GATE ME 2022 Set 2 — Question 37
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Engineering Mechanics → Dynamics → Linear & Angular Impulse-Momentum
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Question
A rigid homogeneous uniform block of mass 1 kg, height m and width m is pinned at one corner and placed upright in a uniform gravitational field ( m/s), supported by a roller in the configuration shown in the figure. A short duration (impulsive) force , producing an impulse , is applied at a height of m from the bottom as shown. Assume all joints to be frictionless. The minimum value of required to topple the block is

Correct answer
(A) 0.953 Ns
Solution
To topple the block, it must rotate about the pin until its center of gravity (CG) is vertically above the pin. The energy required for this is the change in potential energy.
Distance from pin to CG: m.
Height of CG at the toppling point (vertically above pin): m.
Initial kinetic energy after impulse: .
1.Calculate the moment of inertia () about the pin:
For a rectangular block rotating about a corner:2.Calculate the required potential energy change ():
Initial height of CG: m.Distance from pin to CG: m.
Height of CG at the toppling point (vertically above pin): m.
3.Relate the impulse () to initial kinetic energy ():
Angular impulse about the pin: .Initial kinetic energy after impulse: .
4.Equate and for minimum impulse:
Thus, the correct option is (A).Continue learning with Success Tracker
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