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 $v \to 0$, $\theta \to 0$, 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.

Step 1: Dimensional Analysis of the Proposed Relation

To determine the validity of the relation $\tan \theta = v$, we must evaluate the dimensions of both sides of the equation. According to the Principle of Homogeneity of Dimensions, both sides of a physical equation must have identical dimensions.

• Left-Hand Side (LHS): The expression is $\tan \theta$. Since $\theta$ is an angle (ratio of arc length to radius), 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 length per unit time $[M^0 L^1 T^{-1}]$.

[Per the Principle of Dimensional Homogeneity, a dimensionless quantity cannot be equated to a quantity with dimensions of speed.] Thus, the relation $\tan \theta = v$ is dimensionally inconsistent.

Step 2: Deriving the Correct Physical Relation

Let the velocity of the rain be $\vec{v}_r$ (acting vertically downward) and the velocity of the man be $\vec{v}_m$ (acting horizontally). The velocity of the rain relative to the man is given by:

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

To protect himself from the rain, the man must tilt his umbrella at an angle $\theta$ with the vertical, which is the angle of the relative velocity vector $\vec{v}_{rm}$ with the vertical axis.

Step 3: Calculating the Tangent of the Angle

From the velocity vector triangle formed by $\vec{v}_r$ and $\vec{v}_m$:

$\tan \theta = \frac{|\vec{v}_m|}{|\vec{v}_r|}$

Substituting the magnitudes, where $|\vec{v}_m| = v$ (the speed of the man) and $|\vec{v}_r| = u$ (the terminal speed of the rain):

$\tan \theta = \frac{v}{u}$

Step 4: Verification of Consistency

By including the term $u$ (speed of rain), the dimensions are now balanced:

• RHS dimensions: $\frac{[L T^{-1}]}{[L T^{-1}]} = [M^0 L^0 T^0]$ (Dimensionless).

This satisfies the condition of dimensional homogeneity. As $v \to 0$, $\tan \theta \to 0$, which implies $\theta \to 0$, consistent with the observed limit.

Final Solution: The relation $\tan \theta = v$ is incorrect because it is dimensionally inconsistent. The correct relation is $\tan \theta = \frac{v}{u}$, where $v$ is the speed of the man and $u$ is the speed of the rain.