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GATE 2022 ME (ME2) – Question 37

Engineering Mechanics · Impulse and momentum, energy formulations · 2 marks · Multiple choice

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 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

horizontal rightward impulse at height d, pivot at the lower right corner and roller near the lower left.
  1. 0.953 Ns
  2. 1.403 Ns
  3. 0.814 Ns
  4. 1.172 Ns

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Correct answer: (A) 0.953 Ns

Explanation

**Step 1: moment of inertia about the pivot corner.** For a rectangular block of mass $m$, width $b$ and height $h$, about its centroid $I_G=\dfrac m{12}(b^2+h^2)$. About the corner, using the parallel-axis theorem with $r^2=(b/2)^2+(h/2)^2$:
$$I_O=\frac m{12}(b^2+h^2)+m\frac{b^2+h^2}{4}=\frac m3(b^2+h^2)=\frac13(0.09+0.16)=\frac1{12}\text{ kg m}^2 .$$

**Step 2: topple condition.** The impulse gives an angular velocity $\omega$ about the corner: $I_F\,d=I_O\,\omega$. To topple, the centre of mass must rise to the highest point, directly above the pivot. The centre starts at height $h/2=0.20$ m and ends at $r=\sqrt{0.15^2+0.20^2}=0.25$ m, so it rises
$$\Delta h=0.25-0.20=0.05\text{ m}.$$

**Step 3: energy.**
$$\tfrac12I_O\omega^2=mg\,\Delta h\;\Rightarrow\;\omega=\sqrt{\frac{2(1)(9.81)(0.05)}{1/12}}=3.43\text{ rad/s}.$$

**Step 4: impulse.**
$$I_F=\frac{I_O\,\omega}{d}=\frac{(1/12)(3.431)}{0.3}=\mathbf{0.953\ Ns}.$$