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Catalogs Discussion Forums -> Organic Chemistry -> ORG. REAGENTSSSSSS -> Go to message
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5 NH3 is used in prep of amines

6 CO2 along with grignard reagent can form carboxylic acids

10 HNO3 is for introducing nitro (NO2) group in product
 cheers,



Catalogs Discussion Forums -> Trignometry -> SolVE this GUys............ -> Go to message
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This Question is worth asking, man
x2 + 19x+ 88
=x2 +11x+8x+88
=x(x+11) + 8(x+11)
=(x+11) (x+8)
So you not that 8 11
So the given expression is not a perfect square
pls do rate  
Catalogs Discussion Forums -> Organic Chemistry -> NaOH acts as acid ............. ???? HOW ????? -> Go to message
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It does so only in highly acidified state it forms HCl & then undergoes hydrolysis to form Phenol
Catalogs Discussion Forums -> Algebra -> what is remainder when 23^23 is devided by 53? -> Go to message
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9.9811321
Catalogs Discussion Forums -> Lounge -> boyz vs gals -> Go to message
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here +2 from me
70+2=72
Catalogs Discussion Forums -> Organic Chemistry -> What is meant by stereiochemistry of a reaction ?? do all reactions have it?? -> Go to message
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pls ppl
 its urgent !!!

Catalogs Discussion Forums -> Organic Chemistry -> What is meant by stereiochemistry of a reaction ?? do all reactions have it?? -> Go to message
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What is meant by stereiochemistry of a reaction ?? do all reactions have it??
Catalogs Discussion Forums -> Organic Chemistry -> What is meant by stereiochemistry of a reaction ?? -> Go to message
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What is meant by stereiochemistry of a reaction ??
Catalogs Discussion Forums -> Lounge -> boyz vs gals -> Go to message
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76+2=78













Catalogs Discussion Forums -> Physical Chemistry -> WRITE CONJUGATE BASE FOR - HCN , -> Go to message
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remove a proton and u get CN- that's all
cheers,
Catalogs Discussion Forums -> Organic Chemistry -> ORG. REAGENTSSSSSS -> Go to message
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3 RED PHOSPHORUS is reducing agent, often used with HI

5 NHis also another reducing agent used in Birch reduction

7 Br2 is brominating reagent used for giving br+ and br- ions for a reaction

8 AlCl3 used as catalyst for Friedel crafts alkylation

9 H2SO4 is dehydrating reagent(used for removal of water)

11 NaOH is a mild base used for nucleophilic addition reactions

13 Dry ETHER used in Wurtz reaction to form alkanes

14 CCl4 is a chlorinating reagent

15 BaSO4is a poisoning reagent used to poison Pd in Rosenmund's reaction along with sulphur and Quinoline (see,preparation of aldehydes)

16 Pd catalyst or Lindlar's catalyst prepares alkanes from alkenes

17 SnCl2 along with 2HCl can form SnCl4 giving H+ ions

19 HCN is used in preparation of cyanohydrin when reacted with aldehydes
Catalogs Discussion Forums -> Lounge -> Thank you teacher - help them win - Ode to my teachers on Teachers Day!! -> Go to message
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I would love to thank Mrs Opheliya D'Souza who was my English teacher (i know her since i was in seventh standard) She not only helped me with english but also gave advice about anything She was my school teacher and a great personality
She is known to be most strict teacher in the entire school yet she was the best for me as she has made me what i am today.Even during ssc exam result ,when i was worried about my marks,she stood by my side and encouraged me further
thank you mam,once again.
Community shelf Community shelf -> Newtonian mechanics -> Go to message
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Newtonian Mechanics

Inertial frames:

Newton?s first law defines a class of inertial frames.  Inertial frames are reference frames for which the trajectories for force-free motion are solutions to dr/dt = 0.  With respect to inertial frames Newton?s second law has the form
F = dp/dt.  (r = coordinate,  F = force,  p = mv  momentum)
Let Fik be the force that particle i exerts on particle k.  Newton?s third law states that Fik = -Fki.
Newton?s laws are well suited for the study of unconstrained mechanical systems.  Constraints, such as requiring a particle to follow a given curve in space, tell us that there are external forces, but do not tell us what these forces are.  The forces are only known in terms of their effect on the motion.
Conservation laws are very important tools in solving mechanics problems.  
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For a system of particles momentum is conserved if  Fext = 0;  Fext = 0 Û P = constant.
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Angular momentum (L = r´p) is conserved if the torque text = 0;  text = 0 Û L = constant.
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Energy E = T + U is conserved if all forces are conservative;  òF× dr = 0 Û T + U = constant.

