VMOU MT-05 SOLVED ASSIGNMENT | MA/M.SC. MT- 05(Mechanics) | July-2024 & January-2025

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.
  1. (i). Write the moment of Inertia of a circular disc of mass M M MMM and radius ‘ a a aaa ‘ 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.
  1. State and prove D D D^(‘)D^{\prime}D Alembert’s principle.
  2. A uniform rod of mass m m mmm is placed at right angles to a smooth plane of inclination α α alpha\alphaα with one end in contact with it. The rod is then released. Show that when the inclination to the plane is ϕ ϕ phi\phiϕ, the reaction of the plane will be { 3 ( 1 sin ϕ ) 2 + 1 ( 1 + 3 cos 2 ϕ ) 2 } m g cos α 3 ( 1 sin ϕ ) 2 + 1 1 + 3 cos 2 ϕ 2 m g cos α {(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{3(1sinϕ)2+1(1+3cos2ϕ)2}mgcosα.
  3. 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 14 25 tan^(-1)((14)/(25))\tan ^{-1}\left(\frac{14}{25}\right)tan1(1425) to the axis.
  4. 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.
  1. (a) Three equal uniform rods AB , BC , DC AB , BC , DC AB,BC,DC\mathrm{AB}, \mathrm{BC}, \mathrm{DC}AB,BC,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 3 I 2 40 m 3 I 2 40 m (3I^(2))/(40(m))\frac{3 I^2}{40 \mathrm{~m}}3I240 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.
  1. (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 P t + 1 r ( P u ) θ + ( P v ) z = 0 P t + 1 r ( P u ) θ + ( P v ) 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}=0Pt+1r(Pu)θ+(Pv)z=0. where u , v u , v u,v\mathrm{u}, \mathrm{v}u,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 M M MMM and radius ‘ a a aaa ‘ about a diameter.

Answer:

The moment of inertia of a circular disc of mass M M MMM and radius a a aaa about a diameter is:
I diameter = 1 4 M a 2 . I diameter = 1 4 M a 2 . I_(“diameter”)=(1)/(4)Ma^(2).I_{\text{diameter}} = \frac{1}{4} M a^2.Idiameter=14Ma2.

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 perpendicular = 1 2 M a 2 . I perpendicular = 1 2 M a 2 . I_(“perpendicular”)=(1)/(2)Ma^(2).I_{\text{perpendicular}} = \frac{1}{2} M a^2.Iperpendicular=12Ma2.
  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,Iz=Ix+Iy,
    where I z I z I_(z)I_zIz 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_yIx,Iy 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 diameter . I x = I y = I diameter . I_(x)=I_(y)=I_(“diameter”).I_x = I_y = I_{\text{diameter}}.Ix=Iy=Idiameter.
    Substituting into the perpendicular axis theorem:
    1 2 M a 2 = I diameter + I diameter . 1 2 M a 2 = I diameter + I diameter . (1)/(2)Ma^(2)=I_(“diameter”)+I_(“diameter”).\frac{1}{2} M a^2 = I_{\text{diameter}} + I_{\text{diameter}}.12Ma2=Idiameter+Idiameter.
  3. Solve for I diameter I diameter I_(“diameter”)I_{\text{diameter}}Idiameter:
    2 I diameter = 1 2 M a 2 , 2 I diameter = 1 2 M a 2 , 2I_(“diameter”)=(1)/(2)Ma^(2),2 I_{\text{diameter}} = \frac{1}{2} M a^2,2Idiameter=12Ma2,
    I diameter = 1 4 M a 2 . I diameter = 1 4 M a 2 . I_(“diameter”)=(1)/(4)Ma^(2).I_{\text{diameter}} = \frac{1}{4} M a^2.Idiameter=14Ma2.

Final Result:

The moment of inertia of a circular disc about a diameter is:
I diameter = 1 4 M a 2 . I diameter = 1 4 M a 2 . I_(“diameter”)=(1)/(4)Ma^(2).I_{\text{diameter}} = \frac{1}{4} M a^2.Idiameter=14Ma2.

