chapter 14.4(a) - Angle between tangential vectors

In the previous section we saw the need to find the angle between two tangential velocity vectors in circular motion. We will now calculate it using graphical methods.

We have the situation shown in figs (a) and (b) below:

• We want to find the ∠G in fig.b
We will write the steps:
1. Extend $\mathbf\small{\vec{v}}$ upwards to any convenient point Q. This is shown in fig.c
• Extend $\mathbf\small{\vec{v'}}$ downwards to any convenient point R
2. Let the extensions meet at S
• Now the required angle ∠G is same as ∠P'SQ 
• So our aim is to find ∠P'SQ
3. Consider the quadrilateral OPSP'
• The sum of interior angles of any quadrilateral is 360^o
• So we can write: ∠O + ∠P + ∠S + ∠P' = 360
4. We know that ∠P = ∠P' = 90 (∵ tangent is perpendicular to radius)
• This is shown in fig.d
• Also we have ∠O = Δθ 
5. Substituting in (3), we get:
• Δθ + 90 + ∠S + 90 = 360^o
⟹ Δθ + ∠S = 180^o
⟹ ∠S = (180-Δθ)
• That is., ∠PSP' = (180-Δθ)
6. But ∠PSP' = ∠QSR (∵ they are opposite angles)
So we get: ∠QSR = (180-Δθ)
7. ∠QSR and ∠P'SQ form a linear pair
• That is., ∠QSR + ∠P'SQ = 180^o
• Substituting from (6), we get:
(180-Δθ) + ∠P'SQ = 180^o
⟹ ∠P'SQ = Δθ.
8. In (2), we saw that, ∠P'SQ is our required angle between the two vectors at G in fig.b
So we can write:
■ In fig.b, the angle at G between the two vectors is Δθ.  

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