PHY
UNIT-IVFRAMES OF REFERENCES AND RIGID BODY DYNAMICSQUESTION BANK Q 1) Discuss Angular velocity and angular acceleration of a rigid body?A 1) Motion Relative to Translating AxesParticle motion has been described using position vectors that were referred to fixed reference frames. The positions, velocities and accelerations determined in this way are referred to as absolute. Often it isn’t possible or convenient to use a fixed set of axes for the observation of motion. Many problems are simplified considerably by the use of a moving reference frame.In the following we will restrict our attention to moving reference frames that translate but do not rotate.
Figure 12: Observation of particle motion using a translating referenceConsider two particles A and B moving along independent trajectories in the plane, and a fixed reference O. Let rA and rB be the positions of particles A and B in the fixed reference. Instead of observing the motion of particle A relative to the fixed reference as we have done in the past, we will attach a non-rotating reference to particle B and observe the motion of A relative to the moving reference at B. Let i and j be basis vectors of the moving reference, then the position vector of A relative to the reference at B, denoted rA/B is,rA/B =xi +yj where the subscript stands for "A with respect to B" or "A relative to B". Observe that, as the moving frame does not rotate, basis vectors i and j do not change in time. Therefore, taking time derivatives, we obtain simply,
which can be interpreted as the velocity of A relative to B. Now we can express the absolute position vector of A as,
Differentiating the equation in time to obtain expressions for the absolute velocity and acceleration of particle A:
or the absolute velocity of A equals the absolute velocity of B plus the velocity of A relative to B, vA/B , and similarly for the acceleration. The relative terms are the velocity or acceleration measured by an observer attached to the moving reference at particle B.
Figure 13: Relative velocities under change of translating reference.What would happen if the moving reference were attached to A instead?
By comparison with expressions derived previously
We proceed by demonstrating that every motion of a planar rigid body is associated with a single angular velocity ω and angular acceleration
, describing the angular displacement of an arbitrary line inscribed in the body relative to a fixed direction.
Figure 14: Angular velocity and acceleration of a rigid bodyConsider a rigid body undergoing plane motion. The angular positions of two arbitrary lines 1 and 2 attached to the body are specified by 
1and 
2 measured relative to any convenient fixed reference direction. These are related to the intermediate angle 
shown as,
2 = 
1Observe that as the body is rigid, requiring that the distance between each pair of points on the two lines 1 and 2 is constant, angle 
must be invariant. Differentiating the relation above with this in mind,
These hold for arbitrary lines attached to the body, implying in turn that the body can be associated with a unique angular velocity ω, defined as,
for an arbitrary line attached to the body. Arguing analogously, the body can be associated with a unique angular acceleration 
defined as,
Figure 15: Angular velocity and angular acceleration of a rigid bodyConsequently, we have the property that all lines on a rigid body in its plane of motion have the same angular displacement, the same angular velocity ω and the same angular acceleration. 
Figure 16: Decomposition of rigid body motionNext, consider the motion of a rigid body over the interval t0 - t1 as shown, with arbitrary point B0 taken as reference. Clearly, the motion can be consider to occur in two stages: a translation with reference B0 taking arbitrary line A0 B0 to an intermediate position A’B1; and a rotation about point B1 taking A’B1 to its final position A1B1. This corresponds to a decomposition of the motion into the sum of a translation and a rotation. While the translational motion is described by the velocity 
and acceleration 
of the reference point, the rotational motion is characterized by the unique angular velocity ω and angular acceleration 
associated with the body. Thus, we have the property that the motion of a rigid body can be decomposed into a translation of an arbitrary point within the body, followed by a rigid rotation of the body about this point. Further, the motion of an arbitrary point within the body is determined completely once the translational quantities 
and 
and rotational quantities ω and 
are known.Characteristics of rigid body motion: All lines on a rigid body have the same angular velocity ω and the same angular acceleration
. Rigid motion can be decomposed into the translation of an arbitrary point, followed by a rotation about the point. Q 2) Derive equations that describe the motion of a rigid body?A 2)
Figure 17: Definition of a translating reference attached to point BConsider a rigid body moving in the plane with angular velocity ω and angular acceleration 
, and two arbitrary points A and B of the body. We will examine the motion of this body in both, the fixed reference O shown, as well as relative to a non-rotating reference attached to point B. Proceeding, we express the absolute position of point A in terms of the absolute position of point B as,
where rA/B is the position of A relative to B. An analogous expression for absolute velocities follows by taking time derivatives,
where vA/B is the velocity of point A relative to the reference at B,
Figure 18: Motion of point A as seen by a translating observer at BNow, as the body moves, point A traces a circular path of radius rA/B = 
relative to point B, keeping the distance between the two unchanged. The angular velocity of this motion is simply the angular velocity ω of the rigid body. Then, using results derived previously for the time derivatives of rotating vectors we have:
or,
Observe that the expression reflects the decomposition of rigid body motion referred to previously. With B chosen as reference, the velocity of A is the vector sum of a translational portion vB and a rotational portion ω
rA/B .Proceeding to derive expressions for the acceleration of an arbitrary point of a rigid body, we differentiate the equation for velocities to obtain,
Where 
A/B , the acceleration of point A relative to the reference at B, is simply
or,
Thus, like the absolute velocity, the absolute acceleration of point A is the vector sum of a translational portion 
B and a rotational portion 
.
