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Vertical Height of the Center of Mass:

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Energy Equation:

• $v^2={\dot{y}}^2=(a\cos \varphi \dot{\varphi }{)}^2$$v^2={\dot{y}}^2=(a\cos \varphi \dot{\varphi }{)}^2$v^(2)=y^(˙)^(2)=(a cos phiphi^(˙))^(2)v^2 = \dot{y}^2 = (a \cos \phi \dot{\phi})^2$v^2={\dot{y}}^2=(a\cos \varphi \dot{\varphi }{)}^2$,
• $k^2{\dot{\varphi }}^2=\frac{a^2}{3}{\dot{\varphi }}^2$$k^2{\dot{\varphi }}^2=\frac{a^2}{3}{\dot{\varphi }}^2$k^(2)phi^(˙)^(2)=(a^(2))/(3)phi^(˙)^(2)k^2 \dot{\phi}^2 = \frac{a^2}{3} \dot{\phi}^2$k^2{\dot{\varphi }}^2=\frac{a^2}{3}{\dot{\varphi }}^2$ is the rotational kinetic energy.

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Angular Acceleration:

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Reaction Force $R$$R$RR$R$:

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Final Results:

1. Angular Velocity:
   $${\dot{\varphi }}^2=\frac{6g}{a}\frac{(1-\sin \varphi )}{(1+3{\cos }^2\varphi )}\cos \alpha$$$${\dot{\varphi }}^2=\frac{6g}{a}\frac{(1-\sin \varphi )}{(1+3{\cos }^2\varphi )}\cos \alpha$$phi^(˙)^(2)=(6g)/(a)((1-sin phi))/((1+3cos^(2)phi))cos alpha\dot{\phi}^2 = \frac{6g}{a} \frac{(1 – \sin \phi)}{(1 + 3 \cos^2 \phi)} \cos \alpha$${\dot{\varphi }}^2=\frac{6g}{a}\frac{(1-\sin \varphi )}{(1+3{\cos }^2\varphi )}\cos \alpha$$
2. Angular Acceleration:
   $$\ddot{\varphi }=-\frac{3g\cos \alpha \cos \varphi }{a(1+3{\cos }^2\varphi {)}^2}[1+3(1-\sin \varphi {)}^2]$$$$\ddot{\varphi }=-\frac{3g\cos \alpha \cos \varphi }{a(1+3{\cos }^2\varphi {)}^2}\left[1+3(1-\sin \varphi {)}^2\right]$$phi^(¨)=-(3g cos alpha cos phi)/(a(1+3cos^(2)phi)^(2))[1+3(1-sin phi)^(2)]\ddot{\phi} = -\frac{3g \cos \alpha \cos \phi}{a (1 + 3 \cos^2 \phi)^2} \left[1 + 3 (1 – \sin \phi)^2 \right]$$\ddot{\varphi }=-\frac{3g\cos \alpha \cos \varphi }{a(1+3{\cos }^2\varphi {)}^2}[1+3(1-\sin \varphi {)}^2]$$
3. Normal Reaction Force:
   $$R=mg\left\{\frac{3(1-\sin \varphi {)}^2+1}{(1+3{\cos }^2\varphi {)}^2}\right\}\cos \alpha$$$$R=mg\left\{\frac{3(1-\sin \varphi {)}^2+1}{(1+3{\cos }^2\varphi {)}^2}\right\}\cos \alpha$$R=mg{(3(1-sin phi)^(2)+1)/((1+3cos^(2)phi)^(2))}cos alphaR = mg \left\{\frac{3 (1 – \sin \phi)^2 + 1}{(1 + 3 \cos^2 \phi)^2} \right\} \cos \alpha$$R=mg\left\{\frac{3(1-\sin \varphi {)}^2+1}{(1+3{\cos }^2\varphi {)}^2}\right\}\cos \alpha$$

Question:-04

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}\left(\frac{14}{25}\right)$${\tan }^{-1}\left(\frac{14}{25}\right)$tan^(-1)((14)/(25))\tan ^{-1}\left(\frac{14}{25}\right)${\tan }^{-1}\left(\frac{14}{25}\right)$ to the axis.

