Chapter 4.16 - Relative velocity in Two Dimensions

In our present chapter, we are discussing about motions in two dimensions. In the previous section we completed a discussion on uniform circular motion. In this section, we will see relative velocity in two dimensions.

We saw 'relative velocity in the case of motion in one dimension' in a previous chapter. Our present discussion is based on that.

1. Consider two objects A and B shown in fig.4.41(a) below:

Fig.4.41

• They are moving in two different directions and with two different magnitudes
    ♦ A is moving with a velocity of $\mathbf\small{\vec {v_A}}$
    ♦ B is moving with a velocity of $\mathbf\small{\vec {v_B}}$
2. We can write two cases:
(i) Case 1: When viewed from A, The other object B will be moving with a velocity of [$\mathbf\small{\vec{v_B}-\vec{v_A}}$]
• That is., Relative velocity of B with respect to A = $\mathbf\small{\vec{v_{BA}}=\vec{v_B}-\vec{v_A}}$  
(ii) Case 2: When viewed from B, The other object A will be moving with a velocity of [$\mathbf\small{\vec{v_A}-\vec{v_B}}$]
• That is., Relative velocity of A with respect to B = $\mathbf\small{\vec{v_{AB}}=\vec{v_A}-\vec{v_B}}$  
3. So we need to do vector subtraction to find relative velocities. We already know the methods to do such subtractions. We have both graphical and analytical methods.
■ Here we will use analytical method: 
• We use the horizontal and vertical components. Fig.4.41(b) shows those components. 
4. Consider the resultant $\mathbf\small{\vec{v_{BA}}}$ in case 1
    ♦ The x component of this resultant will be: $\mathbf\small{\vec{v_{Bx}}-\vec{v_{Ax}}}$
    ♦ The y component of this resultant will be: $\mathbf\small{\vec{v_{By}}-\vec{v_{Ay}}}$
Once the components are obtained, we can easily calculate the resultant
• Consider the resultant $\mathbf\small{\vec{v_{AB}}}$ in case 2
    ♦ The x component of this resultant will be: $\mathbf\small{\vec{v_{Ax}}-\vec{v_{Bx}}}$
    ♦ The y component of this resultant will be: $\mathbf\small{\vec{v_{Ay}}-\vec{v_{By}}}$
Once the components are obtained, we can easily calculate the original  

Solved example 4.16

Rain is falling vertically with a speed of 35 ms^-1. A woman rides a bicycle with a speed of 12 ms^-1 in east to west direction. What is the direction in which she should hold her umbrella ?  

Solution:

1. Let the rain fall be denoted by $\mathbf\small{\vec {v_R}}$ 

• Let the motion of bicycle be denoted by $\mathbf\small{\vec {v_B}}$ 

• The two vectors are shown in the fig.4.42(a) below:

Fig.4.42

2. We want to find 'how the velocity of the rain will appear, when viewed from the bicycle'

    ♦ That is., we want to find the velocity of the rain with respect to the bicycle. 

    ♦ That is., we want to find $\mathbf\small{\vec {v_{RB}}}$

    ♦ That is., we want to find $\mathbf\small{\vec{v_R}-\vec{v_B}}$   

3. For that, we need to find the following two:

(i) x component of ($\mathbf\small{\vec{v_R}-\vec{v_B}}$)

    ♦ It is given by: $\mathbf\small{\vec{v_{Rx}}-\vec{v_{Bx}}}$ 

(ii) y component of ($\mathbf\small{\vec{v_R}-\vec{v_B}}$)

    ♦ It is given by: $\mathbf\small{\vec{v_{Ry}}-\vec{v_{By}}}$ 

4. Let us write the values:

• $\mathbf\small{\vec{v_{Rx}}}$ = 0 (∵ the rain is falling vertically and so has no horizontal component)

• $\mathbf\small{\vec{v_{Bx}}}$ = -12 $\mathbf\small{\hat{i}}$ (The travel is from left to right. So it is taken as negative)

• $\mathbf\small{\vec{v_{Ry}}}$ = -35 $\mathbf\small{\hat{j}}$ (The travel is from top to bottom. So it is taken as negative)

• $\mathbf\small{\vec{v_{By}}}$ = 0 (∵ the bicycle is travelling horizontally and so has no vertical component)

5. Substituting the values in 3(i), we get:

• x component of ($\mathbf\small{\vec{v_R}-\vec{v_B}}$) = $\mathbf\small{\vec{v_{RBx}}}$ = [0 - (-12 $\mathbf\small{\hat{i}}$)] = 12 $\mathbf\small{\hat{i}}$
    ♦ We get a positive value. So this vector is directed towards the positive side of the x axis.
    ♦ This is shown in fig.b

Substituting the values in 3(ii), we get:
• y component of ($\mathbf\small{\vec{v_R}-\vec{v_B}}$) = $\mathbf\small{\vec{v_{RBy}}}$ = [(-35 $\mathbf\small{\hat{i}}$ - 0)] = -35 $\mathbf\small{\hat{j}}$
    ♦ We get a negative value. So this vector is directed towards the negative side of the y axis
    ♦ This is also shown in fig.b