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 xxx, y y yyy, and z z zzz 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:
x L x = y L y = z L z x L x = y L y = z L z (x)/(L_(x))=(y)/(L_(y))=(z)/(L_(z))\frac{x}{L_x}=\frac{y}{L_y}=\frac{z}{L_z}xLx=yLy=zLz
Here, L x , L y , L z L x , L y , L z L_(x),L_(y),L_(z)L_x, L_y, L_zLx,Ly,Lz are constant because the angular momentum vector L L L\mathbf{L}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 , , q n } q 1 , q 2 , , q n {q_(1),q_(2),dots,q_(n)}\left\{q_1, q_2, \ldots, q_n\right\}{q1,q2,,qn} are independent variables that uniquely specify the configuration of a mechanical system with n n nnn degrees of freedom.

Question:-02

State and prove D D D^(‘)D^{\prime}D 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:
F applied m a = 0 , F applied m a = 0 , sumF_(“applied”)-sum ma=0,\sum \mathbf{F}_\text{applied} – \sum m \mathbf{a} = 0,Fappliedma=0,
or equivalently:
( F applied + F inertial ) = 0 , F applied + F inertial = 0 , sum(F_(“applied”)+F_(“inertial”))=0,\sum \left( \mathbf{F}_\text{applied} + \mathbf{F}_\text{inertial} \right) = 0,(Fapplied+Finertial)=0,
where:
  • F applied F applied F_(“applied”)\mathbf{F}_\text{applied}Fapplied is the vector sum of all applied forces acting on the system,
  • F inertial = m a F inertial = m a F_(“inertial”)=-ma\mathbf{F}_\text{inertial} = -m \mathbf{a}Finertial=ma is the inertial force (negative of mass times acceleration),
  • m m mmm is the mass of the particle, and a a a\mathbf{a}a is its acceleration.
D’Alembert’s principle essentially transforms a dynamic problem into a static equilibrium problem by introducing inertial forces.

Proof

Step 1: Newton’s Second Law

For a system of n n nnn particles, the motion of each particle is governed by Newton’s second law:
F i = m i a i , i = 1 , 2 , , n , F i = m i a i , i = 1 , 2 , , n , F_(i)=m_(i)a_(i),quad i=1,2,dots,n,\mathbf{F}_i = m_i \mathbf{a}_i, \quad i = 1, 2, \ldots, n,Fi=miai,i=1,2,,n,
where:
  • F i F i F_(i)\mathbf{F}_iFi is the total force acting on the i i iii-th particle,
  • m i m i m_(i)m_imi is the mass of the i i iii-th particle,
  • a i a i a_(i)\mathbf{a}_iai is the acceleration of the i i iii-th particle.
Rewriting:
(1) F i m i a i = 0. (1) F i m i a i = 0. {:(1)F_(i)-m_(i)a_(i)=0.:}\mathbf{F}_i – m_i \mathbf{a}_i = 0. \tag{1}(1)Fimiai=0.

Step 2: Include Inertial Forces

Introduce the inertial force acting on the i i iii-th particle:
F inertial,i = m i a i . F inertial,i = m i a i . F_(“inertial,i”)=-m_(i)a_(i).\mathbf{F}_\text{inertial,i} = -m_i \mathbf{a}_i.Finertial,i=miai.
Substitute this into equation (1):
(2) F i + F inertial,i = 0. (2) F i + F inertial,i = 0. {:(2)F_(i)+F_(“inertial,i”)=0.:}\mathbf{F}_i + \mathbf{F}_\text{inertial,i} = 0. \tag{2}(2)Fi+Finertial,i=0.
This equation shows that the total applied force F i F i F_(i)\mathbf{F}_iFi and the inertial force F inertial,i F inertial,i F_(“inertial,i”)\mathbf{F}_\text{inertial,i}Finertial,i are in equilibrium.