Answer:

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Moments of Inertia:

1. Moment of Inertia About $AX$$AX$AXAX$AX$:

2. Moment of Inertia About $AZ$$AZ$AZAZ$AZ$:

3. Moment of Inertia About $AY$$AY$AYAY$AY$:

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Products of Inertia:

1. Products of Inertia $D$$D$DD$D$, $E$$E$EE$E$, and $F$$F$FF$F$:

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Using Euler’s Equations for Impulsive Forces:

1. For rotation about the $AX$$AX$AXAX$AX$-axis:
   $$\frac{16}{15}\rho a^4{w}_x^{\prime }-\frac{2}{3}{a}^4\rho w_z^{\prime }=-P\cdot 2a$$$$\frac{16}{15}\rho a^4{w}_x^{\prime }-\frac{2}{3}{a}^4\rho w_z^{\prime }=-P\cdot 2a$$(16)/(15)rhoa^(4)w_(x)^(‘)-(2)/(3)a^(4)rhow_(z)^(‘)=-P*2a\frac{16}{15} \rho a^4 w_x’ – \frac{2}{3} a^4 \rho w_z’ = -P \cdot 2a$$\frac{16}{15}\rho a^4{w}_x^{\prime }-\frac{2}{3}{a}^4\rho w_z^{\prime }=-P\cdot 2a$$
2. For rotation about the $AY$$AY$AYAY$AY$-axis:
   $$\frac{172}{105}\rho a^4{w}_y^{\prime }=0$$$$\frac{172}{105}\rho a^4{w}_y^{\prime }=0$$(172)/(105)rhoa^(4)w_(y)^(‘)=0\frac{172}{105} \rho a^4 w_y’ = 0$$\frac{172}{105}\rho a^4{w}_y^{\prime }=0$$
   Since $w_y^{\prime }=0$$w_y^{\prime }=0$w_(y)^(‘)=0w_y’ = 0$w_y^{\prime }=0$, there is no rotation about the $AY$$AY$AYAY$AY$-axis.
3. For rotation about the $AZ$$AZ$AZAZ$AZ$-axis:
   $$\frac{4}{7}\rho a^4{w}_z^{\prime }-\frac{2}{3}\rho a^4{w}_x^{\prime }=P\cdot a$$$$\frac{4}{7}\rho a^4{w}_z^{\prime }-\frac{2}{3}\rho a^4{w}_x^{\prime }=P\cdot a$$(4)/(7)rhoa^(4)w_(z)^(‘)-(2)/(3)rhoa^(4)w_(x)^(‘)=P*a\frac{4}{7} \rho a^4 w_z’ – \frac{2}{3} \rho a^4 w_x’ = P \cdot a$$\frac{4}{7}\rho a^4{w}_z^{\prime }-\frac{2}{3}\rho a^4{w}_x^{\prime }=P\cdot a$$

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Eliminating $P$$P$PP$P$:

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Angle of Rotation:

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Question:-05

Use Lagrange’s equations to find the equation of motion of a simple pendulum.

Answer:

Equation of Motion of a Simple Pendulum Using Lagrange’s Equations

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Step 1: Generalized Coordinate

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Step 2: Kinetic Energy ($T$$T$TT$T$)

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Step 3: Potential Energy ($V$$V$VV$V$)

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Step 4: Lagrangian ($L$$L$LL$L$)

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Step 5: Lagrange’s Equation

1. Compute $\frac{\partial L}{\partial \dot{\theta }}$$\frac{\partial L}{\partial \dot{\theta }}$(del L)/(del(theta^(˙)))\frac{\partial L}{\partial \dot{\theta}}$\frac{\partial L}{\partial \dot{\theta }}$:

