Question

A ladder of length L is slipping with its ends against a vertical wall and a horizontal floor. At a certain moment, the speed of the end in contact with the horizontal floor is v and the ladder makes an angle $\alpha$ = 30° with the horizontal.

The speed of the ladder's center must be :-

• A. $\frac{2v}{\sqrt{3}}$
• B. v/2
• C. v
• D. None

Answer 1

Answer: (C)

v

Answer 2

The problem involves the kinematics of a rigid body (the ladder) under constrained motion.

1. Define Coordinates and Constraints:
   Let the length of the ladder be $L$.
   Let the end of the ladder on the horizontal floor be at $(x, 0)$ and the end in contact with the vertical wall be at $(0, y)$.
   The origin $(0,0)$ is the corner where the floor and wall meet.
   The length of the ladder is constant, so it satisfies the equation:
   $x^2 + y^2 = L^2$

2. Velocity of the Ends:
   Differentiate the constraint equation with respect to time $t$:
   $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
   $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$

We are given that the speed of the end in contact with the horizontal floor is $v$. Let's assume the ladder is slipping outwards, so $x$ is increasing: $\frac{dx}{dt} = v$.
The ladder makes an angle $\alpha = 30^\circ$ with the horizontal. From trigonometry:
$x = L \cos \alpha$
$y = L \sin \alpha$

Substitute these into the differentiated equation:
$(L \cos \alpha) v + (L \sin \alpha) \frac{dy}{dt} = 0$
$L v \cos \alpha + L \frac{dy}{dt} \sin \alpha = 0$
$v \cos \alpha + \frac{dy}{dt} \sin \alpha = 0$
$\frac{dy}{dt} = -v \frac{\cos \alpha}{\sin \alpha} = -v \cot \alpha$
The negative sign indicates that the top end is moving downwards.

3. Velocity of the Center of the Ladder:
   Let the center of the ladder be $(x_c, y_c)$. Since it's the midpoint, its coordinates are:
   $x_c = \frac{x + 0}{2} = \frac{x}{2}$
   $y_c = \frac{y + 0}{2} = \frac{y}{2}$

The velocity components of the center of the ladder are:
$\frac{dx_c}{dt} = \frac{1}{2} \frac{dx}{dt} = \frac{v}{2}$
$\frac{dy_c}{dt} = \frac{1}{2} \frac{dy}{dt} = \frac{1}{2} (-v \cot \alpha)$

The speed of the ladder's center, $v_c$, is the magnitude of its velocity vector:
$v_c = \sqrt{\left(\frac{dx_c}{dt}\right)^2 + \left(\frac{dy_c}{dt}\right)^2}$
$v_c = \sqrt{\left(\frac{v}{2}\right)^2 + \left(-\frac{v \cot \alpha}{2}\right)^2}$
$v_c = \frac{v}{2} \sqrt{1 + \cot^2 \alpha}$

4. Substitute Trigonometric Identity and Given Angle:
   Using the trigonometric identity $1 + \cot^2 \alpha = \csc^2 \alpha$:
   $v_c = \frac{v}{2} \sqrt{\csc^2 \alpha} = \frac{v}{2} |\csc \alpha|$

Given $\alpha = 30^\circ$:
$\sin 30^\circ = \frac{1}{2}$
$\csc 30^\circ = \frac{1}{\sin 30^\circ} = 2$

Substitute this value into the expression for $v_c$:
$v_c = \frac{v}{2} (2)$
$v_c = v$

The speed of the ladder's center is $v$.