Formulas:

Laws:
 
Newton's 2nd law:
F = dp/dt
Newton's third law:
Fik = -Fki
Forces:
 
Static and kinetic friction:
fs £ msN,    fk = mkN
Gravity:
Uniform circular motion:
F = mv2/r
Hooke's law:
F = -kr,  Fx= -kx
Concepts:
 
Work:
W = F×d
Kinetic energy:
K = (1/2)mv2
Work-kinetic energy theorem:
Wnet = DK = (1/2)m(vf2-vi2)
Elastic potential energy:
U = (1/2)kx2
Gravitational potential energy:
Conservative systems:
E = K + U,  Fx = -dU/dx
Power:
P = F·v  or  P = dW/dt
Momentum:
p = mv
Impulse:
I = Dp = FavgDt
Angular momentum:
L = r´  
Torque
t = r´F
Angular momentum and torque:
, ,  
dL = tdt

Impulsive forces

Forces and torques that act so powerfully but so briefly that they produce finite changes in linear and angular momentum while the system undergoes negligible displacement are said to be impulsive.
Linear Impulse: dp = Fdt,  Dp = òFdt, Dp = FavgDt
The integral of force over time as
Dt approaches 0 is called the impulse of the force.
Angular impulse: dL = tdt,  DL = òtdt, DL = tavgDt
The integral of torque over time as
Dt approaches 0 is called the angular impulse of the torque.

Collisions

In collisions, it is assumed that the colliding particles interact for such a short time, that the impulse due to external forces is negligible.  Thus the total momentum of the system just before the collision is the same as the total momentum just after the collision.
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Elastic collision: momentum is conserved, mechanical energy is conserved
bullet
Inelastic collisions: momentum is conserved, mechanical energy is not conserved
 

Lagrangian Mechanics

Constraints

Imposing constraints on a system is simply another way of stating that there are forces present in the problem that cannot be specified directly, but are known in term of their effect on the motion of the system.  Holonomic constraints are constraints of the form
fm(r1,r2,r3,¼,rn,t) = 0,   m = 1,2,3, ? ,k.
They reduce the number of degrees of freedom of the system; k equations of constraints reduce the number of the degree of freedom of an n-particle system from 3n to 3n - k, if the constraints are holonomic.
If the constraints are holonomic, then the forces of constraints do no virtual work.
Consider a virtual displacement of the system, i.e. an infinitesimal change in the coordinates of the system, denoted by dri, consistent with the constraints imposed on the system at a given instant t.  The work done by the force in the virtual displacement dri is called the virtual work.  For holonomic constraints, the forces of constraints are perpendicular to the virtual displacements and do no virtual work.

Generalized coordinates

Any set of independent quantities q1, q2, ? , qs, which completely define the position of the system with s degrees of freedom, are called generalized coordinates of the system, and the derivatives  are called generalized velocities.
Examples:
bullet
A particle is constraint to move in the x-y plane, the equation of constraint is z = 0, the constraint is holonomic.  Possible generalized coordinates for the system with two degrees of freedom are x, y;  r, f;  ? .
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A particle is constraint to move on a circle in the x-y plane, the equations of constraints are z = 0,  x+ y2 - r= 0.  The constraints are holonomic.  Possible generalized coordinates for the system with 1 degree of freedom are f ; f3, ? .
The generalized coordinates q1, ? , qs can be expressed in terms of the Cartesian coordinates the system.
q1 = q1(r1, r2, ? , rn ), ? , qn = qn(r1, r2, ? , rn ).
These equations, together with the equation of constraints, can be inverted to find the r?s in terms of the q?s.

Lagrangian Mechanics

Assume a system has n independent generalized coordinates {qi}.  Assume that the generalized applied forces
are given by
,
with U some scalar function, i.e. the generalized applied forces are derivable from a potential.  Then the equations of motion may be obtained from Lagrange?s equations,
,
where  L = T - U  is the Lagrangian of the system.  L is a function of the coordinates and the velocities.
If not all the forces acting on the system are derivable from a potential, then Lagrange's equations can be written in the form
,
where L contains the potential of the conservative forces and Qj represents the generalized forces not arising from a potential.
Define the generalized momentum or conjugate momentum or canonical momentum through
 .
If the Lagrangian does not contain a given coordinate qj then the coordinate is said to be cyclic and the corresponding conjugate momentum pj is conserved.
The Hamiltonian H of a system is given by
.
H is a function of the generalized coordinates and momenta of the system.  The equations of motion can be obtained from Hamilton?s equations,
.
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If the Lagrangian does not explicitly depend on time, then the Hamiltonian does not explicitly depend on time and H is a constant of motion.  [If H does explicitly depend on time, H = H(t), then H is not a constant of motion.]
bullet
If the generalized coordinates do not explicitly depend on time, then  H = T + U = E, the total energy of the system.  [If the generalized coordinates do explicitly depend on time, then  H is not the total energy of the system.]
bullet
So only if Lagrangian does not explicitly depend on time and the generalized coordinates do not explicitly depend on time, then H = T + U = E and the energy is a constant of motion.