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Figure 12: Observation of particle motion using a translating referenceConsider two particles A and B moving along independent trajectories in the plane, and a fixed reference O. Let rA and rB be the positions of particles A and B in the fixed reference. Instead of observing the motion of particle A relative to the fixed reference as we have done in the past, we will attach a non-rotating reference to particle B and observe the motion of A relative to the moving reference at B. Let i and j be basis vectors of the moving reference, then the position vector of A relative to the reference at B, denoted rA/B is,rA/B =xi +yj where the subscript stands for "A with respect to B" or "A relative to B". Observe that, as the moving frame does not rotate, basis vectors i and j do not change in time. Therefore, taking time derivatives, we obtain simply,
which can be interpreted as the velocity of A relative to B. Now we can express the absolute position vector of A as,Differentiating the equation in time to obtain expressions for the absolute velocity and acceleration of particle A:or the absolute velocity of A equals the absolute velocity of B plus the velocity of A relative to B, vA/B , and similarly for the acceleration. The relative terms are the velocity or acceleration measured by an observer attached to the moving reference at particle B.
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By comparison with expressions derived previouslyWe proceed by demonstrating that every motion of a planar rigid body is associated with a single angular velocity ω and angular acceleration, describing the angular displacement of an arbitrary line inscribed in the body relative to a fixed direction.
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1and 2 measured relative to any convenient fixed reference direction. These are related to the intermediate angle shown as,2 = 1Observe that as the body is rigid, requiring that the distance between each pair of points on the two lines 1 and 2 is constant, angle must be invariant. Differentiating the relation above with this in mind,These hold for arbitrary lines attached to the body, implying in turn that the body can be associated with a unique angular velocity ω, defined as,for an arbitrary line attached to the body. Arguing analogously, the body can be associated with a unique angular acceleration defined as,
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and acceleration of the reference point, the rotational motion is characterized by the unique angular velocity ω and angular acceleration associated with the body. Thus, we have the property that the motion of a rigid body can be decomposed into a translation of an arbitrary point within the body, followed by a rigid rotation of the body about this point. Further, the motion of an arbitrary point within the body is determined completely once the translational quantities and and rotational quantities ω and are known.Characteristics of rigid body motion: .
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, and two arbitrary points A and B of the body. We will examine the motion of this body in both, the fixed reference O shown, as well as relative to a non-rotating reference attached to point B. Proceeding, we express the absolute position of point A in terms of the absolute position of point B as,where rA/B is the position of A relative to B. An analogous expression for absolute velocities follows by taking time derivatives,where vA/B is the velocity of point A relative to the reference at B,
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relative to point B, keeping the distance between the two unchanged. The angular velocity of this motion is simply the angular velocity ω of the rigid body. Then, using results derived previously for the time derivatives of rotating vectors we have:or,Observe that the expression reflects the decomposition of rigid body motion referred to previously. With B chosen as reference, the velocity of A is the vector sum of a translational portion vB and a rotational portion ω rA/B .Proceeding to derive expressions for the acceleration of an arbitrary point of a rigid body, we differentiate the equation for velocities to obtain,Where A/B , the acceleration of point A relative to the reference at B, is simplyor,Thus, like the absolute velocity, the absolute acceleration of point A is the vector sum of a translational portion B and a rotational portion . Q 3) What is the value of Kinetic energy of rigid body in rotation?A 3) Kinetic energy of rigid body in rotation Kinetic energy of a particle or body represents the form of energy that arises from motion. We are aware that kinetic energy of a particle in translation is given by the expression: K=
mv2In pure rotation, however, the rigid body has no "over all" translation of the body. However, the body in rotation must have kinetic energy as it involves certain motion. A closer look on the rotation of rigid body reveals that though we may not be able to assign translation to the rigid body as a whole, but we can recognize translation of individual particles as each of them rotate about the axis in circular motion with different linear speeds. The speed of a particle is given by:
Figure 20: Kinetic energy of rigid body in rotation Pure rotational motion Each particle of the body follows a circular path about axis in pure rotational motion. vi= ωriThus, kinetic energy of an individual particle is : Ki=
