A rigid homogeneous uniform block of mass 1 kg, height h = 0.4 m and width b = 0.3 m is pinned at one corner and placed upright in a uniform gravitational field (g = 9.81 m/s²), supported by a roller in the configuration shown in the figure. A short duration (impulsive) force F, producing an impulse I_F is applied at a height of d = 0.3 m from the bottom as shown. Assume all joints to be frictionless. The minimum value of I_F required to topple the block is

The correct option/answer is 0.953 Ns.

1. Understanding the Condition for Toppling:
To topple the block, it must be rotated about the pin joint at the bottom-right corner (point O) such that its center of gravity (G) passes over the vertical line passing through O.
Initially, the center of gravity G is located at the geometric center of the block. From the schematic in the image, the dimensions of the block are:
Height, $h=0.4\text{m}$
Width, $b=0.3\text{m}$
Mass, $m=1\text{kg}$
Acceleration due to gravity, $g=9.81{\text{m/s}}^2$

2. Finding the Change in Height of the Center of Gravity:
Initially, the height of the center of gravity from the bottom edge of the block is:
$y_i=\frac{h}{2}$
The distance from the pin joint O to the center of gravity G is:
$\mathrm{OG}=\sqrt{{\left(\frac{h}{2}\right)}^2+{\left(\frac{b}{2}\right)}^2}=\sqrt{{0.2}^2+{0.15}^2}=0.25\text{m}$
At the critical point of toppling, the line OG becomes vertical. In this position, the height of the center of gravity above the pin joint O is:
$y_f=\mathrm{OG}=0.25\text{m}$
Therefore, the minimum vertical height increase of the center of gravity required for the block to topple is:
$h^{\text{'}}=y_f-y_i=0.25-0.2=0.05\text{m}$

3. Calculating the Moment of Inertia of the Block about Point O:
Using the parallel axis theorem, the moment of inertia $I_O$ of the rectangular block of mass m about the corner pin joint O is:
$I_O=\frac{1}{12}m(h^2+b^2)+m{\mathrm{OG}}^2$
Substitute ${\mathrm{OG}}^2={\left(\frac{h}{2}\right)}^2+{\left(\frac{b}{2}\right)}^2=\frac{1}{4}(h^2+b^2)$:
$I_O=\frac{1}{3}m(h^2+b^2)$
For the given block:
$I_O=\frac{1}{3}\times 1\times ({0.4}^2+{0.3}^2)=\frac{0.25}{3}=\frac{1}{12}{\text{kg m}}^2$

4. Applying Conservation of Mechanical Energy:
Immediately after the impulsive force is applied, let the block acquire an initial angular velocity $\omega$. For the block to just reach the critical upright position and topple, its final angular velocity at that maximum height position will be zero. By conservation of energy:
${\mathrm{K.E.}}_{\mathrm{initial}}={\mathrm{P.E.}}_{\mathrm{gain}}$
$\frac{1}{2}{I}_O{\omega }^2=mg{h}^{\text{'}}$
Substituting the values:
$\frac{1}{2}\times \frac{1}{12}\times {\omega }^2=1\times 9.81\times 0.05$
${\omega }^2=2\times 12\times 9.81\times 0.05=11.772$
$\omega =\sqrt{11.772}\approx 3.431\text{rad/s}$

5. Applying Angular Impulse-Momentum Theorem:
The impulsive force F is applied at a height $d=0.3\text{m}$ from the bottom. The angular impulse about the pin joint O is related to the change in angular momentum by:
$\text{Angular Impulse}=I_F\times d=I_O(\omega -0)$
Substituting the values:
$I_F\times 0.3=\frac{1}{12}\times 3.431$
$I_F=\frac{3.431}{12\times 0.3}=\frac{3.431}{3.6}\approx 0.953\text{Ns}$

Thus, the minimum impulse $I_F$ required to topple the block is 0.953 Ns.