6. Now we can find the magnitude:
Magnitude of $\mathbf\small{\vec{v_R}-\vec{v_B}}$ = $\mathbf\small{|\vec{v_R}-\vec{v_B}|}$ = $\mathbf\small{|\vec{v_{RB}}|}$ = $\mathbf\small{\sqrt{12^2+(-35)^2}}$ = 37 ms^-1.
7. Next we find the direction:
• We have: $\mathbf\small{\tan \theta =\frac{|\vec{v{RBy}}|}{|\vec{v{RBx}}|}=\frac{35}{12}}$
• Thus we get: θ = 71.076^o
■ We must note the following points 3 while using this formula:
(i) Magnitude of the y component is in the numerator
(ii) Magnitude of the x component is in the denominator
(iii) The angle 'θ' thus obtained will always be the angle between the resultant and the x component
• This 'θ' is marked in the fig.c
8. But what we want is 'φ' which the resultant makes with the vertical
• It can be easily obtained because, in fig.c, we can see that φ = (90-θ) 
• Thus we get: φ = (90-71.076) = 18.924^o.
9. So we can write the result:
• When viewed from the bicycle, rain falls towards the cyclist at an angle of 18.924^o with the vertical
• So the cyclist must hold the umbrella at an angle of 18.924^o with the vertical,
10. We must note the difference between this example and example 4.1.
• In that example, the boy was stationary. The rain was falling at an angle because of the wind
• But here, the rain is falling vertically. But to the cyclist, it appears to be falling at an angle 

Solved example 4.17
In a harbour, wind is blowing at a speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?
Solution:
1. The speed and direction of the wind is given:
• Speed is 72 km/h and direction is N-E
2. Let us set up the coordinate axes as follows:
    ♦ E-W direction is the x axis
    ♦ N-S direction is the y axis
• Then the N-E direction (in which the wind is blowing) will be inclined at 45^o with the x axis.
• This is shown in fig.4.43(a) below:

Fig.4.43

3. Now we can write the details about the velocity vector of wind
• Let us denote it as $\mathbf\small{\vec {v_W}}$
• We can write:
    ♦ $\mathbf\small{|\vec {v_W}|}$ = 72 km/h
    ♦ $\mathbf\small{\vec {v_W}}$ makes an angle 45^o with the x axis
• This is also shown in fig.a
4. This velocity of the wind will not change
• But when viewed from a moving boat, the wind will appear to have a different magnitude and different direction

• We want to find 'how the velocity of the wind will appear, when viewed from the moving boat'

    ♦ That is., we want to find the velocity of the wind with respect to the moving boat. 

    ♦ That is., we want to find $\mathbf\small{\vec {v_{WB}}}$

    ♦ That is., we want to find $\mathbf\small{\vec{v_W}-\vec{v_B}}$ 
Where $\mathbf\small{\vec {v_B}}$ is the velocity of the boat

5. For that, we need to find the following two:

(i) x component of ($\mathbf\small{\vec{v_W}-\vec{v_B}}$)

    ♦ It is given by: $\mathbf\small{\vec{v_{Wx}}-\vec{v_{Bx}}}$ 

(ii) y component of ($\mathbf\small{\vec{v_W}-\vec{v_B}}$)

    ♦ It is given by: $\mathbf\small{\vec{v_{Wy}}-\vec{v_{By}}}$ 

6. Let us write the values:

• $\mathbf\small{\vec{v_{Wx}}}$ = 72 cos 45 $\mathbf\small{\hat{i}}$ = 50.912 $\mathbf\small{\hat{i}}$ km/h

• $\mathbf\small{\vec{v_{Bx}}}$ = 0 (∵ the boat is travelling north and so has no component towards east)
• $\mathbf\small{\vec{v_{Wy}}}$ = 72 sin 45 $\mathbf\small{\hat{j}}$ = 50.912 $\mathbf\small{\hat{i}}$ km/h
• $\mathbf\small{\vec{v_{By}}}$ = 51 $\mathbf\small{\hat{j}}$

7. Substituting the values in 5(i), we get:

• x component of ($\mathbf\small{\vec{v_W}-\vec{v_B}}$) = $\mathbf\small{\vec{v_{WBx}}}$ = [50.912 $\mathbf\small{\hat{i}}$ - 0] = 50.912 $\mathbf\small{\hat{i}}$
    ♦ We get a positive value. So this vector is directed towards the positive side of the x axis.
    ♦ This is shown in fig.b

Substituting the values in 5(ii), we get:
• y component of ($\mathbf\small{\vec{v_W}-\vec{v_B}}$) = $\mathbf\small{\vec{v_{WBy}}}$ = [(72 sin 45 $\mathbf\small{\hat{j}}$ - 51 $\mathbf\small{\hat{j}}$)] = -0.0883
    ♦ We get a negative value. So this vector is directed towards the negative side of the y axis
    ♦ This is also shown in fig.b
8. Now we can find the magnitude. But it is not asked in the question
9. But we do have to find the direction
• We have: $\mathbf\small{\tan \theta =\frac{|\vec{v_{WBy}}|}{|\vec{v_{WBx}}|}=\frac{0.0883}{50.912}}$
• Thus we get: θ = 0.0997^o
■ We must note the following points 3 while using this formula:
(i) Magnitude of the y component is in the numerator
(ii) Magnitude of the x component is in the denominator
(iii) The angle 'θ' thus obtained will always be the angle between the resultant and the x component
• This 'θ' is marked in the fig.c
10. We see that, the relative velocity vector falls between the east and south directions
• But the deviation from east is only 0.0997^o, which is a very small quantity
• So we can write:
■ The flag flutters in the east direction approximately