Step 3: Principle of Virtual Work

Consider a virtual displacement δ r i δ r i deltar_(i)\delta \mathbf{r}_iδri for the i i iii-th particle, which is consistent with the constraints of the system. The principle of virtual work states:
(3) i = 1 n ( F i + F inertial,i ) δ r i = 0. (3) i = 1 n F i + F inertial,i δ r i = 0. {:(3)sum_(i=1)^(n)(F_(i)+F_(“inertial,i”))*deltar_(i)=0.:}\sum_{i=1}^n \left( \mathbf{F}_i + \mathbf{F}_\text{inertial,i} \right) \cdot \delta \mathbf{r}_i = 0. \tag{3}(3)i=1n(Fi+Finertial,i)δri=0.
Substitute F inertial,i = m i a i F inertial,i = m i a i F_(“inertial,i”)=-m_(i)a_(i)\mathbf{F}_\text{inertial,i} = -m_i \mathbf{a}_iFinertial,i=miai into equation (3):
(4) i = 1 n ( F i m i a i ) δ r i = 0. (4) i = 1 n F i m i a i δ r i = 0. {:(4)sum_(i=1)^(n)(F_(i)-m_(i)a_(i))*deltar_(i)=0.:}\sum_{i=1}^n \left( \mathbf{F}_i – m_i \mathbf{a}_i \right) \cdot \delta \mathbf{r}_i = 0. \tag{4}(4)i=1n(Fimiai)δri=0.

Step 4: Generalize for the System

Equation (4) implies that for the entire system, the total virtual work done by the applied forces and inertial forces vanishes:
i = 1 n ( F i + F inertial,i ) δ r i = 0. i = 1 n F i + F inertial,i δ r i = 0. sum_(i=1)^(n)(F_(i)+F_(“inertial,i”))*deltar_(i)=0.\sum_{i=1}^n \left( \mathbf{F}_i + \mathbf{F}_\text{inertial,i} \right) \cdot \delta \mathbf{r}_i = 0.i=1n(Fi+Finertial,i)δri=0.
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

  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.

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 m m mmm is placed at right angles to a smooth plane of inclination α α alpha\alphaα with one end in contact with it. The rod is then released. Show that when the inclination to the plane is ϕ ϕ phi\phiϕ, the reaction of the plane will be { 3 ( 1 sin ϕ ) 2 + 1 ( 1 + 3 cos 2 ϕ ) 2 } m g cos α 3 ( 1 sin ϕ ) 2 + 1 1 + 3 cos 2 ϕ 2 m g cos α {(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{3(1sinϕ)2+1(1+3cos2ϕ)2}mgcosα.

Answer:

We are analyzing the motion of a uniform rod B C B C BCBCBC of mass m m mmm and length 2 a 2 a 2a2a2a, placed at right angles to a smooth inclined plane A B A B ABABAB. The inclined plane is at an angle α α alpha\alphaα with the horizontal. The rod is initially at rest, and its center of mass G G GGG is located at a distance B G = a B G = a BG=aBG = aBG=a from point B B BBB. The vertical height of the center of mass G G GGG above the inclined plane is:
G H = a cos α (from B G H ) . G H = a cos α (from B G H ) . GH=a cos alphaquad(from (/_\BGH)”)”.GH = a \cos \alpha \quad \text{(from \(\triangle BGH\))}.GH=acosα(from BGH).
When the rod is released, it begins to rotate and slide. After a time t t ttt, let the rod make an angle ϕ ϕ phi\phiϕ with the inclined plane A B A B ABABAB. We analyze the motion of the center of mass G G GGG and the forces acting on the rod.
original image

Coordinates of the Center of Mass:

We define the axes with B A B A BABABA as the horizontal axis and the perpendicular to B A B A BABABA as the vertical axis. The coordinates of the center of mass G G GGG are:
x = B N = a cos ϕ , y = G N = a sin ϕ (from B G N ) . x = B N = a cos ϕ , y = G N = a sin ϕ (from B G N ) . 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 \(\triangle BGN\))}.x=BN=acosϕ,y=GN=asinϕ(from BGN).

