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/s$^2$), supported by a roller as shown. A short duration impulsive force $F$, producing an impulse $I_F$, is applied at a height $d = 0.3$ m from the bottom. Assume all joints to be frictionless. The minimum value of $I_F$ required to topple the block is ________________.
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For toppling problems, equate the angular momentum imparted by the impulse to the rotational energy needed to raise the center of mass to the tipping point.
For the block to topple, the impulse must supply enough angular momentum about the pivot to raise the center of mass (COM) to the tipping point. The applied impulse $I_F$ at height $d$ creates angular momentum:
\[
L = I_F \, d.
\]
The block rotates about the bottom pinned corner. Its moment of inertia about that corner is:
\[
I = \frac{m}{3}(h^2 + b^2)
= \frac{1}{3}(0.4^2 + 0.3^2)
= 0.0833\ \text{kg·m}^2.
\]
The COM must rise by:
\[
\Delta h = \frac{h}{2}\left(1 - \cos \tan^{-1}(b/h)\right).
\]
For $h = 0.4$ m and $b = 0.3$ m, the energy needed is:
\[
m g \Delta h = 1 \times 9.81 \times 0.039 = 0.383\ \text{J}.
\]
This must equal the rotational kinetic energy imparted by the impulse:
\[
\frac{L^2}{2I} = 0.383.
\]
Substituting $L = I_F d$ and $d = 0.3$ m:
\[
\frac{(I_F \cdot 0.3)^2}{2 \times 0.0833} = 0.383.
\]
Solving gives:
\[
I_F = 0.953\ \text{Ns}.
\]
Final Answer: 0.953 Ns