Solved example 4.18
When a man walks at a rate of 3 km/h, the rain appears to fall vertically. When he walks at the rate of 6 km/h, it appears to fall at an angle of 45° with the vertical. What is the original magnitude and direction of the rain?
Solution:
1. Consider the simple case when the rain falls in the exact vertical direction
■ For a man moving from right to left, the rain will obviously appear to be falling at a slope
• This slope will be towards the right.
• That is., $\mathbf\small{\vec{v_{RM}}}$ will slope towards the right
    ♦ Where $\mathbf\small{\vec{v_{RM}}}$ is the relative velocity of the rain with respect to the man
    ♦ This is shown in fig.4.44(a) below:

Fig.4.44

2. But in our present case, even when there is motion at 3 km/h, the rain appears to be exact vertical
■ So it is clear that, the original rain is sloping towards the left. This is shown in fig.b
• We are asked to find this original magnitude and direction
• That is., we need to find the details of $\mathbf\small{\vec{v_{R}}}$ in fig.b
3. We know this:
• The two components of $\mathbf\small{\vec{v_{RM}}}$ in fig.b are:
(i) $\mathbf\small{\vec{v_{RMx}}}$
(ii) $\mathbf\small{\vec{v_{RMy}}}$
4. So we will first find those components:
(i) We have: $\mathbf\small{\vec{v_{RMx}}=\vec{v_{Rx}}-\vec{v_{Mx}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Rx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}}$
    ♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
    ♦ This is shown in fig.c
(b) $\mathbf\small{\vec{v_{Mx}}=3 \hat{i}}$
• Substituting in 4(i), we get: $\mathbf\small{\vec{v_{RMx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}-3 \hat{i}}$
(ii) We have: $\mathbf\small{\vec{v_{RMy}}=\vec{v_{Ry}}-\vec{v_{My}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Ry}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
    ♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
    ♦ This is shown in fig.c
(b) $\mathbf\small{\vec{v_{My}}=0}$
• Substituting in 4(ii), we get: $\mathbf\small{\vec{v_{RMy}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
5. Now consider the $\mathbf\small{\vec{v_{RM}}}$ in fig.c
• It is perfect vertical. That means, it has no horizontal component
• That means, $\mathbf\small{\vec{v_{RMx}}}$ = 0
• From 4(i)b, we can write: $\mathbf\small{\vec{v_{RMx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}-3 \hat{i}}$ = 0
■ From this we get: $\mathbf\small{|\vec{v_{R}}|\cos \theta=3}$
This is one of the two equations which we will need to find $\mathbf\small{|\vec{v_{R}}|}$ and θ.
6. To get the other equation, we consider the motion at 6 km/h
• This time, the $\mathbf\small{\vec{v_{RM}}}$ is indeed at a slope. This is shown in fig.d
    ♦ Given that, it makes an angle of 45° with the vertical
    ♦ So it will make the same angle of 45° with the horizontal also
Note the two points:
(i) $\mathbf\small{\vec{v_{R}}}$ in fig.d is same as that in fig.c. This is because, the original speed and direction of rain does not change 
(ii) $\mathbf\small{\vec{v_{M}}}$ in fig.d has the same direction as that in fig.c But the magnitude changed from 3 to 6 km/h
7. We know this:
• The two components of $\mathbf\small{\vec{v_{RM}}}$ in fig.d are:
(i) $\mathbf\small{\vec{v_{RMx}}}$
(ii) $\mathbf\small{\vec{v_{RMy}}}$
8. So we will first find those components:
(i) We have: $\mathbf\small{\vec{v_{RMx}}=\vec{v_{Rx}}-\vec{v_{Mx}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Rx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}}$
    ♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
    ♦ This is shown in fig.d
(b) $\mathbf\small{\vec{v_{Mx}}=6 \hat{i}}$
• Substituting in 8(i), we get: $\mathbf\small{\vec{v_{RMx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}-6 \hat{i}}$
(ii) We have: $\mathbf\small{\vec{v_{RMy}}=\vec{v_{Ry}}-\vec{v_{My}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Ry}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
    ♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
    ♦ This is shown in fig.d
(b) $\mathbf\small{\vec{v_{My}}=0}$
• Substituting in 8(ii), we get: $\mathbf\small{\vec{v_{RMy}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
9. Using the 'formula for direction of the resultant', we can write:
$\mathbf\small{\tan 45=\frac{|\vec{v_{RMy}}|}{|\vec{v_{RMx}}|}=\frac{||\vec{v_R}|\sin \theta| }{|(|\vec{v_R}|\cos \theta)-6|}}$
10. But in (5), we obtained: $\mathbf\small{|\vec{v_{R}}|\cos \theta=3}$
Substituting this in (9), we get: $\mathbf\small{\tan 45=\frac{||\vec{v_R}|\sin \theta| }{|(3)-6|}=\frac{||\vec{v_R}|\sin \theta| }{|(-3)|}=\frac{|\vec{v_R}|\sin \theta }{3}}$
11. But tan 45 = $\mathbf\small{\frac{1}{\sqrt{2}}}$
Thus we get: $\mathbf\small{|\vec{v_R}|\sin \theta =\frac{3}{\sqrt{2}}}$
12.So we have two results:
(i) $\mathbf\small{|\vec{v_{R}}|\cos \theta=3}$
(ii) $\mathbf\small{|\vec{v_R}|\sin \theta =\frac{3}{\sqrt{2}}}$
Dividing (ii) by (i), we get: $\mathbf\small{\tan \theta =\frac{1}{\sqrt{2}}}$
Thus we get θ = 45°.
13. Substituting this value of θ in 12(i), we get:
$\mathbf\small{|\vec{v_{R}}|\cos 45=3}$
$\mathbf\small{\Rightarrow |\vec{v_{R}}|=\frac{3}{\frac{1}{\sqrt{2}}}=3\sqrt{2}=4.243\: km/h}$
14. So we can write:
• The rain is actually falling at an angle of 45° towards the left
• It has a magnitude of 4.243 km/h
• But when the man walks ar the rate of 6 km/h, the rain appears to be falling at an angle of 45° towards the right