2. Compute $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta }}\right)$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta }}\right)$(d)/(dt)((del L)/(del(theta^(˙))))\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right)$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta }}\right)$:

3. Compute $\frac{\partial L}{\partial \theta }$$\frac{\partial L}{\partial \theta }$(del L)/(del theta)\frac{\partial L}{\partial \theta}$\frac{\partial L}{\partial \theta }$:

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Step 6: Lagrange’s Equation

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Step 7: Small Angle Approximation (Optional)

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Final Equation:

Question:-06(a)

Three equal uniform rods $\mathrm{AB},\mathrm{BC},\mathrm{DC}$$\mathrm{AB},\mathrm{BC},\mathrm{DC}$AB,BC,DC\mathrm{AB}, \mathrm{BC}, \mathrm{DC}$\mathrm{AB},\mathrm{BC},\mathrm{DC}$ are smoothly jointed at B and C and the ends A and D are fastened to smooth fixed points whose distance apart is equal to the length of either rod. The frame being at rest in the form of a square. A blow $I$$I$II$I$ is given perpendicular to AB at its middle point and in the plane of the square. Show that the energy set up is $\frac{3I^2}{40\mathrm{m}}$$\frac{3I^2}{40\mathrm{m}}$(3I^(2))/(40(m))\frac{3 I^2}{40 \mathrm{~m}}$\frac{3I^2}{40\mathrm{m}}$, where $m$$m$mm$m$ is the mass of each rod. Find also the blows at the joints A and C.

Answer:

Motion of the Rods:

• After the blow, the velocities of the rods are as follows:
  • Rod $AB$$AB$ABAB$AB$: Velocity = $a\dot{\theta }$$a\dot{\theta }$atheta^(˙)a \dot{\theta}$a\dot{\theta }$
  • Rod $CD$$CD$CDCD$CD$: Velocity = $a\dot{\theta }$$a\dot{\theta }$atheta^(˙)a \dot{\theta}$a\dot{\theta }$
  • Rod $BC$$BC$BCBC$BC$: Velocity = $2a\dot{\theta }$$2a\dot{\theta }$2atheta^(˙)2a \dot{\theta}$2a\dot{\theta }$

Kinetic Energy of the System:

Kinetic Energy of Rod $AB$$AB$ABAB$AB$:

• Translational K.E. = $\frac{m}{2}(a\dot{\theta }{)}^2$$\frac{m}{2}(a\dot{\theta }{)}^2$(m)/(2)(atheta^(˙))^(2)\frac{m}{2} (a \dot{\theta})^2$\frac{m}{2}(a\dot{\theta }{)}^2$
• Rotational K.E. = $\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{m(2a{)}^2}{3}\right){\dot{\theta }}^2=\frac{m{a}^2}{3}{\dot{\theta }}^2$$\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{m(2a{)}^2}{3}\right){\dot{\theta }}^2=\frac{m{a}^2}{3}{\dot{\theta }}^2$(1)/(2)Iomega^(2)=(1)/(2)((m(2a)^(2))/(3))theta^(˙)^(2)=(ma^(2))/(3)theta^(˙)^(2)\frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{m (2a)^2}{3}\right) \dot{\theta}^2 = \frac{m a^2}{3} \dot{\theta}^2$\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{m(2a{)}^2}{3}\right){\dot{\theta }}^2=\frac{m{a}^2}{3}{\dot{\theta }}^2$

Kinetic Energy of Rod $BC$$BC$BCBC$BC$:

Total Kinetic Energy:

Virtual Work Done by the Impulse:

Lagrange’s Equation:

Energy Setup by the Blow:

Impulses at Points $B$$B$BB$B$ and $C$$C$CC$C$:

For Rod $AB$$AB$ABAB$AB$:

For Rod $CD$$CD$CDCD$CD$:

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Final Results:

1. Angular velocity after the blow: $\dot{\theta }=\frac{3I}{20ma}$$\dot{\theta }=\frac{3I}{20ma}$theta^(˙)=(3I)/(20 ma)\dot{\theta} = \frac{3I}{20ma}$\dot{\theta }=\frac{3I}{20ma}$
2. Kinetic energy of the system: $T=\frac{3I^2}{40m}$$T=\frac{3I^2}{40m}$T=(3I^(2))/(40 m)T = \frac{3I^2}{40m}$T=\frac{3I^2}{40m}$
3. Impulses:
   • $I_B=\frac{2}{5}I$$I_B=\frac{2}{5}I$I_(B)=(2)/(5)II_B = \frac{2}{5} I$I_B=\frac{2}{5}I$
   • $I_C=\frac{I}{10}$$I_C=\frac{I}{10}$I_(C)=(I)/( 10)I_C = \frac{I}{10}$I_C=\frac{I}{10}$

Question:-06(b)

Deduce the Lagrange’s equations from Hamilton’s Principle.

Answer:

• $T$$T$TT$T$ represents the kinetic energy of the system.
• $V$$V$VV$V$ represents the potential energy of the system.
• The Lagrangian function $L$$L$LL$L$ is defined as:$$L=T-V$$$$L=T-V$$L=T-VL = T – V$$L=T-V$$

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Hamilton’s Principle:

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Integration by Parts:

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Lagrange’s Equations of Motion:

Question:-07(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:

Answer:

• $PQ=\delta r$$PQ=\delta r$PQ=delta rPQ = \delta r$PQ=\delta r$ (along the radial direction),
• $PS=r\delta \theta$$PS=r\delta \theta$PS=r delta thetaPS = r \delta \theta$PS=r\delta \theta$ (along the angular direction),
• $P{P}^{\prime }=\delta z$$P{P}^{\prime }=\delta z$PP^(‘)=delta zPP’ = \delta z$P{P}^{\prime }=\delta z$ (along the axial direction).

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Flow Analysis:

1. Excess Flow Along $PQ$$PQ$PQPQ$PQ$:

2. Excess Flow Along $PS$$PS$PSPS$PS$:

3. Excess Flow Along $P{P}^{\prime }$$P{P}^{\prime }$PP^(‘)PP’$P{P}^{\prime }$:

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Mass of the Fluid in the Parallelepiped:

Rate of Change of Mass:

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Continuity Equation:

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Question:-07(b)

Find the Cauchy-Riemann equations in polar coordinates.

Answer:

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Step 1: Define the Complex Function

• $u(x,y)$$u(x,y)$u(x,y)u(x, y)$u(x,y)$ is the real part,
• $v(x,y)$$v(x,y)$v(x,y)v(x, y)$v(x,y)$ is the imaginary part,
• $z=x+iy$$z=x+iy$z=x+iyz = x + i y$z=x+iy$ is the complex variable.

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Step 2: Transform to Polar Coordinates

• $x=r\cos \theta$$x=r\cos \theta$x=r cos thetax = r \cos\theta$x=r\cos \theta$,
• $y=r\sin \theta$$y=r\sin \theta$y=r sin thetay = r \sin\theta$y=r\sin \theta$,
• $z=x+iy=r{e}^{i\theta }$$z=x+iy=r{e}^{i\theta }$z=x+iy=re^(i theta)z = x + iy = r e^{i\theta}$z=x+iy=r{e}^{i\theta }$.

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Step 3: Cauchy-Riemann Equations in Cartesian Coordinates

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Step 4: Partial Derivatives in Polar Coordinates

