Chapter 7.10 - Basics of Vector Cross Product

In the previous section, we saw the examples which demonstrates the uniform motion of 'C'. Next we have to learn about angular velocity. For that, first we must be familiar with vector cross products. So we will see that in this section

1. Fig.7.53(a) below shows a point P in space

Fig.7.53

• There can be infinite number of planes on which 'P' can lie. This is shown in the fig.b
2. Those planes in fig.b come in all possible orientations
• That means, we can take a plane of any orientation. That plane can be positioned so that 'P' will lie on it
• We get this freedom because there is only one point 'P'
3. What if there is a second point 'Q'?
• This is shown in fig.7.54(a) below:

Fig.7.54

• Note that, a straight line can be drawn between any two points in space
• If the 'line connecting P and Q' lies on a plane, we can say that, both P and Q lies on that plane
4. How many such planes are possible?
Ans: Infinite number of such planes are possible. Fig.7.54(b) shows three of them
5. However, the following 6 facts should be noted:
(i) Though infinite number of planes are possible in this case also, 'any orientation' cannot be allowed
(ii) We first fix up one plane which contains the line PQ
(iii) Then we rotate that plane by considering the line PQ as the 'axis of that rotation'
(iv) All planes thus obtained by rotating through various angles will contain both P and Q
(v) Thus infinite number of planes are possible
(vi) But 'all possible orientations' that we saw in fig.7.53(b) is not possible
6. Thus we find that, 2 points also give as a large amount of freedom with regard to the 'choice of planes'
• What if there is a third point R?
• This is shown in fig.7.55(a) below:

Fig.7.55

(i) We first fix up a plane containing line PQ
(ii) Then we rotate that plane by considering the line PQ as the axis
(iii) Just when the third point R falls on that plane, we stop the rotation
(iv) That is the plane we want
• No other plane will contain all the three point P, Q and R
• That means, there is one and only one plane which contains 3 given points in space
• This is shown in fig.7.55(b)
7. What if there is a fourth point 'S'?
Ans: If there are a total of four given points, there is no guarantee that, we will ever find a plane which contains all four of them
• It depends on the positions of those points in space
• We can randomly take any three of them. There sure will be a plane containing those three
• But accommodating the fourth point also in that plane may not be possible

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Now we can start the discussion on vector products

• We have seen scalar product of two vectors in chapter 6
■ The result that we obtain when performing 'scalar multiplication' of two given vectors will be a scalar
    ♦ Eg: We multiply force vector and displacement vector to obtain 'work', which is a scalar
■ The result that we obtain when performing 'vector multiplication' of two given vectors will be a vector
• We will see examples in later sections
• First we will see the features of vector products. We will write them in steps:
1. Consider the two vectors $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$ shown in fig.7.56(a) below:

Fig.7.56

• To find their vector product, both of them must lie on a plane
2. Now, $\mathbf\small{\vec{a}}$ has two well defined points: It's tail end and head end
• Similarly, $\mathbf\small{\vec{b}}$ has two well defined points: It's tail end and head end
• So there are a total of four distinct points
3. If both $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$ are to lie on a single plane, all those four point must lie on that plane
• But we have seen that any four points need not lie on a single plane
4. So we make an adjustment:
• We shift one of the vectors so that, the tail ends of both the vectors coincide 
• Then there will only be three points to consider. We can sure find the plane which contains those three points
• In fig.7.56(b), $\mathbf\small{\vec{b}}$ has been shifted so that it's tail coincide with that of $\mathbf\small{\vec{a}}$
5. Now, we are going to work on the 'plane which contains both $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
• In our present case, it is convenient to make that plane coincide with our familiar xy-plane
• For that, we rearrange the frame of reference
• The rearrangement is a simple process. Only 3 steps are required:
(i) Take the reference frame and put it in such a way that, the 'O' is in the 'plane of $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
(ii) With the 'O' as pivot, rotate the frame in a suitable direction so that, the x-axis falls on that 'plane of $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
(iii) With the x-axis as the 'axis of rotation', rotate the frame in a suitable direction so that, the y-axis also falls on that 'plane of $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
• The final position is shown in fig.7.57(a) below:

Fig.7.57

6. Now, according to the 'definition of vector product', the 'product vector' that we are seeking, will be perpendicular to both $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$
• That means: The 'product vector' will be perpendicular to the 'plane of $\mathbf\small{\vec{a}}$ and $\mathbf\small{\vec{b}}$'
• For convenience, we will draw a line (perpendicular to the plane) through the tail ends of the vectors. It is shown in magenta color in fig.7.57(b)
• We can assume that, the product vector that we are seeking, lies along that magenta line
7. We see that the magenta line is parallel to the z axis. 
• Now, the next question arises:
■ What is the direction of the vector that we are seeking?
• We know that, every vector has a direction
• So just saying that 'the product vector lies along the magenta line in fig.7.57(b)' is not good enough
• We want the answer for these questions as well:
Is the product vector directed towards the positive side of the z-axis?
OR, Is it directed towards the negative side of the z-axis?
8. To find the answer, we use the right hand screw rule. Let us see how this rule is applied:
• To learn about this rule, first we must know the difference between these two:
(i) A right handed screw (ii) A left handed screw
9. In fig.7.58(a) below, a screw is about to be driven into a wooden block

Fig.7.58

• The direction in which the head of the screw is turned is important
• In the fig.a, the direction of turning is shown by the brown curved arrow
10. Is this brown arrow showing clockwise direction?
OR, Is it showing anti-clockwise direction?
Ans: The answer depends on the direction in which we look at the screw
• If we look at the screw (along it's axis) from it's head towards the tail, the arrow can be said to be clockwise
• If we look at the screw (along it's axis) from it's tail towards the head, the arrow will become anti-clockwise
11. Normally, while turning a screw, we will be looking at it from the head towards the tail
• So we need not specify the direction in which we are looking
• All we need to mention is one of the two below (whichever is appropriate):
(i) Turning a screw in clockwise direction
(ii) Turning a screw in anti-clockwise direction
• Both will imply that, we are looking from the head towards the tail
12. So, the brown screw in fig.7.58(a) above, is being turned in the clockwise direction
• That screw will advance into the block
13. The cyan screw in fig.7.58(b) above, is also being turned in the clockwise direction
• That screw will not advance into the block  
• In fact, the opposite will happen. That is., the cyan screw will back out from the block
14. To put it in simple words, we can write the following 3 statements:
(i) Both brown and cyan screws are turned in the clockwise direction
(ii) The brown screw will tighten into the block
(iii) The cyan screw will loosen away from the block
■ The brown screw in fig.a is a right handed screw
■ The cyan screw in fig.b is a left handed screw
• Normally, those we buy from hardware stores are right handed screws
• Left handed screws are manufactured only in small quantities. They are used for some special purposes

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Now that we know the difference between right handed and left handed screws, we can see how the right hand screw rule is applied in vector multiplication. We will see that in the next section

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Chapter 7.9 - Frame of Reference at Center of Mass

In the previous section, we saw that, if the external force acting on a system of particles is zero, then the 'C' of that system will be moving with uniform velocity. In this section we will see some examples for this type of motion

Example 1:
We will write it in steps:

1. Consider a moving radium nucleus (Ra)
• The radium nucleus is an unstable particle. It disintegrates into two other particles: 
(i) A radon nucleus (Ra)
(ii) An alpha particle (He)
2. The forces leading to the decay are internal. Those forces do not contribute to the $\mathbf\small{\vec{F}_{system}}$
• The other external forces acting on the system are negligible
• So we can write: $\mathbf\small{\vec{F}_{system}=0}$  
3. Then what happens to the 'C' of the system?
• We have seen that, if $\mathbf\small{\vec{F}_{system}=0}$, the 'C' of that system will travel at a constant speed along a straight line
• Consider fig.7.47(a) below:

Fig.7.47

• The Ra was initially travelling along the x-axis. So the 'C' of the system is on the x-axis
• After disintegration, there is no Ra. It has changed to Rn and He
4. So what happened to the 'C'?
• Ans: Since $\mathbf\small{\vec{F}_{system}=0}$, the 'C' will continue to be on the x-axis
• It travels at a constant speed along the x-axis
5. Take any instant. Measure the distances of Rn and He (from the x-axis) at that instant
• Upon calculations on those distances, we will see that, the 'C' (of the Rn-He system) will be on the x-axis
6. In the above analysis, the observer is on the ground and watching the disintegration of Ra
• A special case occurs when the observer is at the center of 'C', travelling with it. Let us see it's details:
7. Initially, the observer is at the center of Ra nucleus
• After disintegration, the observer finds that, the Ra nucleus which was covering him, has disappeared
• Instead, two new particles (Rn and He) have formed
• From his view point, 
    ♦ What will be the direction of Rn?
    ♦ What will be the direction of He?
8. Let us analyse:
• We have learned about relative velocity (Details here)
• Relative velocity of Rn with respect to 'C' is given by:
$\mathbf\small{\vec{v}_{(Rn)(C)}=\vec{v}_{(Rn)}-\vec{v}_{(C)}}$
• When viewed from 'C', the Rn will appear to be moving with the above $\mathbf\small{\vec{v}_{(Rn)(C)}}$
9. Now, we can find $\mathbf\small{\vec{v}_{(C)}}$ using Eq.7.7: $\mathbf\small{\vec{v}_{C(t)}=\frac{\sum {m_i\;\vec{v}_{i(t)}}}{M}}$
• Substituting the known values, we get:
$\mathbf\small{\vec{v}_{C}=\frac{m_{Rn}\;\vec{v}_{Rn}+m_{He}\;\vec{v}_{He}}{m_{Rn}+m_{He}}}$
10. Now the equation in (8) becomes:
$\mathbf\small{\vec{v}_{(Rn)(C)}=\vec{v}_{(Rn)}-\frac{m_{Rn}\;\vec{v}_{Rn}+m_{He}\;\vec{v}_{He}}{m_{Rn}+m_{He}}}$
$\mathbf\small{\Rightarrow \vec{v}_{(Rn)(C)}=\frac{m_{Rn}\;\vec{v}_{Rn}+m_{He}\;\vec{v}_{Rn}-m_{Rn}\;\vec{v}_{Rn}-m_{He}\;\vec{v}_{He}}{m_{Rn}+m_{He}}}$
$\mathbf\small{\Rightarrow \vec{v}_{(Rn)(C)}=\frac{m_{He}[\vec{v}_{Rn}-\;\vec{v}_{He}]}{m_{Rn}+m_{He}}}$
11. Similarly, the relative velocity of He with respect to 'C' is given by:
$\mathbf\small{\vec{v}_{(He)(C)}=\vec{v}_{(He)}-\vec{v}_{(C)}}$
• When viewed from 'C', the He will appear to be moving with the above $\mathbf\small{\vec{v}_{(He)(C)}}$
12. We have already calculated $\mathbf\small{\vec{v}_{C}}$  in (9) above
• So the equation in (11) becomes:
$\mathbf\small{\vec{v}_{(He)(C)}=\vec{v}_{(He)}-\frac{m_{Rn}\;\vec{v}_{Rn}+m_{He}\;\vec{v}_{He}}{m_{Rn}+m_{He}}}$
$\mathbf\small{\Rightarrow \vec{v}_{(He)(C)}=\frac{m_{Rn}\;\vec{v}_{He}+m_{He}\;\vec{v}_{He}-m_{Rn}\;\vec{v}_{Rn}-m_{He}\;\vec{v}_{He}}{m_{Rn}+m_{He}}}$
$\mathbf\small{\Rightarrow \vec{v}_{(He)(C)}=\frac{m_{Rn}[\vec{v}_{He}-\;\vec{v}_{Rn}]}{m_{Rn}+m_{He}}}$
13. So we have two results:
(i) When viewed from 'C', the Rn will appear to be moving with a velocity given by:
$\mathbf\small{\vec{v}_{(Rn)(C)}=\frac{m_{He}[\vec{v}_{Rn}-\;\vec{v}_{He}]}{m_{Rn}+m_{He}}}$
(ii) When viewed from 'C', the He will appear to be moving with a velocity given by:
$\mathbf\small{\vec{v}_{(He)(C)}=\frac{m_{Rn}[\vec{v}_{He}-\;\vec{v}_{Rn}]}{m_{Rn}+m_{He}}}$
14. We see that, the denominators are the same for both results
• In the numerator, the 'quantity with in the brackets' are also the same, except for the sign
• So the two velocities in (13) will be having opposite directions
■ That means, when viewed from 'C', the Rn will appear to be moving exactly opposite to He
15. But that is not all:
• $\mathbf\small{\vec{v}_{(Rn)(C)}}$  starts from 'C'
• $\mathbf\small{\vec{v}_{(He)(C)}}$ also starts from 'C'
• So the two vectors have a common point
■ That means, the two vectors will be lying on the same line
16. So we can write:
■When viewed from 'C', both Rn and He will appear to be moving back to back along the same straight line. This is shown in fig.7.47(b) above