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In the next section, we will see a few more solved examples.

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Chapter 4.14- Uniform Circular Motion

In the previous section we completed a discussion on projectile motion. In this section, we will see Uniform circular motion.

If we want to say that an 'object is in uniform circular motion', the following two conditions should be satisfied:
(i) The object should be moving in a circular path
    ♦ The center of that circle should not change 
    ♦ The radius of that circle should not change    
(ii) The speed of travel along that path should be constant
• We know that, when a particle travels along a circular path, the direction of velocity at any point is tangential to the circle at that point
• So we can say: when a particle travels along a circular path, it's velocity changes continuously  
■ If there is change in velocity, there must be an acceleration.
• We want to find the magnitude and direction of this acceleration
We will write the steps:
1. Consider fig.4.36(a) below

Fig.4.36

An object is travelling along the circumference of a circle shown in green colour.
2. The origin O of the coordinate axes is made to coincide with the center C of the circle
• This enables us to draw the position vectors easily
3. Let at any instant when the stop watch shows 't' s, the position of the object be P
• Let after an interval of  'Δt' s, the position of the object be P' 
Then we can write the following 2 points:
(i) Position vector of the object at P is $\mathbf\small{\overrightarrow{OP}}$
    ♦ It is denoted as $\mathbf\small{\vec{r}}$
(ii) Position vector of the object at P' is $\mathbf\small{\overrightarrow{OP'}}$
    ♦ It is denoted as $\mathbf\small{\vec{r'}}$
4. When we know the initial and final postion vectors, we can find the displacement vector
• We have seen the method for doing it in  a previous section.
• Using that method, the displacement vector for our present case is $\mathbf\small{\overrightarrow{PP'}}$
    ♦ It is denoted as $\mathbf\small{\overrightarrow{\Delta r}}$
5. So we obtained the displacement vector. We now consider velocities:
• Let the velocity at P be $\mathbf\small{\vec{v}}$  
• Let the velocity at P' be $\mathbf\small{\vec{v'}}$
• They are marked in the fig.a
We see the following 3 points:
(i) $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$ have the same length
    ♦ This is because, though they have different directions, magnitudes are the same
(ii) $\mathbf\small{\vec{v}}$ is tangential to the circle at P
    ♦ So obviously, $\mathbf\small{\vec{v}}$ is perpendicular to $\mathbf\small{\vec{r}}$    
(iii) $\mathbf\small{\vec{v'}}$ is tangential to the circle at P'
    ♦ So obviously, $\mathbf\small{\vec{v'}}$ is perpendicular to $\mathbf\small{\vec{r'}}$    
6. Next we consider 'change in velocity':
■ Change in velocity is (final velocity - initial velocity)
• We have seen how to find it in the case of velocity vectors. We saw it in a previous section. 
Application of that method is shown in fig.b
(i) We drag $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$ to a convenient place. This is shown in fig.b
(iii) We arrange them in such a way that their tails coincide
(iv) Finally we draw the required $\mathbf\small{\vec{\Delta v}}$ from the head of $\mathbf\small{\vec{v}}$ to head of $\mathbf\small{\vec{v'}}$. This vector is shown in cyan color
7. We obtained $\mathbf\small{\vec{\Delta v}}$ by dragging the concerned vectors away to a convenient place. 
• But where does this $\mathbf\small{\vec{\Delta v}}$ actually act?
    ♦ It cannot act at P. Because at P, $\mathbf\small{\vec{v}}$ is acting 
    ♦ It cannot act at P'. Because at P', $\mathbf\small{\vec{v'}}$ is acting   
• Since the magnitudes of $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$ are equal, their difference $\mathbf\small{\vec{\Delta v}}$ will be acting at a point midway between P and P'
• So we drag $\mathbf\small{\vec{\Delta v}}$ and place it's tail at the midpoint of PP'
■ On doing so, we find that, $\mathbf\small{\vec{\Delta v}}$ is acting towards the center of the circle. It is shown in fig.c
This is a very useful result. We can write:
■ In uniform circular motion, the 'change in velocity vector'  acts towards the center of the circular path
8. Next we consider acceleration
• In the above step, we have calculated the change in velocity $\mathbf\small{\vec{\Delta v}}$.
• We know that, if we divide this $\mathbf\small{\vec{\Delta v}}$ by the time duration Δt, we will get average acceleration $\mathbf\small{\bar{\vec{a}}}$
• That is., $\mathbf\small{\bar{\vec{a}}=\frac{\vec{\Delta v}}{\Delta t}}$
9. We are dividing the vector by a scalar. In such a division, the direction of the vector will not change
• So the direction of average acceleration $\mathbf\small{\bar{\vec{a}}}$ is same as the direction of $\mathbf\small{\vec{\Delta v}}$. We can write:
■ In uniform circular motion, the 'average acceleration'  acts towards the center of the circular path
10. So we get the direction of average acceleration. Next we want it's magnitude
• For that, first we want the angle between $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$
We can find it using the following steps:
(i) The angle between $\mathbf\small{\vec{r}}$ and $\mathbf\small{\vec{r'}}$ is Δθ.
(ii) The angle between $\mathbf\small{\vec{r}}$ and $\mathbf\small{\vec{v}}$ will be always 90°.
(iii) The angle between $\mathbf\small{\vec{r'}}$ and $\mathbf\small{\vec{v'}}$ will be always 90°.
(iv) So the angle between $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{v'}}$ will be Δθ. This is shown in fig.b
[A detailed proof can be seen here]
11. Now, triangle OPP' in fig.a is an isosceles triangle because OP = OP' = radius of the circle
    ♦ So base angles at P and P' are equal
    ♦ We get: ∠P = ∠P' = [0.5 × (180-Δθ)]
• Similarly, triangle IGH in fig.b is an isosceles triangle because GH = GI =  $\mathbf\small{|\vec{v}|}$ = $\mathbf\small{|\vec{v'}|}$ 
    ♦ So base angles at I and H are equal
    ♦ We get: ∠I = ∠H = [0.5 × (180-Δθ)]
12. So all the angles in the two triangles OPP' and IGH are equal. 
■ Thus they are similar triangles. Some notes on similar triangles can be seen here.
• We can take ratios:
$\mathbf\small{\frac{side\: opposite\: \Delta \theta \: in\: OPP' }{side\: opposite\: \Delta \theta \: in\: IGH}}$ = $\mathbf\small{\frac{side\: opposite\: [0.5 \times(180-\Delta \theta)] \: in\: OPP' }{side\: opposite\: [0.5 \times(180-\Delta \theta)] \: in\: IGH}}$
• Thus we get: $\mathbf\small{\frac{PP'}{IH}=\frac{OP}{GI}}$
$\mathbf\small{\Rightarrow \frac{|\vec{\Delta r}|}{|\vec{\Delta v}|}=\frac{|\vec{r}|}{|\vec{v}|}}$
$\mathbf\small{\Rightarrow |\vec{\Delta v}|=\frac{|\vec{v}|\times |\vec{\Delta r}|}{|\vec{r}|}=\frac{|\vec{v}|\times |\vec{\Delta r}|}{R}}$
• Where R is the radius of the circle
13. Thus we get the 'magnitude of change in velocity'
■ If we divide this magnitude by the Δt during which the change occurs, we will get the magnitude of the average acceleration
• So we can write: $\mathbf\small{|\bar{\vec{a}}|=\frac{|\vec{\Delta v}|}{\Delta t}}$
Substituting for $\mathbf\small{|\vec{\Delta v}|}$ from (12), we get Eq.4.26: $\mathbf\small{|\bar{\vec{a}}|=\frac{|\vec{v}|\times |\vec{\Delta r}|}{R\times \Delta t}}$ 
14. Thus we obtained the magnitude of average acceleration. Our next aim is to find $\mathbf\small{|\vec{a}|}$, the magnitude of instantaneous acceleration  
• For that, we make 'Δt' closer and closer to zero.
• So P' will become closer and closer to P
A sample is shown in fig.4.37(a) below:

Fig.4.37

• In fig.a, P' is closer to P than in fig.4.36(a) that we saw at the beginning of this section 
• All the steps are repeated:
    ♦ $\mathbf\small{\vec{\Delta v}}$ is obtained in fig.b
    ♦ It is shown in cyan color
    ♦ In fig.c, the tail of the cyan vector is placed midway between P and P'
• We find that, here also, the cyan vector is directed towards the center of the circle
15. What will happen if we make Δt smaller and smaller
Ans: The interval Δt will become so small that it can no longer be called an 'interval'
• Instead, we will have to call it an 'instant'
• Then, the acceleration calculated in such a small Δt is not average acceleration $\mathbf\small{\bar{\vec{a}}}$
■ It is the instantaneous acceleration $\mathbf\small{\vec{a}}$
• So we can write: $\mathbf\small{|\vec{a}|= \lim \limits_{\Delta t \to 0} \frac{|\vec{\Delta v}|}{\Delta t}}$
• Substituting for $\mathbf\small{|\vec{\Delta v}|}$ from (12), we get:
$\mathbf\small{|\vec{a}|= \lim \limits_{\Delta t \to 0} \frac{|\vec{v}|\times|\vec{\Delta r}|}{R \times \Delta t}}$
16. This limit can be worked out as follows:
(i) Consider the travel from P to P' in fig.4.37(a).
• When Δt tends to zero, P' is very close to P
• Then Δθ is very small
(ii) The object is travelling with a uniform speed of $\mathbf\small{|\vec{v}|}$ along the circular path   
• So it will cover a circular distance of [$\mathbf\small{|\vec{v}|}$ × Δt] in Δt seconds
• Obviously, this distance will be equal to arc length PP'
• So we can write: arc length PP' = [$\mathbf\small{|\vec{v}|}$ × Δt]
(iii) When Δt tends to zero, that is., when Δt is very small, Δθ is also very small and so, arc length PP' will be nearly equal to chord length PP'
• But chord length PP' = $\mathbf\small{\vec{\Delta r}}$
(iv) Using the results in (ii) and (iii), we can write:
chord length PP' = $\mathbf\small{\vec{\Delta r}}$ = [$\mathbf\small{|\vec{v}|}$ × Δt] 
(v) So at the limiting state shown in (15), we can use [$\mathbf\small{|\vec{v}|}$ × Δt] instead of $\mathbf\small{\vec{\Delta r}}$
• So the result in (15) becomes: $\mathbf\small{|\vec{a}|=\frac{|\vec{v}|\times [|\vec{v}|\times \Delta t]}{R \times \Delta t}}$
■ Thus we get Eq.4.27: $\mathbf\small{|\vec{a}|=\frac{|\vec{v}|^2}{R}}$
17. In the limiting state, $\mathbf\small{\vec{r}}$ nearly coincides with $\mathbf\small{\vec{r'}}$. This is shown in fig.4.37(c)
• $\mathbf\small{\vec{r}}$ is perpendicular to $\mathbf\small{\vec{v}}$ because, $\mathbf\small{\vec{v}}$ is tangential to the circle at P
• The instantaneous acceleration at P is directed towards the center from P
19. This acceleration is called centripetal acceleration. 
• The word 'centripetal' comes from the Greek term which means 'center seeking'.
• This term is very suitable because, as we saw earlier, this acceleration is always directed towards the center.
• It may be noted that the instantaneous acceleration $\mathbf\small{\vec{a}}$ is not a constant vector because, it's direction changes continuously.

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In the next section, we will see another expression for centripetal acceleration.

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Chapter 4.10 - Projectile Motion

In the previous section we saw 2-dimensional motion under constant acceleration. In this section, we will see Projectile motion.
1. Consider fig.4.29(a) below:
 