Equation of Motion Perpendicular to the Plane:

The motion of the center of mass G G GGG is constrained to be perpendicular to the inclined plane A B A B ABABAB. The equation of motion along the y y yyy-direction is:
m y ¨ = m ( a cos ϕ ϕ ¨ a sin ϕ ϕ ˙ 2 ) = R m g cos α m y ¨ = m a cos ϕ ϕ ¨ a sin ϕ ϕ ˙ 2 = R m g cos α 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 \alphamy¨=m(acosϕϕ¨asinϕϕ˙2)=Rmgcosα
where:
  • R R RRR is the normal reaction force exerted by the inclined plane on the rod,
  • m g cos α m g cos α mg cos alphamg \cos \alphamgcosα is the component of the gravitational force perpendicular to the plane.

Moment About the Center of Mass G G GGG:

Taking moments about the center of mass G G GGG, the equation of rotational motion is:
m a 2 3 ϕ ¨ = R a cos ϕ m a 2 3 ϕ ¨ = R a cos ϕ m(a^(2))/(3)phi^(¨)=-Ra cos phim \frac{a^2}{3} \ddot{\phi} = -R a \cos \phima23ϕ¨=Racosϕ
where a 2 3 a 2 3 (a^(2))/(3)\frac{a^2}{3}a23 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 G G GGG above the inclined plane was:
G H = a cos α G H = a cos α GH=a cos alphaGH = a \cos \alphaGH=acosα
At any time t t ttt, the vertical height of G G GGG above the inclined plane is:
G Q = G N cos α = ( a sin ϕ ) cos α G Q = G N cos α = ( a sin ϕ ) cos α GQ=GN cos alpha=(a sin phi)cos alphaGQ = GN \cos \alpha = (a \sin \phi) \cos \alphaGQ=GNcosα=(asinϕ)cosα
The distance fallen by G G GGG in the downward direction is:
Distance fallen = G H G Q = a cos α a sin ϕ cos α = a cos α ( 1 sin ϕ ) Distance fallen = G H G Q = a cos α a sin ϕ cos α = a cos α ( 1 sin ϕ ) “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)Distance fallen=GHGQ=acosαasinϕcosα=acosα(1sinϕ)

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 ϕ ˙ 2 ] = Work done by gravity m 2 v 2 + k 2 ϕ ˙ 2 = Work done by gravity (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}m2[v2+k2ϕ˙2]=Work done by gravity
where:
  • v 2 = y ˙ 2 = ( a cos ϕ ϕ ˙ ) 2 v 2 = y ˙ 2 = ( a cos ϕ ϕ ˙ ) 2 v^(2)=y^(˙)^(2)=(a cos phiphi^(˙))^(2)v^2 = \dot{y}^2 = (a \cos \phi \dot{\phi})^2v2=y˙2=(acosϕϕ˙)2,
  • k 2 ϕ ˙ 2 = a 2 3 ϕ ˙ 2 k 2 ϕ ˙ 2 = a 2 3 ϕ ˙ 2 k^(2)phi^(˙)^(2)=(a^(2))/(3)phi^(˙)^(2)k^2 \dot{\phi}^2 = \frac{a^2}{3} \dot{\phi}^2k2ϕ˙2=a23ϕ˙2 is the rotational kinetic energy.
Substituting these expressions, the energy equation becomes:
m 2 [ a 2 cos 2 ϕ ϕ ˙ 2 + a 2 3 ϕ ˙ 2 ] = m g a cos α ( 1 sin ϕ ) m 2 a 2 cos 2 ϕ ϕ ˙ 2 + a 2 3 ϕ ˙ 2 = m g a cos α ( 1 sin ϕ ) (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)m2[a2cos2ϕϕ˙2+a23ϕ˙2]=mgacosα(1sinϕ)
Simplifying:
m a 2 6 [ 3 cos 2 ϕ + 1 ] ϕ ˙ 2 = m g a cos α ( 1 sin ϕ ) m a 2 6 3 cos 2 ϕ + 1 ϕ ˙ 2 = m g a cos α ( 1 sin ϕ ) (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)ma26[3cos2ϕ+1]ϕ˙2=mgacosα(1sinϕ)
Solving for ϕ ˙ 2 ϕ ˙ 2 phi^(˙)^(2)\dot{\phi}^2ϕ˙2:
ϕ ˙ 2 = 6 g a ( 1 sin ϕ ) ( 1 + 3 cos 2 ϕ ) cos α ϕ ˙ 2 = 6 g a ( 1 sin ϕ ) ( 1 + 3 cos 2 ϕ ) cos α 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 \alphaϕ˙2=6ga(1sinϕ)(1+3cos2ϕ)cosα