mivi2where " Ki" is the kinetic energy of "i" th particle having a speed " vi". In terms of angular speed, the kinetic energy of an individual particle is: Ki=
mi(ωri2Ki=
mi(ωri)2Now, the kinetic energy of the rigid body is sum of the kinetic energies of the particles constituting the rigid body: K=∑Ki=∑
mi(ωri)2We note here that angular speeds of all particles constituting the body are same. Hence, the constant "1/2" and " ω2" can be taken out of the summation sign : K =
ω2∑mi(ri)2However, we know that: I=∑mi(ri)2Combining two equations, we have: K= 
Iω2This is the desired expression of kinetic energy of a rigid body rotating about a fixed axis i.e. in pure rotational motion. The form of expression of the kinetic energy here emphasizes the correspondence between linear and angular quantities. Comparing with the expression of kinetic energy for translational motion, we find that "moment of inertia (I)" corresponds to "mass (m)" and "linear speed (v)" corresponds to "angular speed (ω)". We can also interpret the result obtained above from a different perspective. We could have directly inferred that expression of kinetic energy in rotation should have an equivalent form as: K (Kinetic energy)= 
x(inertia)x(speed)2In rotation, inertia to the rotation is "moment of inertia (I)" and speed of the rigid body is "angular speed (ω)". Substituting for the quantities, we have the expression for kinetic energy of rigid body in rotation as : K= 
Iω2Comparing this equation with the expression of the sum of kinetic energy of individual particles as derived earlier: K =
ω2∑mi(ri)2Clearly, I= ∑mi(ri)2This conclusion, thus, clearly establishes that the expression as given by ∑mi(ri)2 indeed represents the inertia of the rigid body in rotation. Q4) Explain the rotation in plane also derive the conditions of the instantaneous velocity vector of any particle is always perpendicular to its instantaneous position vector?A 4) When all parts of a rigid body move parallel to a fixed plane, then the motion of the object is referred to as plane motion. There are two types of plane motion, which are given as follows:The pure rotational motion: The rigid body in such a motion rotates about a fixed axis that is perpendicular to a fixed plane. In other words, the axis is fixed and does not move or change its direction relative to an inertial frame of reference. The general plane motion: The motion here can be considered as a combination of pure translational motion parallel to a fixed plane in addition to a pure rotational motion about an axis that is perpendicular to that plane. Suppose a rigid body of an arbitrary shape is in pure rotational motion about the z-axis (see Figure 19(a)). Let us analyze the motion of a particle that lies in a slice of the body in the x-y plane as in Figure 19(b). This particle (at point P) will rotate in a circle of fixed radius r which represents the perpendicular distance from P to the axis of rotation. If you look at any other particle in the object you will see that every particle will rotate in its own circle that has the axis of rotation at its center. In other words, different particles move in different circles but the center of all of these circles lies on the rotational axis. Suppose the particle moves through an arc length s starting at the positive x-axis. Its angular position is then given byθ=s/rr and θ are the polar coordinates of a point in a plane where θ is always measured from the positive x-axis. Because θ is the ratio of the arc length to the radius, it is a pure (dimensionless) number.
When a rigid body moves in such a way that one point in it remains fixed, then motion of the rigid body is called rotation about a point. The point which remains fixed during motion is called the centre of rotation.Consider a body which is rotating about a point O. All the particles of the body will rotate about it. Since the body is rigid, the distance between all pairs remains fixed during the motion. Let us consider a particles P of the rigid body having position vector 
. Let b be its distance from the origin O. The rigidity condition requires that 
= bWhere b is constant.Squaring the above both sides we obtain
= 
.
=
Differentiate both sides with respect to time to obtain
.
+
= 0
.
+
= 0
.
= 0This equation shown that whenever a rigid body rotates about a fixed point, the instantaneous velocity vector of any particle is always perpendicular to its instantaneous position vector. Q 5) Derive relation of angular momentum about a point of a rigid body in planar motion?A 5) Consider a rigid body which a rotating about a fixed point with an angular velocity 
. Choose the origin O at the fixed point and axes as shown in Figure 21The velocity of particle i having position vector
is 
= 
…………(1)If mi is the mass of particle i then its angular momentum about the fixed point O is
= 
…………(2)Substituting the value of 
, it takes the form 
= 
…………(3)
Figure 21: Angular momentum about a point of a rigid body in planar motionThe angular momentum of rigid body can be obtained by summing the above equation over all the particles of the rigid body. Thus 
= 
= 
…………(4)Using the vector relation
…………(5)Eq. (4) may be written as
= 
= 
…………(6)Let (xi, yi, zi) be the cartesian coordinates of particle i, then 
…………(7)If ωx, ωy and ωz are the components of angular velocity along coordinate axes then
…………(8)Now take the dot product of 
with 
to obtain 
. 