1. For $x=r\cos \theta$$x=r\cos \theta$x=r cos thetax = r \cos\theta$x=r\cos \theta$ and $y=r\sin \theta$$y=r\sin \theta$y=r sin thetay = r \sin\theta$y=r\sin \theta$:
   $$\frac{\partial }{\partial x}=\frac{\cos \theta }{r}\frac{\partial }{\partial r}-\frac{\sin \theta }{r^2}\frac{\partial }{\partial \theta },$$$$\frac{\partial }{\partial x}=\frac{\cos \theta }{r}\frac{\partial }{\partial r}-\frac{\sin \theta }{r^2}\frac{\partial }{\partial \theta },$$(del)/(del x)=(cos theta)/(r)(del)/(del r)-(sin theta)/(r^(2))(del)/(del theta),\frac{\partial}{\partial x} = \frac{\cos\theta}{r} \frac{\partial}{\partial r} – \frac{\sin\theta}{r^2} \frac{\partial}{\partial \theta},$$\frac{\partial }{\partial x}=\frac{\cos \theta }{r}\frac{\partial }{\partial r}-\frac{\sin \theta }{r^2}\frac{\partial }{\partial \theta },$$
   $$\frac{\partial }{\partial y}=\frac{\sin \theta }{r}\frac{\partial }{\partial r}+\frac{\cos \theta }{r^2}\frac{\partial }{\partial \theta }.$$$$\frac{\partial }{\partial y}=\frac{\sin \theta }{r}\frac{\partial }{\partial r}+\frac{\cos \theta }{r^2}\frac{\partial }{\partial \theta }.$$(del)/(del y)=(sin theta)/(r)(del)/(del r)+(cos theta)/(r^(2))(del)/(del theta).\frac{\partial}{\partial y} = \frac{\sin\theta}{r} \frac{\partial}{\partial r} + \frac{\cos\theta}{r^2} \frac{\partial}{\partial \theta}.$$\frac{\partial }{\partial y}=\frac{\sin \theta }{r}\frac{\partial }{\partial r}+\frac{\cos \theta }{r^2}\frac{\partial }{\partial \theta }.$$
2. Transform the partial derivatives of $u$$u$uu$u$ and $v$$v$vv$v$ using these relations.

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Step 5: Transform Cauchy-Riemann Equations