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Example 2:

1. Consider a system of binary (double) stars shown in the animation in fig.7.48 below:

Fig.7.48

• The yellow and pink spheres represent two stars P and Q, having equal masses
• The magenta line is an imaginary line joining the centers of P and Q
2. 'C' of the system is shown as a small white sphere
• Since the masses are equal, 
    ♦ The 'C' of the system will lie on the magenta line
    ♦ Also, 'C' will be exact midway between P and Q
• Both P and Q rotate about 'C'
3. In the fig.7.48 above, the 'C' is stationary on the cyan line
• The animation in fig.7.49 below, shows the same system. But this time, 'C' is moving along the cyan line

Fig.7.49

4. So we have a system which has rotation as well as translation
• If we trace the paths of the two stars, they will appear as shown in fig.7.50 below:

Fig.7.50

• The series of pink spheres indicate the path followed by P
• The series of yellow spheres indicate the path followed by Q
5. When the crest of Q and trough of P occur together,
    ♦ Q spheres are further apart
    ♦ P spheres are closer together
• When the crest of P and trough of Q occur together,
    ♦ P spheres are further apart
    ♦ Q spheres are closer together
■ So we see that the trajectories are complicated. We will try to obtain simplified trajectories
6. Generally, there is no external force acting on the binary star system
• So the 'C' in fig.7.49 will be moving with a uniform velocity
7. As in the previous example, let the observer be at the center of 'C'
• From his view point, 
    ♦ What will be the direction of P?
    ♦ What will be the direction of Q?
8. Let us analyse:
• We have learned about relative velocity (Details here)
• Relative velocity of P with respect to 'C' is given by:
$\mathbf\small{\vec{v}_{(P)(C)}=\vec{v}_{(P)}-\vec{v}_{(C)}}$
• When viewed from 'C', the star P will appear to be moving with the above $\mathbf\small{\vec{v}_{(P)(C)}}$
9. Now, we can find $\mathbf\small{\vec{v}_{(C)}}$ using Eq.7.7: $\mathbf\small{\vec{v}_{C(t)}=\frac{\sum {m_i\;\vec{v}_{i(t)}}}{M}}$
• Substituting the known values, we get:
$\mathbf\small{\vec{v}_{C}=\frac{m_{P}\;\vec{v}_{P}+m_{Q}\;\vec{v}_{Q}}{m_{P}+m_{Q}}}$
10. Now the equation in (8) becomes:
$\mathbf\small{\vec{v}_{(P)(C)}=\vec{v}_{(P)}-\frac{m_{P}\;\vec{v}_{P}+m_{Q}\;\vec{v}_{Q}}{m_{P}+m_{Q}}}$
$\mathbf\small{\Rightarrow \vec{v}_{(P)(C)}=\frac{m_{P}\;\vec{v}_{P}+m_{Q}\;\vec{v}_{P}-m_{P}\;\vec{v}_{P}-m_{Q}\;\vec{v}_{Q}}{m_{P}+m_{Q}}}$
$\mathbf\small{\Rightarrow \vec{v}_{(P)(C)}=\frac{m_{Q}[\vec{v}_{P}-\;\vec{v}_{Q}]}{m_{P}+m_{Q}}}$
11. Similarly, the relative velocity of Q with respect to 'C' is given by:
$\mathbf\small{\vec{v}_{(Q)(C)}=\vec{v}_{(Q)}-\vec{v}_{(C)}}$
• When viewed from 'C', the star Q will appear to be moving with the above $\mathbf\small{\vec{v}_{(Q)(C)}}$
12. We have already calculated $\mathbf\small{\vec{v}_{C}}$  in (9) above
• So the equation in (11) becomes:
$\mathbf\small{\vec{v}_{(Q)(C)}=\vec{v}_{(Q)}-\frac{m_{P}\;\vec{v}_{P}+m_{Q}\;\vec{v}_{Q}}{m_{P}+m_{Q}}}$
$\mathbf\small{\Rightarrow \vec{v}_{(Q)(C)}=\frac{m_{P}\;\vec{v}_{Q}+m_{Q}\;\vec{v}_{Q}-m_{P}\;\vec{v}_{P}-m_{Q}\;\vec{v}_{Q}}{m_{P}+m_{P}}}$
$\mathbf\small{\Rightarrow \vec{v}_{(Q)(C)}=\frac{m_{P}[\vec{v}_{Q}-\;\vec{v}_{P}]}{m_{P}+m_{Q}}}$
13. So we have two results:
(i) When viewed from 'C', the star P will appear to be moving with a velocity given by:
$\mathbf\small{\vec{v}_{(P)(C)}=\frac{m_{Q}[\vec{v}_{P}-\;\vec{v}_{Q}]}{m_{P}+m_{Q}}}$
(ii) When viewed from 'C', the star Q will appear to be moving with a velocity given by:
$\mathbf\small{\vec{v}_{(Q)(C)}=\frac{m_{P}[\vec{v}_{Q}-\;\vec{v}_{P}]}{m_{P}+m_{Q}}}$
14. We see that, the denominators are the same for both results
• In the numerator, the 'quantity with in the brackets' are also the same, except for the sign
• So the two velocities in (13) will be having opposite directions
• Also, since m_P = m_Q, the magnitudes will be equal
■ That means, when viewed from 'C', the P will appear to be moving exactly opposite to Q
15. But that is not all:
• In the previous example 1, the two vectors had a common point
• But here, there is no such common point
• So here, the two vectors must satisfy the following three conditions:
(i) They are opposite to each other
(ii) They have same magnitudes
(iii) They have no common point
16. The three conditions can be simultaneously satisfied only if:
■ The two velocities are parallel
• For that, the two velocities must be at the ends of a diameter
• This is shown in fig.7.51(a) below:

Fig.7.51

That means, when viewed from 'C', the stars P and Q will be diametrically opposite
17. The situation 'where they are not diametrically opposite' is shown in fig.7.51(b) 
• It is clear that, in such a situation, the two vectors are not parallel. Their paths will have a common point. They are not opposite to each other

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• From the two examples, it is clear that, from the view point of a person at the 'C', the analysis becomes simpler
• For obtaining such a view point, we attach the frame of reference to the 'C'
• That is., the origin of the frame of reference is placed exactly at the 'C'

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Now we will see some solved examples
Solved example 7.9:
A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and run about on the trolley in any manner, What is the speed of the CM of the (trolley + child) system?
Solution:
1. The trolley is initially moving with uniform speed
• The floor is smooth. So there is no friction
• That means, there is no external net force on the system
2. When the child begins to run, the forces produced are internal
• Those forces do not contribute any thing towards the external force
• That means, the external force remains zero
■ So, the CM continues to move with the same speed V

Solved example 7.10
A large boat and a small boat are at rest on the surface of a lake. The distance between the two boats is 300 m. A man in the large boat, begins to pull the small boat towards him using a light inextensible rope. When the small boat reaches him, the large boat has moved 45 m towards the small boat. If the large boat has a mass of 5400 kg, what is the mass of the small boat? Neglect the friction offered by the water
Solution:
1. Given:
• Mass of large boat = m_L = 5400 kg
• Initial distance between the two boats = 300 m   
• Final distance between the two boats = 45 m
• We are asked to find the mass of the small boat m_S
2. Initial position of the 'C' of the two-boat system can be calculated as follows:
• The 'C' is some where on the line joining the two boats
• Assume that the origin O is at the center of the small boat
• This is shown in fig.7.52 below:

Fig.7.52

• We have: $\mathbf\small{X=\frac{m_S x_S+m_L x_L}{m_S + m_L}}$   
• Substituting the known values, we get: $\mathbf\small{X=\frac{m_S \times 0+5400 \times 300}{m_S + 5400}}$
$\mathbf\small{\Rightarrow X=\frac{1620000}{m_S + 5400}}$
3. When the man pulls the boat, the forces involved are internal
• So there is no external force
• So the velocity of the 'C' does not change
• The 'C' was initially at rest (∵ both boats are initially at rest)
• So the 'C' continues to remain at rest. In other words, there is no change in the position of 'C'
4. Finally, when large boat has traveled 45 m, the two boats touch each other
• When the touching happens, both the boats are at 'C'
• That means, the large boat was initially 45 m away from 'C'
• From the fig.7.52, we get: X = (300-45) = 255 m
5. Substituting this value of X in (2), we get: $\mathbf\small{255=\frac{1620000}{m_S + 5400}}$
• Thus we get: m_S = 953 kg

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Next we have to learn about angular velocity. For that, first we must be familiar with vector cross products. So we will see that in the next section

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