…………(9)Substitutes these values in eq. (6) to obtain
Q 6) Transform rectangular coordinates in spherical coordinates?A 6) Spherical coordinates are similar to the way we describe a point on the surface of the earth using latitude and longitude. By specifying the radius of a sphere and the latitude and longitude of a point on the surface of that sphere, we can describe any point in R3. To describe the latitude and longitude, we use two angles: θ (the angle from the positive x axis) and ϕ (the angle from the positive z axis). We therefore have three coordinates (ρ,θ,ϕ), where ρ is the radius of the sphere.
Figure 7: Spherical coordinateswhere the first component is the distance from the origin to the point, the second component is the polar angle and the third component is the altitude angle measured from the positive z-axis to the line connecting the origin and the point. Note that 0≤θ<2π0≤ϕ≤πρ≥0The formulas that we need in order to convert between rectangular and spherical coordinates are given below, without derivation ρ=x2+y2+z2x=ρcosθsinϕy=ρsinθsinϕz=ρcosϕ Q 7) Convert (3, π/4, 4) in cylindrical coordinates to rectangular coordinates.A 7) Use the formulas, noting that r=3, θ=π/4, and z=4.x=rcosθ=3cosπ/4= 
y=rsinθ=3sinπ/4= 
z=4Therefore point is (
) in rectangular coordinates.To go in the other direction (from rectangular coordinates to cylindrical), use the following set of formulas (again, the first two are exactly what we use to convert from rectangular to polar in R2:r=x2+y2θ=tan−1
z=z Q 8) Plot the point with spherical coordinates (8, π/3, π/6) and express its location in both rectangular and cylindrical coordinates.A 8) Use the equations in Note to translate between spherical and cylindrical coordinates ρ=x2+y2+z2x=ρcosθsinϕ = 8sin(π/6)cos(π/3) = 8(1/2)1/2=2y=ρsinθsinϕ =8sin(π/6)sin(π/3)=8(1/2)
=2
z=ρcosϕ =ρcosφ =8cos(π/6)= 8(
=4
The point with spherical coordinates (8,π/3,π/6) has rectangular coordinates (2, 2
4
)Finding the values in cylindrical coordinates is equally straightforward:r=ρsinφ=8sinπ/6= 4θ= θz=ρcosφ=8cosπ/6= 4
Thus, cylindrical coordinates for the point are (4,π/3, 4
).
Figure 8: Coordinate System Q 9) Difference between inertial & non-inertial frames?A 9) Inertial frame of reference An inertial frame is defined as one in which Newton’s law of inertia holds that is anybody which isn’t being acted on by an outside force stays at rest if it is initially at rest, or continues to move at a constant velocity if that’s what it was doing to begin with. Non-inertial frame of referenceA frame that is accelerated with respect to the assumed inertial frame of reference. Newton’s law will not hold true in these frames. Q 10) What is Galilean Transformations? How velocity and acceleration transform?A 10)Galilean Coordinate Transformations: For simplicity: - Let coordinates in both references equal at (t = 0). Use Cartesian coordinate systems.Newton’s laws are only applicable at inertial reference frames. According to Galilean transformations, Newton’s laws are invariant at any inertial reference frame. In other words, the mechanical movement of a particle is exactly the same at two different reference frames, If both of them are inertial (i.e. one moves with constant velocity w.r.t the other & vice versa). There is no mechanical experiment by which one can distinguish whether a system is at rest or is moving with a constant speed in a straight line (Galilean relativity).Suppose we know from experiment that these laws of mechanics are true in one frame of reference. How do they look in another frame, moving with respect to the first frame? To find out, we have to figure out how to get from position, velocity and acceleration in one frame to the corresponding quantities in the second frame. Obviously, the two frames must have a constant relative velocity, otherwise the law of inertia won’t hold in both of them. Let’s choose the coordinates so that this velocity is along the x-axis of both of them.