Lagrange multipliers

Assume you have chosen coordinate for a system that are not independent, but are connected by m equations of constraints of the form
.
Then the equations of motion can be obtained from
(n equations),  and   (m equations).
We have m + n equations and m + n unknowns, the n coordinates and the m l?s.  The ll are called the undetermined Lagrange multipliers.   is the generalized force of constraint associated with the coordinate qk

Small oscillations

Let
.
Then solutions of the form qj = Re(Ajeiw t) can be found.  We can find the w2 from det (kij-w 2Tij) = 0.  For a system with n degrees of freedom, n characteristic frequencies wa can be found.  Some frequencies may be degenerate.
For a particular frequency
wa we solve
to find the Aja
[While the secular equation det(kij-w 2Tij) = 0 can in principle always be solved, it is often simpler to find the normal modes by using physical insight and noting the symmetries of the system.]
The most general solution for each coordinate qj is a sum of simple harmonic oscillations in all of the frequencies wa .

Rigid body motion

A rigid body is a system of particles in which the distances between the particles do not vary.  To describe the motion of a rigid body we use two systems of coordinates, a space-fixed system X, Y, Z, and a moving system x, y, z, which is rigidly fixed in the body and participates in its motion.
Let the origin of the body-fixed system be the body?s center of mass (CM).  The orientation of the axes of that system relative to the axes of the space-fixed system is given by three independent angles.  The vector R points from the origin of the spaced-fixed system to the CM of the body.  Thus a rigid body is a mechanical system with six degrees of freedom.
Let r denote the position of an arbitrary point P in the body-fixed system.  In the space fixed system its position is given by r + R, and its velocity is
v = d(R + r)/dt = dR/dt + dr/dt = V + W ´ r.
Here V is the velocity of the CM and W is the angular velocity of the body.  The direction of W is along the axis of rotation and W = df/dt.
The kinetic energy of the body is
T = (1/2)Smivi2 = (1/2)Smi(V + W ´ r)2.
We rewrite
T = (1/2)MV2 + (1/2)Smi(W2ri2 - (ri)2),  M = SmiSmiri = 0.
We find T = TCM + Trot, i.e. the kinetic energy is the sum of the kinetic energy of the motion of the CM and the kinetic energy of the rotation about the CM.   In component form we write
.
Here
is the inertia tensor.  The Wi are the components of W along the axis of the body fixed system.  For a continuous system .
By appropriate choice of the orientation of the body-fixed coordinate system the inertia tensor can be reduced to diagonal form.  The directions of the axes xi are then called the principal axes of inertia and the diagonal components of the tensor are then called the principal moments of inertia. Then
.
Definitions:
Asymmetrical top:
 
Symmetrical top:
 
Spherical top:
 
Let L denote the angular momentum about the CM of the body.
,
which in component form yields
.
If x1, x2, and x3 are the principal axes of inertia, then
L1 = I1W1,  L2 = I2W2,  L3 = I3W3.
The Lagrangian of a rigid body is
.

The equations of motion

The equations of motion of a rigid body are dP/dt = F, where P = total momentum and F = external forces, and dL/dt = t , where L = angular momentum about CM and t = total torque produced by external forces.
Let A be an arbitrary vector and dA/dt its rate of change with respect to the space fixed axes, d'A/dt its rate of change with respect to the body fixed axes.
dA/dt = d'A/dt + W ´ A.
Therefore
dP/dt = d'P/dt + W ´ P = F    and    dL/dt = d'L/dt + W ´ L = t .
Let the body fixed axes be the principal axes of inertia of the body.  Then
Li = IiWi  and  d'Li/dt = IidWi/dt,
and we have Euler?s equations:

The Eulerian angles

The orientation of the body-fixed coordinate system with respect to the space-fixed coordinate system is described by three angles.  These angles are often taken as the Eulerian angles, defined in the figure.
We can express the components of the angular velocity W along the body-fixed axes x, y, z in terms of the Eulerian angles and their derivatives.

Formulas

Center of mass:

Motion in a non-inertial frame

In non-inertial frames fictitious forces appear.  Consider a particle moving with velocity v in a reference frame K which moves with velocity V(t) relative to the inertial frame K0 and rotates with angular velocity W(t).
The Lagrangian of the particle is
L = (1/2)mv2 + mv × (W ´ r) + (1/2)m(W ´ r)2 - mW×r - U, with W = dV(t)/dt.
L/v = mv + m(W ´ r),  L/r = m(v ´ W) + m(W ´ r) ´ W - mW - U/r.
The equations of motion are
mdv/dt = -U/r - mdV/dt  + mr ´ dW/dt - 2mW ´ v - mW ´ (W ´ r).
Here
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