Angular Acceleration:

Differentiating ϕ ˙ 2 ϕ ˙ 2 phi^(˙)^(2)\dot{\phi}^2ϕ˙2 with respect to time t t ttt, we get:
2 ϕ ˙ ϕ ¨ = 6 g a cos α [ ( 1 + 3 cos 2 ϕ ) ( cos ϕ ϕ ˙ ) ( 1 sin ϕ ) ( 6 cos ϕ sin ϕ ϕ ˙ ) ( 1 + 3 cos 2 ϕ ) 2 ] 2 ϕ ˙ ϕ ¨ = 6 g a cos α ( 1 + 3 cos 2 ϕ ) ( cos ϕ ϕ ˙ ) ( 1 sin ϕ ) ( 6 cos ϕ sin ϕ ϕ ˙ ) ( 1 + 3 cos 2 ϕ ) 2 2phi^(˙)phi^(¨)=(6g)/(a)cos alpha[((1+3cos^(2)phi)(-cos phi(phi^(˙)))-(1-sin phi)(-6cos phi sin phi(phi^(˙))))/((1+3cos^(2)phi)^(2))]2 \dot{\phi} \ddot{\phi} = \frac{6g}{a} \cos \alpha \left[\frac{(1 + 3 \cos^2 \phi)(-\cos \phi \dot{\phi}) – (1 – \sin \phi)(-6 \cos \phi \sin \phi \dot{\phi})}{(1 + 3 \cos^2 \phi)^2} \right]2ϕ˙ϕ¨=6gacosα[(1+3cos2ϕ)(cosϕϕ˙)(1sinϕ)(6cosϕsinϕϕ˙)(1+3cos2ϕ)2]
Dividing through by 2 ϕ ˙ 2 ϕ ˙ 2phi^(˙)2 \dot{\phi}2ϕ˙ and simplifying:
ϕ ¨ = 3 g cos α cos ϕ a ( 1 + 3 cos 2 ϕ ) 2 [ 1 + 3 ( 1 sin ϕ ) 2 ] ϕ ¨ = 3 g cos α cos ϕ a ( 1 + 3 cos 2 ϕ ) 2 1 + 3 ( 1 sin ϕ ) 2 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]ϕ¨=3gcosαcosϕa(1+3cos2ϕ)2[1+3(1sinϕ)2]

Reaction Force R R RRR:

Substituting the value of ϕ ¨ ϕ ¨ phi^(¨)\ddot{\phi}ϕ¨ into the moment equation, we find the normal reaction force R R RRR:
R = m g { 3 ( 1 sin ϕ ) 2 + 1 ( 1 + 3 cos 2 ϕ ) 2 } cos α R = m g 3 ( 1 sin ϕ ) 2 + 1 ( 1 + 3 cos 2 ϕ ) 2 cos α 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 \alphaR=mg{3(1sinϕ)2+1(1+3cos2ϕ)2}cosα

Final Results:

  1. Angular Velocity:
    ϕ ˙ 2 = 6 g a ( 1 sin ϕ ) ( 1 + 3 cos 2 ϕ ) cos α ϕ ˙ 2 = 6 g a ( 1 sin ϕ ) ( 1 + 3 cos 2 ϕ ) cos α 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 \alphaϕ˙2=6ga(1sinϕ)(1+3cos2ϕ)cosα
  2. Angular Acceleration:
    ϕ ¨ = 3 g cos α cos ϕ a ( 1 + 3 cos 2 ϕ ) 2 [ 1 + 3 ( 1 sin ϕ ) 2 ] ϕ ¨ = 3 g cos α cos ϕ a ( 1 + 3 cos 2 ϕ ) 2 1 + 3 ( 1 sin ϕ ) 2 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]