Fig.4.29

• A stone is thrown into the air. 
• It is not thrown straight up. But at an angle (less than 90^o) with the horizontal. 
2. The stop watch is turned on at the instant when the stone is thrown. 
• The position of the stone at that instant is taken as the origin ‘O’. 
• A horizontal line through O is taken as the x axis.
• A vertical line through O is taken as y axis.
• The velocity of the stone at O is called initial velocity of the projectile. It is denoted as $\mathbf\small{\vec v_0}$   
3. Once the stone is thrown, it is on it’s own. That is., once it is thrown, no propelling force acts on it.
(In a rocket, it’s engine produces exhaust gas which propels it forward. Such a motion is not considered as projectile motion)
4. We see that the stone is thrown at an angle.
• So the initial velocity will have a vertical component $\mathbf\small{\vec v_{0y}}$ and a horizontal component $\mathbf\small{\vec v_{0x}}$
5. The vertical component is responsible for taking the stone ‘vertically away’ from O
• But this vertical component will be affected by the acceleration due to gravity ‘g’
    ♦ This is the acceleration vector. We can denote it as (g)$\mathbf\small{\hat j}$
• As a result, magnitude of the vertical velocity component will go on decreasing.   
6. The horizontal component is responsible for taking the stone ‘horizontally away’ from the origin
• This component is not affected by ‘g’
• So the horizontal component of will remain constant during the entire journey.
■ Note that, the air resistance can cause opposition to the projectile motion. But for our present discussion, air resistance is considered to be negligible. So we will not take it into account here.
7. At O, let $\mathbf\small{\theta_0}$ be the angle made by $\mathbf\small{\vec v_0}$ with the horizontal. Then at O:
• The horizontal component of $\mathbf\small{\vec v_0}$ is given by: $\mathbf\small{\vec v_{0x}}$ = $\mathbf\small{(\left | \vec{v_0} \right |\cos \theta _0)}\hat{i}$ 
• The vertical component of $\mathbf\small{\vec v_0}$ is given by: $\mathbf\small{\vec v_{0y}}$ = $\mathbf\small{(\left | \vec{v_0} \right |\sin \theta _0)}\hat{j}$
8. The stone was thrown when the stop watch showed '0' s. What happens to these components when the stop watch shows a reading of 't' seconds?
Ans: The horizontal component will remain the same because, there is no acceleration in the horizontal direction
■ The vertical component will have a smaller value because there is negative acceleration (due to gravity) in the vertical direction
• We can find it's exact value at time = 't' s
• For that, we use the familiar equation:  v = v₀ + at
• Thus we can write: $\mathbf\small{\vec{v_y}=\vec{v_{0y}}+\vec{a_y}\,t}$
$\mathbf\small{\Rightarrow \vec{v_y}=(\left | \vec v_0 \right |\sin\theta _0)\hat{j}-(g)\hat{j}t}$    
$\mathbf\small{\Rightarrow \vec{v_y}=(\left | \vec v_0 \right |\sin\theta _0-gt)\hat{j}}$
• So we can write:
At any time 't', after the beginning of the journey, the magnitude of the vertical component of velocity is given by Eq.4.12: $\mathbf\small{\left | \vec{v_y} \right |=(\left | \vec v_0 \right |\sin\theta _0-gt)}$
9. At time = 't' seconds:
• The magnitude of the horizontal component remains the same
• The vertical component has a lower magnitude as given by Eq.4.12 above.
■ As a result, the resultant velocity $\mathbf\small{\vec v}$ (which is the resultant of the horizontal and vertical components) will have a smaller magnitude than $\mathbf\small{\vec v_0}$. This is shown in fig.4.29(b). We see the following:
• At time = 't' seconds:
    ♦ The stone has reached P
    ♦ $\mathbf\small{\vec v}$ has a smaller length than $\mathbf\small{\vec v_0}$
    ♦ $\mathbf\small{\theta}$ is different from $\mathbf\small{\theta_0}$
10. We saw how the 'velocity of the stone' varies during it's travel. Next we will see how 'it's distance from O' varies
■ First we will see the horizontal travel
(i) We have seen that the horizontal velocity remains the same.
• So we can use the familiar 'equation for uniform motion': s = vt
(ii) Thus we get:
Horizontal displacement in time 't' s = $\mathbf\small{\vec{\Delta r_x}=(\left | \vec{v_0} \right |\cos\theta_0 )\hat{i}\times t}$
