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Q2.25:
A man walking briskly in rain with speed (v) must slant his umbrella forward making an angle (\theta) with the vertical. A student derives the following relation between (\theta) and (v) : (\tan \theta = v) and checks that the relation has a correct limit: as (v \rightarrow 0), (\theta \rightarrow 0), as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct ? If not, guess the correct relation.

Solution :

Step 1: Dimensional Analysis of the Proposed Relation

The student proposes the relation: $\tan \theta = v$.

To evaluate the validity of this equation, we must perform a dimensional analysis on both sides of the equation [Per the Principle of Homogeneity of Dimensions].

  • Left-Hand Side (LHS): The term $\tan \theta$ represents a trigonometric function of an angle. Since an angle is the ratio of arc length to radius ($s/r$), it is a dimensionless quantity ($[M^0 L^0 T^0]$). Therefore, $\tan \theta$ is dimensionless.
  • Right-Hand Side (RHS): The variable $v$ represents the speed of the man, which has dimensions of distance per unit time ($[L T^{-1}]$).

Comparing the two: $[M^0 L^0 T^0] \neq [L T^{-1}]$. Since the dimensions do not match, the relation $\tan \theta = v$ is physically incorrect, despite satisfying the boundary condition as $v \to 0$.

Step 2: Deriving the Correct Physical Relation

Let $v_r$ be the velocity of the rain falling vertically relative to the ground, and $v_m$ be the velocity of the man walking horizontally. To stay dry, the man must point his umbrella in the direction of the relative velocity of the rain with respect to himself ($\vec{v}_{rm}$).

$\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$

Vectorially, if $\vec{v}_r$ is directed downward and $\vec{v}_m$ is directed forward, the resultant vector $\vec{v}_{rm}$ makes an angle $\theta$ with the vertical. From the geometry of the velocity vector triangle:

θ vᵣ vₘ vᵣₘ

Using trigonometry in the right-angled triangle formed by these vectors [By Trigonometric Ratios in Right Triangles]:

$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{v_m}{v_r}$

Step 3: Justification and Verification

The correct relation is $\tan \theta = \frac{v}{v_r}$, where $v_r$ is the constant speed of the falling rain.

Condition Analysis
Dimensionality $\frac{[L T^{-1}]}{[L T^{-1}]} = [M^0 L^0 T^0]$. The equation is now dimensionally consistent.
Limit: $v \to 0$ $\tan \theta \to 0 \implies \theta \to 0$. The condition holds.
Limit: $v \to \infty$ $\tan \theta \to \infty \implies \theta \to 90^\circ$. The umbrella must be held horizontally for very high speeds.

Final Solution: The relation $\tan \theta = v$ is incorrect because it is not dimensionally consistent. The correct relation is $\tan \theta = \frac{v}{v_r}$, where $v_r$ is the constant speed at which the rain falls.


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