1. From $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$(del u)/(del x)=(del v)/(del y)\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$:
   $$\frac{\cos \theta }{r}\frac{\partial u}{\partial r}-\frac{\sin \theta }{r^2}\frac{\partial u}{\partial \theta }=\frac{\sin \theta }{r}\frac{\partial v}{\partial r}+\frac{\cos \theta }{r^2}\frac{\partial v}{\partial \theta }.$$$$\frac{\cos \theta }{r}\frac{\partial u}{\partial r}-\frac{\sin \theta }{r^2}\frac{\partial u}{\partial \theta }=\frac{\sin \theta }{r}\frac{\partial v}{\partial r}+\frac{\cos \theta }{r^2}\frac{\partial v}{\partial \theta }.$$(cos theta)/(r)(del u)/(del r)-(sin theta)/(r^(2))(del u)/(del theta)=(sin theta)/(r)(del v)/(del r)+(cos theta)/(r^(2))(del v)/(del theta).\frac{\cos\theta}{r} \frac{\partial u}{\partial r} – \frac{\sin\theta}{r^2} \frac{\partial u}{\partial \theta} = \frac{\sin\theta}{r} \frac{\partial v}{\partial r} + \frac{\cos\theta}{r^2} \frac{\partial v}{\partial \theta}.$$\frac{\cos \theta }{r}\frac{\partial u}{\partial r}-\frac{\sin \theta }{r^2}\frac{\partial u}{\partial \theta }=\frac{\sin \theta }{r}\frac{\partial v}{\partial r}+\frac{\cos \theta }{r^2}\frac{\partial v}{\partial \theta }.$$
   Multiply through by $r$$r$rr$r$:
   $$\cos \theta \frac{\partial u}{\partial r}-\frac{\sin \theta }{r}\frac{\partial u}{\partial \theta }=\sin \theta \frac{\partial v}{\partial r}+\frac{\cos \theta }{r}\frac{\partial v}{\partial \theta }.$$$$\cos \theta \frac{\partial u}{\partial r}-\frac{\sin \theta }{r}\frac{\partial u}{\partial \theta }=\sin \theta \frac{\partial v}{\partial r}+\frac{\cos \theta }{r}\frac{\partial v}{\partial \theta }.$$cos theta(del u)/(del r)-(sin theta)/(r)(del u)/(del theta)=sin theta(del v)/(del r)+(cos theta)/(r)(del v)/(del theta).\cos\theta \frac{\partial u}{\partial r} – \frac{\sin\theta}{r} \frac{\partial u}{\partial \theta} = \sin\theta \frac{\partial v}{\partial r} + \frac{\cos\theta}{r} \frac{\partial v}{\partial \theta}.$$\cos \theta \frac{\partial u}{\partial r}-\frac{\sin \theta }{r}\frac{\partial u}{\partial \theta }=\sin \theta \frac{\partial v}{\partial r}+\frac{\cos \theta }{r}\frac{\partial v}{\partial \theta }.$$
2. From $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$(del u)/(del y)=-(del v)/(del x)\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$:
   $$\frac{\sin \theta }{r}\frac{\partial u}{\partial r}+\frac{\cos \theta }{r^2}\frac{\partial u}{\partial \theta }=-(\frac{\cos \theta }{r}\frac{\partial v}{\partial r}-\frac{\sin \theta }{r^2}\frac{\partial v}{\partial \theta }).$$$$\frac{\sin \theta }{r}\frac{\partial u}{\partial r}+\frac{\cos \theta }{r^2}\frac{\partial u}{\partial \theta }=-\left(\frac{\cos \theta }{r}\frac{\partial v}{\partial r}-\frac{\sin \theta }{r^2}\frac{\partial v}{\partial \theta }\right).$$(sin theta)/(r)(del u)/(del r)+(cos theta)/(r^(2))(del u)/(del theta)=-((cos theta)/(r)(del v)/(del r)-(sin theta)/(r^(2))(del v)/(del theta)).\frac{\sin\theta}{r} \frac{\partial u}{\partial r} + \frac{\cos\theta}{r^2} \frac{\partial u}{\partial \theta} = -\left(\frac{\cos\theta}{r} \frac{\partial v}{\partial r} – \frac{\sin\theta}{r^2} \frac{\partial v}{\partial \theta}\right).$$\frac{\sin \theta }{r}\frac{\partial u}{\partial r}+\frac{\cos \theta }{r^2}\frac{\partial u}{\partial \theta }=-(\frac{\cos \theta }{r}\frac{\partial v}{\partial r}-\frac{\sin \theta }{r^2}\frac{\partial v}{\partial \theta }).$$
   Multiply through by $r$$r$rr$r$:
   $$\sin \theta \frac{\partial u}{\partial r}+\frac{\cos \theta }{r}\frac{\partial u}{\partial \theta }=-\cos \theta \frac{\partial v}{\partial r}+\frac{\sin \theta }{r}\frac{\partial v}{\partial \theta }.$$$$\sin \theta \frac{\partial u}{\partial r}+\frac{\cos \theta }{r}\frac{\partial u}{\partial \theta }=-\cos \theta \frac{\partial v}{\partial r}+\frac{\sin \theta }{r}\frac{\partial v}{\partial \theta }.$$sin theta(del u)/(del r)+(cos theta)/(r)(del u)/(del theta)=-cos theta(del v)/(del r)+(sin theta)/(r)(del v)/(del theta).\sin\theta \frac{\partial u}{\partial r} + \frac{\cos\theta}{r} \frac{\partial u}{\partial \theta} = -\cos\theta \frac{\partial v}{\partial r} + \frac{\sin\theta}{r} \frac{\partial v}{\partial \theta}.$$\sin \theta \frac{\partial u}{\partial r}+\frac{\cos \theta }{r}\frac{\partial u}{\partial \theta }=-\cos \theta \frac{\partial v}{\partial r}+\frac{\sin \theta }{r}\frac{\partial v}{\partial \theta }.$$

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Step 6: Simplify the Equations

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Final Result