$\mathbf\small{\Rightarrow \vec{\Delta r_x}=[(\left | \vec{v_0} \right |\cos\theta_0 )t]\hat{i}}$  
(iii) That means, the magnitude of $\mathbf\small{\vec{\Delta r_x}}$ = $\mathbf\small{\left | \vec{\Delta r_x} \right |=\left ( \left | \vec{v_0} \right | \cos \theta _0 \right )t}$
(iv) This magnitude is the distance OP'. But the distance OP' is the x coordinate of P
• So we can write: At any time 't', after the beginning of the journey, the object will be at a parallel distance of '$\mathbf\small{\left ( \left | \vec{v_0} \right | \cos \theta _0 \right )t}$' from the y axis
• In other words, at any time 't', after the beginning of the journey, the x coordinate of the object is given by Eq.4.13: x = $\mathbf\small{\left ( \left | \vec{v_0} \right | \cos \theta _0 \right )t}$ 
■ Now we will see the vertical travel
(i) The vertical travel is affected by an acceleration 'g'. So we will use the familiar equation: $\mathbf\small{s=v_0 t+\frac{1}{2}at^2}$
• Thus we can write: 
Vertical displacement in time 't' s = $\mathbf\small{\left | \vec{v_{0y}} \right |t-\frac{1}{2}g t^2}$
(iv) This magnitude is the distance P'P. But the distance PP' is the y coordinate of P
• So we can write: At any time 't', after the beginning of the journey, the object will be at a parallel distance of '$\mathbf\small{\left | \vec{v_{0y}} \right |t-\frac{1}{2}g t^2}$' from the x axis
• In other words, at any time 't', after the beginning of the journey, the y coordinate of the object is given by Eq.4.14: y = $\mathbf\small{\left | \vec{v_{0y}} \right |t-\frac{1}{2}g t^2}$
11. So we are now able to specify the position of a projectile at any time 't'.
• We are able to do it by using x and y coordinates.
■ If we can eliminate 't' from Eqs.4.13 and 4.14, we will get a direct relation between x and y.
Let us try:
(i) From Eq.4.13, we get: $\mathbf\small{t=\frac{x}{\left |\vec{v_0}  \right |\cos \theta _0}}$
• We can use this instead of 't' in Eq.4.14. 
• We get Eq.4.15: $\mathbf\small{y=\left [ \tan\theta_0  \right ]x-\left [ \frac{g}{2\left ( \left | \vec{v_0} \right |\cos \theta_0  \right )^2} \right ]x^2}$
(ii) Consider the quantity inside the first pair of square brackets
• $\mathbf\small{\theta_0}$ is the initial angle with which the stone is thrown at the beginning
• Once the stone is thrown at a particular initial angle, it will not be altered
• That is., $\mathbf\small{\theta_0}$ is a constant. 
• So $\mathbf\small{\tan \theta_0}$ is a constant. We will denote it as 'a'   
(iii) Consider the quantities inside the second pair of square brackets
    ♦ 'g' and '2' are constants
    ♦ $\mathbf\small{\theta_0}$ is a constant as we saw above
• Now  $\mathbf\small{\left | \vec{v_0} \right |}$ remains
    ♦ Once the stone is thrown at a particular initial velocity, it will not be altered
    ♦ That is., $\mathbf\small{\left | \vec{v_0} \right |}$ is a constant
• So every thing inside the second pair are constants
• So the final result inside that second pair is a constant. We will denote it as 'b'
(iv) Eq.4.15 becomes: $\mathbf\small{y=ax+bx^2}$
Where $\mathbf\small{a=\left [ \tan\theta_0  \right ]\: \: \text{and}\; \; b=-\left [ \frac{g}{2\left ( \left | \vec{v_0} \right |\cos \theta_0  \right )^2} \right ]}$
(v) But $\mathbf\small{y=ax+bx^2}$ is the equation of a parabola. So we can write:
■ The path of a projectile is a parabola
• This is shown in fig.c
12. Time required to reach the maximum height:
• Consider the path of the projectile shown in fig.c
• We see a peak point M. After M, we see no further upward motion
• That means, at this peak point M, the magnitude of the vertical component is zero
(i) Let $\mathbf\small{t_m}$ be the time required to reach M
(ii) Consider the vertical component of the velocity. We can use Eq.4.12: $\mathbf\small{\left | \vec{v_y} \right |=(\left | \vec v_0 \right |\sin\theta _0-gt_m)}$
(iii) Substituting the known values, we get: $\mathbf\small{0=(\left | \vec v_0 \right |\sin\theta _0-gt_m)}$   
■ From this we get Eq.4.16: $\mathbf\small{t_m=\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$
13. Maximum height reached by the stone (h_m):
(i) For this, we need the height of M from the x axis