Notice we also throw in a clock with each frame. Suppose S′ is moving relative to S at speed v along the x-axis. For convenience, let us label the moment when O′ passes O as the zero point of timekeeping. That mean at t=0 O and O’ coincide with each other.Now let us try to find the coordinates of the event (x, y, z, t) in S′ It’s easy to see t′ = tWe synchronized the clocks when O′ passed O. Also, evidently, y′ = y and z′ = z, from the figure. We can also see that x = x′ +vt. Thus (x, y, z, t) in S corresponds to (x′, y′, z′, t′) in S′, wherex = x′ +vt. y = y’z = z’t =t’That’s how positions transform; these are known as the Galilean Transformations.Velocity The velocity in S′ in the x′ direction is given by u’x 
This is obvious anyway it’s just the addition of velocities formula.ux = u’x +vAcceleration 
Since v is constant. That is to say, a’x =axThe acceleration is the same in both frames. This again is obvious as the acceleration is the rate of change of velocity, and the velocities of the same particle measured in the two frames differ by a constant factor-the relative velocity of the two frames.Q 11) Derive Lorentz Transformation equation?A 11)Lorentz transformation is the relationship between two different coordinate frames that move at a constant velocity and are relative to each other. The name of the transformation comes from a Dutch physicist Hendrik Lorentz.The Galilean transformation nevertheless violates Einstein’s postulates, because the velocity equations state that a pulse of light moving with speed c along the x-axis would travel at speed c−v in the other inertial frame. Specifically, the spherical pulse has radius r=ct at time t in the unprimed frame, and also has radius r′=ct′ at time t' in the primed frame. Expressing these relations in Cartesian coordinates givesx2+y2+z2−c2t2 =0 ………..(1)x′2+y′2+z′2−c2t′2=0 ………..(2)The left-hand sides Equations and can be set equal because both are zero. Because y=y′ and z=z′ so, we obtainx2−c2t2 =x′2−c2t′2 ………..(3)However this cannot be satisfied for nonzero relative velocity v of the two frames if we assume the Galilean transformation results in t=t′ with x=x′+vt′.To find the correct set of transformation equations, assume the two coordinate systems S and S′ The laws of mechanics are invariant under Galilean transformations, whereas electrodynamics and Maxwell’s equations are varying under this transformation. This shows that the velocity of light will have different values for different observers moving with different uniform velocities. But the speed of light is invariant in all inertial frames of reference; hence, Galilean transformations need to be modified. Lorentz deduced the transformation equations, which are in agreement with the results of Michelson–Morley experiment. The Lorentz transformations are deduced as given below.Lorentz transformation equations are developed. Let x, y, z and t be the space and time coordinate in rest system or standard system. The system with x', y', z' moves with velocity 'v' along the x-axis. The origins at t=t'=0 coincide. The most general transformation equations relating the coordinates of an event in two systems can be written as, x’ = Ax +Bt ………..(4)y’ = y ………..(5)z’ = z ………..(6)t’ = Cx +Dt ………..(7)
The transformations are linear.If they are not linear one system would predict acceleration while in other system velocity was even constant. Because of symmetry, y' and z' axes are left unchanged by the transformation.Constant can be calculated by using the following boundary conditions.Observers in S and S' see (O') the origin of S' as x = vtx’ = 0Substituting values of x and x' in equation (4) We get0 = Avt +Bti.e. B =- Av ………..(8) Observer in S' and S see the origin (O) of S as x' = vtx = 0Substituting values of x and x' in equation (4) we get-vt’ =Bt ………..(9)Substituting the value of B from equation (8), we get-vt’ =- Avt ………..(10)In equation (7) put x=0 we gett’ = Dt So equation (10) become-v Dt =- AvtA =D ………..(11)A light pulse is sent out from the origin of S towards x at t=0, its location is given by x= ctx’ = ct’Substituting values of x and x' in equation (4), we getct’ = Act +BtSubstituting the value of B from equation (8), we getct’ = Act +(- Avt) ………..(12)Substituting values of x and D from eq no (11) in eq no.(7) , we gett’ = Cct +AtSubstituting above value of t' in eq no.(12) , we getc(Cct +At) = Act - AvtorCc2t +Act =Act - AvtCc2t =AvtOrC =-A 
………..(13)A light pulse is emitted along y axis in S system at t=0, its coordinate in both systems are
y= ct, x=0 r’=ct’ where using equation (5) r’2 = x’2+y’2using equation (5)y’= y= ctand from equation (4) and (7)x’ = Bt = -Avtt’= Dt =-AtTherefore r’2 = x’2+y’2 becomec2t’2 = (-Avt)2 +(ct)2Substituting value of t' above equation becamec2A2t2 = A2v2t2 +c2t2(c2 - v2) A2t2 = c2t2A2 = 
A = 
A = 
………..(14)Substituting the Values of A, B, C, D in equation (1) and (4)So the Lorentz transformation equations would be, x’= 
- 
x’= 
(x-vt)x’=
(x-vt)Where 
And t’ = - 
+ 
t’= 
( t - 
)Lorentz transformation equations therefore arex’= 
(x - vt)y’ =yz’ = zt’= 
( t - 
)Or these can be written asx’= 
(x-vt)y’ =yz’ = zt’= 
( t - 
)Where 
This set of equations, relating the position and time in the two inertial frames, is known as the Lorentz transformation. They are named in honour of H.A. Lorentz (1853–1928), who first proposed them. Interestingly, he justified the transformation on what was eventually discovered to be a fallacious hypothesis.The reverse transformation expresses by the following equations. Simply interchanging the primed and unprimed variables and substituting gives:x= 
(x’+ vt)y =y’z = z’t= 
(t’ + 
) Q 12) Derive equation of motion and also explain Coriolis acceleration and Centripetal acceleration? A 12)Time derivatives in fixed and rotating coordinates
