(a) A heavy carriage wheel of weight W, and radius r is to be dragged over an obstacle of height h by a horizontal force P applied to the centre of the wheel. Show that P is slightly greater than (W · √(2rh - h²))/(r - h). (10 marks)

(b) A disc moun

Question

(a) A heavy carriage wheel of weight W, and radius r is to be dragged over an obstacle of height h by a horizontal force P applied to the centre of the wheel. Show that P is slightly greater than (W · √(2rh - h²))/(r - h). (10 marks)

(b) A disc mounted on a shaft is having three masses of 6 kg, 5 kg and 4 kg, which are attached at a radial distance of 70 mm, 80 mm and 50 mm at the angular positions of 45°, 135° and 240° respectively. The angular positions are measured counter-clockwise from the reference line along the x-axis. Calculate the amount of countermass at the radial distance of 85 mm required for the static balance. (10 marks)

(c) From a balloon ascending with a velocity of 25 m/s above the surface of a lake, a stone is let fall and the sound of the splash is heard 5 seconds later. Find the height of the balloon when the stone was dropped, assuming that the velocity of sound is 340 m/s. (10 marks)

(d) What is the importance of the atomic packing factor ? Compute the atomic packing factor for the FCC crystal structure. (10 marks)

(e) A 1·5 m long shaft having diameter 2·0 cm is held at the ends by long bearings. The weight of a disc at the centre of the shaft is 20 kg. If the modulus of elasticity of the material of shaft is 2 × 10⁶ kg/cm², then calculate the critical speed of the shaft in cycles per minute. (10 marks)

UPSC Answer Check · Accepted Answer

(a) Let C be the point where the wheel touches the obstacle, O the centre of the wheel. At the instant the wheel is just about to lift over the obstacle, the reaction at C passes through O, so it has no moment about C. The vertical distance from C to O is r − h. Let the horizontal distance from C to the vertical line through O be x. Since OC = r, x² + (r − h)² = r² x² = r² − (r − h)² = 2rh − h² x = √(2rh − h²) Taking moments about C: Moment of horizontal force P = P(r − h) Moment of weight W = Wx For just lifting, P(r − h) = Wx P = W√(2rh − h²)/(r − h) For actual dragging over the obstacle, P must be slightly larger than this value, because the above condition is only the verge of lifting. Hence P is slightly greater than W√(2rh − h²)/(r − h) valid for 0 < h < r. (b) For static balance, the vector sum of the mass-radius products must be zero. Let the reference line be the +x-axis. Given: m₁r₁ = 6 × 70 = 420 kg·mm at θ₁ = 45° m₂r₂ = 5 × 80 = 400 kg·mm at θ₂ = 135° m₃r₃ = 4 × 50 = 200 kg·mm at θ₃ = 240° Resolve along x and y: x = 420 cos45° + 400 cos135° + 200 cos240° x = 210√2 − 200√2 − 100 = 10√2 − 100 = −85.858 kg·mm y = 420 sin45° + 400 sin135° + 200 sin240° y = 210√2 + 200√2 − 100√3 = 410√2 − 100√3 = 406.622 kg·mm Resultant of the existing masses: R = √(x² + y²) = √((−85.858)² + (406.622)²) R ≈ 415.588 kg·mm The countermass is placed at radius 85 mm. If its mass is m_c, m_c × 85 = 415.588 m_c = 415.588/85 = 4.889 kg Angle of the resultant of existing masses: θ_R = tan⁻¹(406.622 / −85.858) = 101.92° So the countermass must be opposite this direction: θ_c = θ_R + 180° = 281.92° Countermass required ≈ 4.89 kg at 281.92° from the reference line, or equivalently 78.08° clockwise from the +x-axis. (c) Let H be the height of the balloon when the stone is dropped. The balloon is ascending at 25 m/s, so the stone initially has velocity 25 m/s upward. Take upward positive and g = 9.8 m/s². Let t₁ be the time for the stone to reach the lake surface. −H = 25t₁ − (1/2)(9.8)t₁² H = 4.9t₁² − 25t₁ For H > 0, this requires t₁ > 25/4.9 ≈ 5.102 s. But the total time from drop to hearing the splash is 5 s, so the sound travel time t₂ = 5 − t₁ must be positive, giving t₁ < 5 s. These two conditions cannot both hold. Hence the given data are inconsistent, and no physically admissible positive height exists. If the balloon velocity were ignored and the stone assumed to start from rest relative to the ground, the height would be about 107.5 m, but that contradicts the stated ascending motion. (d) The atomic packing factor (APF) measures how densely atoms are packed in a crystal structure. It is important because it influences theoretical density, slip behaviour, ductility, diffusion, phase stability, mechanical properties, and alloy design. A higher APF generally means a more closely packed structure. For FCC: Atoms per unit cell, n = 4 Atoms touch along the face diagonal, so 4r = a√2 Thus a = 2√2 r Volume of unit cell: V_cell = a³ = (2√2 r)³ = 16√2 r³ Volume occupied by atoms: V_atoms = 4 × (4/3)πr³ = (16/3)πr³ Therefore, APF = V_atoms / V_cell APF = [(16/3)πr³] / [16√2 r³] APF = π/(3√2) = π√2/6 ≈ 0.74048 APF for FCC = π√2/6 ≈ 0.7405. (e) Since the shaft is held in long bearings, the ends are effectively fixed in bending. For a fixed-fixed beam with a central load, the static central deflection is δ = W L³/(192 E I) Given: L = 1.5 m = 150 cm d = 2.0 cm W = 20 kg E = 2 × 10⁶ kg/cm² I = πd⁴/64 = π(2)⁴/64 = π/4 cm⁴ Substitute: δ = 20 × 150³ / [192 × 2 × 10⁶ × (π/4)] δ = 67,500,000 / (96,000,000π) δ = 0.703125/π ≈ 0.2238 cm Critical angular speed: ω_n = √(g/δ) = √(981/0.2238) ≈ 66.20 rad/s Critical speed in cycles/min: N_c = (60/2π)ω_n = (30/π) × 66.20 N_c ≈ 632.2 cycles/min Critical speed ≈ 632 rpm.

(a) A heavy carriage wheel of weight W, and radius r is to be dragged over an obstacle of height h by a horizontal force P…

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