(ii) Let us use Eq.4.14: y = $\mathbf\small{\left | \vec{v_{0y}} \right |t-\frac{1}{2}g t^2}$
• In this equation, if we put t = t_m, we will get the vertical distance traveled during 't_m'
(iii) But the 'vertical distance traveled during t_m' is the height of M. So we get:
$\mathbf\small{y=h_m=\left ( \left |\vec v_0 \right | \sin \theta_0 \right )\left ( \frac{\left |\vec v_0 \right | \sin \theta_0}{g} \right )-\frac{g}{2}\left ( \frac{\left |\vec v_0 \right | \sin \theta_0}{g} \right )^2}$
• From this, we get Eq.4.17: $\mathbf\small{h_m=\frac{\left (\left | \vec v_0 \right| \sin \theta_0  \right )^2}{2g}}$
14. Time required for the whole flight (T_f):
(i) After M, the stone continues the flight for some more time. In the end, it falls back to the ground.
• Let us consider the vertical motion after M. We want the time 't' required for this motion.
• The vertical distance traveled in this motion is h_m.
(ii) We can use the familiar equation: $\mathbf\small{s=v_0 t+\frac{1}{2}at^2}$ 
• In this motion, the initial velocity is zero. It is like the stone just dropped from a height of h_m
• So we can put v₀ = 0
• We get: $\mathbf\small{h_m=0 \times t+\frac{1}{2}g{t}^2}$
$\mathbf\small{\Rightarrow \frac{\left (\left | \vec v_0 \right| \sin \theta_0  \right )^2}{2g}=\frac{1}{2}gt^2}$
$\mathbf\small{\Rightarrow \frac{\left (\left | \vec v_0 \right| \sin \theta_0  \right )^2}{g^2}=t^2}$
$\mathbf\small{\Rightarrow t=\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$
• So total time of flight = T_f = (t_m+t) = $\mathbf\small{\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}+\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$ 
• So we get Eq.4.18: $\mathbf\small{T_f=\frac{2\left | \vec{v_0} \right |\sin \theta_0}{g}}$
■ Note:
• From Eq.4.16, we have: Time required for the upward travel from O to M = $\mathbf\small{\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$   
• In the above step (13), we have: Time required for the downward travel from M to the ground = $\mathbf\small{\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$   
• So the times for upward travel and downward travel are the same
15. Horizontal range of a projectile ($\mathbf\small{|\vec R|}$):
(i) For this we consider the horizontal motion
• The horizontal component of the velocity (which is a constant value) will be effective for the entire time (T_f) of the flight 
(ii) So the horizontal distance = Horizontal component of velocity × time
= $\mathbf\small{\vec v_{0x}}$ × $\mathbf\small{T_f}$ = $\mathbf\small{(\left | \vec{v_0} \right |\cos \theta _0)}\hat{i}$ × $\mathbf\small{\frac{2\left | \vec{v_0} \right |\sin \theta_0}{g}}$ =$\mathbf\small{\frac{\left ( \left | \vec{v_0} \right |^2 2 \sin \theta_0 \cos \theta_0 \right )\hat{i}}{g}}$
• But from math classes, we have: $\mathbf\small{2 \sin \theta_0 \cos \theta_0}$ = $\mathbf\small{\sin 2\theta_0}$
(iii) So we can write: $\mathbf\small{\vec{R}=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )\hat{i}}{g}}$
(iv) Magnitude of $\mathbf\small{\vec R}$ is the actual distance
■ Thus we get Eq.4.19: Range of the projectile = $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )}{g}}$
16. Maximum possible range for a given velocity:
• Suppose that a machine can throw an object only at a certain 'fixed speed' $\mathbf\small{\left |\vec v_0 \right |}$
    ♦ But the angle of projection can be changed to any value.
[That is., magnitude of $\mathbf\small{\vec v_0}$ is fixed. But the direction can change]
■ Then what angle would we choose to obtain 'maximum range'?
Solution:
1. We have:
Range of the projectile = $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )}{g}}$
2. In our present case, $\mathbf\small{\left |\vec v_0 \right |}$ is a constant
• So the only variable is sin 2θ₀.
3. That means, for maximum range, sin 2θ₀ must be maximum
• The maximum value possible for sin 2θ₀ is '1'. 
4. This '1' is obtained when '2θ₀' is 90^o.
• So θ₀ must be 45^o.
■ We can write:
The maximum range is obtained when the angle of projection θ₀ is 45^o.

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In the next section, we will apply the above equations to an actual projectile.

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