Figure 9: Time derivatives in fixed and rotating coordinates Let 
= position vector from the origin (center of the earth). Then, we can find out that the instantaneous tangential velocity of a particle at any distance 
(
+
where a = mean radius of earth and z = altitude above ground) is given by
……….(1)Here 
is the vector angular velocity of the earth rotation. We can see this by first looking at the magnitude of 
. According to the above figure,
……….(2)Now, let's try to relate 
in 2 reference frames:a) Fixed (absolute) system (
)b) Rotating (
) system (
)
Figure 10: Rotating coordinates Now, a vector 
is the same vector no matter what coordinate system it is viewed from. This
Given that the primed system is rotating, however, the time derivative of 
will be different if viewed from the 2 systems. Mathematically, we have'a' for absolute
……….(3)Note that the primed unit vectors vary with time. Due to rotation of unit vector
Figure 11: Rotation of unit vectorRecall earlier that 
a particle rotating at angular velocity 
. Since
therefore
The above is correct only for a vector whose length does not change. The same is true for the unit vectors, therefore
EQUATION OF MOTION IN ABSOLUTE AND ROTATING COORDINATESLet's now apply concepts from the previous section to any arbitrary velocity vector of a particle assuming that the origins of the rotating and fixed (inertial) frames are the same. In this case, the position vector to any location is the same in both systems.Our goal is to find the equations of motion in the absolute and rotating coordinates.Now, we just showed that
……….(8)
mv2In pure rotation, however, the rigid body has no "over all" translation of the body. However, the body in rotation must have kinetic energy as it involves certain motion. A closer look on the rotation of rigid body reveals that though we may not be able to assign translation to the rigid body as a whole, but we can recognize translation of individual particles as each of them rotate about the axis in circular motion with different linear speeds. The speed of a particle is given by:
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mivi2where " Ki" is the kinetic energy of "i" th particle having a speed " vi". In terms of angular speed, the kinetic energy of an individual particle is: Ki=mi(ωri2Ki=mi(ωri)2Now, the kinetic energy of the rigid body is sum of the kinetic energies of the particles constituting the rigid body: K=∑Ki=∑mi(ωri)2We note here that angular speeds of all particles constituting the body are same. Hence, the constant "1/2" and " ω2" can be taken out of the summation sign : K =ω2∑mi(ri)2However, we know that: I=∑mi(ri)2Combining two equations, we have: K= Iω2This is the desired expression of kinetic energy of a rigid body rotating about a fixed axis i.e. in pure rotational motion. The form of expression of the kinetic energy here emphasizes the correspondence between linear and angular quantities. Comparing with the expression of kinetic energy for translational motion, we find that "moment of inertia (I)" corresponds to "mass (m)" and "linear speed (v)" corresponds to "angular speed (ω)". We can also interpret the result obtained above from a different perspective. We could have directly inferred that expression of kinetic energy in rotation should have an equivalent form as: K (Kinetic energy)= x(inertia)x(speed)2In rotation, inertia to the rotation is "moment of inertia (I)" and speed of the rigid body is "angular speed (ω)". Substituting for the quantities, we have the expression for kinetic energy of rigid body in rotation as : K= Iω2Comparing this equation with the expression of the sum of kinetic energy of individual particles as derived earlier: K =ω2∑mi(ri)2Clearly, I= ∑mi(ri)2This conclusion, thus, clearly establishes that the expression as given by ∑mi(ri)2 indeed represents the inertia of the rigid body in rotation. Q4) Explain the rotation in plane also derive the conditions of the instantaneous velocity vector of any particle is always perpendicular to its instantaneous position vector?A 4) When all parts of a rigid body move parallel to a fixed plane, then the motion of the object is referred to as plane motion. There are two types of plane motion, which are given as follows:The pure rotational motion: The rigid body in such a motion rotates about a fixed axis that is perpendicular to a fixed plane. In other words, the axis is fixed and does not move or change its direction relative to an inertial frame of reference. The general plane motion: The motion here can be considered as a combination of pure translational motion parallel to a fixed plane in addition to a pure rotational motion about an axis that is perpendicular to that plane. Suppose a rigid body of an arbitrary shape is in pure rotational motion about the z-axis (see Figure 19(a)). Let us analyze the motion of a particle that lies in a slice of the body in the x-y plane as in Figure 19(b). This particle (at point P) will rotate in a circle of fixed radius r which represents the perpendicular distance from P to the axis of rotation. If you look at any other particle in the object you will see that every particle will rotate in its own circle that has the axis of rotation at its center. In other words, different particles move in different circles but the center of all of these circles lies on the rotational axis. Suppose the particle moves through an arc length s starting at the positive x-axis. Its angular position is then given byθ=s/rr and θ are the polar coordinates of a point in a plane where θ is always measured from the positive x-axis. Because θ is the ratio of the arc length to the radius, it is a pure (dimensionless) number.
