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Thursday, October 18, 2018

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:
To find relative velocity, we need to calculate the vector difference of the two velocities.
Fig.4.41
• They are moving in two different directions and with two different magnitudes
    ♦ A is moving with a velocity of vA
    ♦ B is moving with a velocity of vB
2. We can write two cases:
(i) Case 1: When viewed from A, The other object B will be moving with a velocity of [vBvA]
• That is., Relative velocity of B with respect to A = vBA=vBvA  
(ii) Case 2: When viewed from B, The other object A will be moving with a velocity of [vAvB]
• That is., Relative velocity of A with respect to B = vAB=vAvB  
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 vBA in case 1
    ♦ The x component of this resultant will be: vBxvAx
    ♦ The y component of this resultant will be: vByvAy
Once the components are obtained, we can easily calculate the resultant
• Consider the resultant vAB in case 2
    ♦ The x component of this resultant will be: vAxvBx
    ♦ The y component of this resultant will be: vAyvBy
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 vR 

• Let the motion of bicycle be denoted by vB 

• 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 vRB

    ♦ That is., we want to find vRvB   

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

(i) x component of (vRvB)

    ♦ It is given by: vRxvBx 

(ii) y component of (vRvB)
    ♦ It is given by: vRyvBy 
4. Let us write the values:
vRx = 0 (∵ the rain is falling vertically and so has no horizontal component)
vBx = -12 ˆi (The travel is from left to right. So it is taken as negative)
vRy = -35 ˆj (The travel is from top to bottom. So it is taken as negative)
vBy = 0 (∵ the bicycle is travelling horizontally and so has no vertical component)
5. Substituting the values in 3(i), we get:
• x component of (vRvB) = vRBx = [0 - (-12 ˆi)] = 12 ˆi
    ♦ 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 (vRvB) = vRBy = [(-35 ˆi - 0)] = -35 ˆj
    ♦ 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 vRvB = |vRvB| = |vRB| = 122+(35)2 = 37 ms-1.
7. Next we find the direction:
• We have: tanθ=|vRBy||vRBx|=3512
• Thus we get: θ = 71.076o
■ 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) = 18.924o.
9. So we can write the result:
• When viewed from the bicycle, rain falls towards the cyclist at an angle of 18.924o with the vertical
• So the cyclist must hold the umbrella at an angle of 18.924o 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 45o 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 vW
• We can write:
    ♦ |vW| = 72 km/h
    ♦ vW makes an angle 45o 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 vWB
    ♦ That is., we want to find vWvB 
Where vB is the velocity of the boat
5. For that, we need to find the following two:
(i) x component of (vWvB)
    ♦ It is given by: vWxvBx 
(ii) y component of (vWvB)
    ♦ It is given by: vWyvBy 
6. Let us write the values:
vWx = 72 cos 45 ˆi = 50.912 ˆi km/h
vBx = 0 (∵ the boat is travelling north and so has no component towards east)
vWy = 72 sin 45 ˆj = 50.912 ˆi km/h
vBy = 51 ˆj
7. Substituting the values in 5(i), we get:
• x component of (vWvB) = vWBx = [50.912 ˆi - 0] = 50.912 ˆi
    ♦ 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 (vWvB) = vWBy = [(72 sin 45 ˆj - 51 ˆj)] = -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: tanθ=|vWBy||vWBx|=0.088350.912
• Thus we get: θ = 0.0997o
■ 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 0.0997o, 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., vRM will slope towards the right
    ♦ Where vRM 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 vR in fig.b
3. We know this:
• The two components of vRM in fig.b are:
(i) vRMx
(ii) vRMy
4. So we will first find those components:
(i) We have: vRMx=vRxvMx
• Let us write the values:
(a) vRx=(|vR|cosθ)ˆi
    ♦ Where θ is the angle made by vR with the horizontal
    ♦ This is shown in fig.c
(b) vMx=3ˆi
• Substituting in 4(i), we get: vRMx=(|vR|cosθ)ˆi3ˆi
(ii) We have: vRMy=vRyvMy
• Let us write the values:
(a) vRy=(|vR|sinθ)ˆj
    ♦ Where θ is the angle made by vR with the horizontal
    ♦ This is shown in fig.c
(b) vMy=0
• Substituting in 4(ii), we get: vRMy=(|vR|sinθ)ˆj
5. Now consider the vRM in fig.c
• It is perfect vertical. That means, it has no horizontal component
• That means, vRMx = 0
• From 4(i)b, we can write: vRMx=(|vR|cosθ)ˆi3ˆi = 0
■ From this we get: |vR|cosθ=3
This is one of the two equations which we will need to find |vR| and θ.
6. To get the other equation, we consider the motion at 6 km/h
• This time, the vRM 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) vR in fig.d is same as that in fig.c. This is because, the original speed and direction of rain does not change 
(ii) vM 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 vRM in fig.d are:
(i) vRMx
(ii) vRMy
8. So we will first find those components:
(i) We have: vRMx=vRxvMx
• Let us write the values:
(a) vRx=(|vR|cosθ)ˆi
    ♦ Where θ is the angle made by vR with the horizontal
    ♦ This is shown in fig.d
(b) vMx=6ˆi
• Substituting in 8(i), we get: vRMx=(|vR|cosθ)ˆi6ˆi
(ii) We have: vRMy=vRyvMy
• Let us write the values:
(a) vRy=(|vR|sinθ)ˆj
    ♦ Where θ is the angle made by vR with the horizontal
    ♦ This is shown in fig.d
(b) vMy=0
• Substituting in 8(ii), we get: vRMy=(|vR|sinθ)ˆj
9. Using the 'formula for direction of the resultant', we can write:
tan45=|vRMy||vRMx|=||vR|sinθ||(|vR|cosθ)6|
10. But in (5), we obtained: |vR|cosθ=3
Substituting this in (9), we get: tan45=||vR|sinθ||(3)6|=||vR|sinθ||(3)|=|vR|sinθ3
11. But tan 45 = 12
Thus we get: |vR|sinθ=32
12.So we have two results:
(i) |vR|cosθ=3
(ii) |vR|sinθ=32
Dividing (ii) by (i), we get: tanθ=12
Thus we get θ = 45°.
13. Substituting this value of θ in 12(i), we get:
|vR|cos45=3
|vR|=312=32=4.243km/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

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

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