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Figure 19(a): A rigid body of an arbitrary shape is in pure rotational motion about the z-axis
| Figure 19(b): The motion of a particle that lies in a slice of the body in the x-y plane
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. Let b be its distance from the origin O. The rigidity condition requires that = bWhere b is constant.Squaring the above both sides we obtain = . =Differentiate both sides with respect to time to obtain. + = 0. + = 0. = 0This equation shown that whenever a rigid body rotates about a fixed point, the instantaneous velocity vector of any particle is always perpendicular to its instantaneous position vector. Q 5) Derive relation of angular momentum about a point of a rigid body in planar motion?A 5) Consider a rigid body which a rotating about a fixed point with an angular velocity . Choose the origin O at the fixed point and axes as shown in Figure 21The velocity of particle i having position vector is = …………(1)If mi is the mass of particle i then its angular momentum about the fixed point O is = …………(2)Substituting the value of , it takes the form = …………(3)
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= = …………(4)Using the vector relation …………(5)Eq. (4) may be written as = = …………(6)Let (xi, yi, zi) be the cartesian coordinates of particle i, then …………(7)If ωx, ωy and ωz are the components of angular velocity along coordinate axes then …………(8)Now take the dot product of with to obtain . …………(9)Substitutes these values in eq. (6) to obtain
In this expansion it is useful to introduce the following abbreviations.
…………(11) These nine quantities Ixx, Ixy,….Izy are the components of the moments of intertia of the body about the fixed x,y,z axes. The components Ixx, Iyy and Izz are called the moments of intertia about x-axis, y-axis and z-axis respectively and the components Ixy,Iyz, …., Izy are called the products of intertia. A quantity such as the moment of intertia is called a tensor of second rank. In terms of the moments of intertia and products of intertia,
…………(12) This equation shows that the angular momentum vector Eq. (11) shows that
Thus the rotational behaviour of a rigid body about a given point it determined by a set of six quantities: the three moments of intertia and the three products of intertia. These moments of intertia and products of intertia will in general be function of time.
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y=rsinθ=3sinπ/4= z=4Therefore point is () in rectangular coordinates.To go in the other direction (from rectangular coordinates to cylindrical), use the following set of formulas (again, the first two are exactly what we use to convert from rectangular to polar in R2:r=x2+y2θ=tan−1 z=z Q 8) Plot the point with spherical coordinates (8, π/3, π/6) and express its location in both rectangular and cylindrical coordinates.A 8) Use the equations in Note to translate between spherical and cylindrical coordinates ρ=x2+y2+z2x=ρcosθsinϕ = 8sin(π/6)cos(π/3) = 8(1/2)1/2=2y=ρsinθsinϕ =8sin(π/6)sin(π/3)=8(1/2)=2z=ρcosϕ =ρcosφ =8cos(π/6)= 8( =4The point with spherical coordinates (8,π/3,π/6) has rectangular coordinates (2, 24)Finding the values in cylindrical coordinates is equally straightforward:r=ρsinφ=8sinπ/6= 4θ= θz=ρcosφ=8cosπ/6= 4Thus, cylindrical coordinates for the point are (4,π/3, 4).
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This is obvious anyway it’s just the addition of velocities formula.ux = u’x +vSince v is constant. That is to say, a’x =axThe acceleration is the same in both frames. This again is obvious as the acceleration is the rate of change of velocity, and the velocities of the same particle measured in the two frames differ by a constant factor-the relative velocity of the two frames.Q 11) Derive Lorentz Transformation equation?A 11)Lorentz transformation is the relationship between two different coordinate frames that move at a constant velocity and are relative to each other. The name of the transformation comes from a Dutch physicist Hendrik Lorentz.The Galilean transformation nevertheless violates Einstein’s postulates, because the velocity equations state that a pulse of light moving with speed c along the x-axis would travel at speed c−v in the other inertial frame. Specifically, the spherical pulse has radius r=ct at time t in the unprimed frame, and also has radius r′=ct′ at time t' in the primed frame. Expressing these relations in Cartesian coordinates givesx2+y2+z2−c2t2 =0 ………..(1)x′2+y′2+z′2−c2t′2=0 ………..(2)The left-hand sides Equations and can be set equal because both are zero. Because y=y′ and z=z′ so, we obtainx2−c2t2 =x′2−c2t′2 ………..(3)However this cannot be satisfied for nonzero relative velocity v of the two frames if we assume the Galilean transformation results in t=t′ with x=x′+vt′.To find the correct set of transformation equations, assume the two coordinate systems S and S′ The laws of mechanics are invariant under Galilean transformations, whereas electrodynamics and Maxwell’s equations are varying under this transformation. This shows that the velocity of light will have different values for different observers moving with different uniform velocities. But the speed of light is invariant in all inertial frames of reference; hence, Galilean transformations need to be modified. Lorentz deduced the transformation equations, which are in agreement with the results of Michelson–Morley experiment. The Lorentz transformations are deduced as given below.Lorentz transformation equations are developed. Let x, y, z and t be the space and time coordinate in rest system or standard system. The system with x', y', z' moves with velocity 'v' along the x-axis. The origins at t=t'=0 coincide. The most general transformation equations relating the coordinates of an event in two systems can be written as, x’ = Ax +Bt ………..(4)y’ = y ………..(5)z’ = z ………..(6)t’ = Cx +Dt ………..(7)
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The transformations are linear.If they are not linear one system would predict acceleration while in other system velocity was even constant. Because of symmetry, y' and z' axes are left unchanged by the transformation.Constant can be calculated by using the following boundary conditions.
………..(13)
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A = A = ………..(14)Substituting the Values of A, B, C, D in equation (1) and (4)So the Lorentz transformation equations would be, x’= - x’= (x-vt)x’=(x-vt)Where And t’ = - + t’= ( t - )Lorentz transformation equations therefore arex’= (x - vt)y’ =yz’ = zt’= ( t - )Or these can be written asx’= (x-vt)y’ =yz’ = zt’= ( t - )Where This set of equations, relating the position and time in the two inertial frames, is known as the Lorentz transformation. They are named in honour of H.A. Lorentz (1853–1928), who first proposed them. Interestingly, he justified the transformation on what was eventually discovered to be a fallacious hypothesis.The reverse transformation expresses by the following equations. Simply interchanging the primed and unprimed variables and substituting gives:x= (x’+ vt)y =y’z = z’t= (t’ + ) Q 12) Derive equation of motion and also explain Coriolis acceleration and Centripetal acceleration? A 12)Time derivatives in fixed and rotating coordinates
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= position vector from the origin (center of the earth). Then, we can find out that the instantaneous tangential velocity of a particle at any distance ( + where a = mean radius of earth and z = altitude above ground) is given by ……….(1)Here is the vector angular velocity of the earth rotation. We can see this by first looking at the magnitude of . According to the above figure, ……….(2)Now, let's try to relate in 2 reference frames:a) Fixed (absolute) system ()b) Rotating ( ) system ()
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is the same vector no matter what coordinate system it is viewed from. ThisGiven that the primed system is rotating, however, the time derivative of will be different if viewed from the 2 systems. Mathematically, we have'a' for absolute
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a particle rotating at angular velocity . SincethereforeThe above is correct only for a vector whose length does not change. The same is true for the unit vectors, therefore
If we now look at (
which is a time rate of change in Thus, from (3), we can write
Let's look at a physical interpretation. If
In another word,
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……….(8)where Apply Eq.(6) to
Apply the Newton's second law
The part in the box is the equation of motion in the rotating coordinate system. It describes the change of (relative) velocity in time subjecting the net force. The forces on the right hand side are real forces, and the second and third term on the left arises because of the coordinate rotation, and there are apparent (not real) forces. We will discuss them in more details in the following. We will look at the last term one the left with respect to an earth oriented coordinate system:
Figure 11: Earth oriented coordinate system
Note that Look at the last term in (10)
We can write it as
Using this we can rewrite equation (10) as
Abs. Accel = Rel. Accel. + Coriolis accel + (-Centripetal accel ) = Net real force This is a very important equation. How do we make use of this equation? Let's look at the forces on the right hand side of (11) in the real atmosphere. They are Gravity = PGF (per unit mass) = - These are real forces (those seen and felt in an inertial / fixed reference frame), so let's put them into the equation:
this is the acceleration that we see and measure in our earth coordinate system – not an inertial frame, but the next 2 terms take that into account.
Rearranging (12)
Equation (13) is "THE EQUATION OF MOTION".
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