Question 1

Dependence of kinetic energy The kinetic energy (energy of motion) of a body __________ upon its mass and velocity.

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Introduction / Context: Kinetic energy appears in collision analysis, vehicle dynamics, and energy balances. Understanding on which variables it depends is essential for predicting impact severity and designing safer systems. Given Data / Assumptions: Standard translational kinetic energy model. Body treated as a particle (no rotation considered here). Concept / Approach: The formula for translational kinetic energy is KE = 1/2 * m * v^2. Clearly, KE scales linearly with mass and quadratically with speed. Doubling velocity increases KE by a factor of 4, while doubling mass doubles KE. Hence KE depends on both m and v. Step-by-Step Solution: Recall formula: KE = 1/2 * m * v^2.Inspect dependence: proportional to m; proportional to v^2.Therefore, KE depends on mass and velocity. Verification / Alternative check: Dimensional analysis: [KE] = kg * (m^2/s^2) = N * m = J, consistent with energy units and confirming the variables that appear in the expression. Why Other Options Are Wrong: “Does not depend” contradicts the explicit formula showing dependence on m and v. Common Pitfalls: Underestimating the effect of speed due to the square term; small speed increases can significantly increase KE. Final Answer: depends

Question 2

Effects of a force acting on a body When a single resultant force acts on a body, which of the following outcomes may occur?

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Introduction / Context: Forces can do more than merely start or stop motion. In structures and machines, forces may alter velocity, balance other forces to produce equilibrium, and induce internal stresses that must be resisted by material strength. This question checks your holistic understanding of force effects. Given Data / Assumptions: Body can be rigid or deformable. Other forces may already act; adding a new force affects the net resultant. Internal stresses arise when materials resist deformation. Concept / Approach: From Newton’s second law, a nonzero resultant force changes motion (acceleration). If an applied force is chosen to cancel the existing resultant, the new net force can become zero, creating static or dynamic equilibrium. Any external force transmitted through material sections generates internal forces and stresses that must satisfy equilibrium and compatibility. Step-by-Step Solution: Option (a): A nonzero net force changes velocity (magnitude and/or direction).Option (b): An appropriately directed force can balance others, making ΣF = 0.Option (c): External loading is resisted by internal stress resultants (tension, compression, shear, bending).Since all statements can be true, choose “all of these.” Verification / Alternative check: Consider a simply supported beam under a central load: supports apply forces that balance the load (equilibrium), yet internal bending stresses arise and if unbalanced, motion would change—illustrating all facets simultaneously. Why Other Options Are Wrong: Selecting only one effect ignores other legitimate consequences of applying a force. Common Pitfalls: Thinking “internal stresses” occur only in failure. They are always present whenever loads are carried. Final Answer: all of these

Question 3

Engineering Mechanics – Units and Definitions What is the correct SI unit for the second moment of area (also called the area moment of inertia) of a plane area?

Accepted Answer

Introduction / Context: In strength of materials and structural analysis, we frequently use the area moment of inertia (also called the second moment of area) to quantify a section’s resistance to bending and deflection. This is not the same as the mass moment of inertia used in dynamics. Knowing the correct SI unit prevents dimensional mistakes in formulas for beam bending and column buckling. Given Data / Assumptions: We are referring to a geometric property of an area, not a mass distribution. Standard SI base units are used (m, kg, s). Common symbols: I, Ix, Iy for area moments about reference axes. Concept / Approach: The area moment of inertia is defined by integrals such as Ix = ∫ y^2 dA and Iy = ∫ x^2 dA. The integrand multiplies an area element by the square of a length. Therefore, its unit must be area (m^2) times length squared (m^2), giving m^4. This is purely geometric and independent of material density. Step-by-Step Solution: Start from definition: I = ∫ (distance)^2 dA. Unit of distance^2 = m^2. Unit of area element dA = m^2. Multiply: m^2 * m^2 = m^4 ⇒ unit of I is m^4. Verification / Alternative check: In beam deflection formulas like δ = (W L^3) / (3 E I) for a cantilever with end load, W has unit N, L^3 has m^3, E has N/m^2, so I must supply m^4 to keep δ in metres. The dimensional balance confirms I in m^4. Why Other Options Are Wrong: kg·m^2: unit of mass moment of inertia (rotational dynamics), not area property. kg·m·s^2 and kg/m^2: unrelated to second moment of area. N·m·s: not relevant; mixes force, length, and time. Common Pitfalls: Confusing area moment (m^4) with mass moment (kg·m^2). Using cm^4 or mm^4 without converting properly to SI. Final Answer: m^4

Question 4

Simple Harmonic Motion (SHM) – Acceleration–Displacement Relation Is the acceleration of a particle executing SHM proportional to its displacement from the mean position (with appropriate sign)?

Accepted Answer

Introduction / Context: Simple harmonic motion (SHM) is a foundational model for vibrations of mechanical systems, oscillations of pendulums (small angles), and electrical LC circuits. A hallmark of SHM is a specific relationship between acceleration and displacement. Given Data / Assumptions: Motion is undamped and unforced. Displacement x is measured from the mean (equilibrium) position. Angular frequency is ω (a constant for a given system). Concept / Approach: The defining equation of SHM is a = -ω^2 x. This means the acceleration magnitude is proportional to the displacement magnitude, and the negative sign indicates it is always directed toward the mean position (restoring). Thus the statement “acceleration is proportional to displacement” is correct, with the implied opposite direction. Step-by-Step Solution: Write the SHM differential equation: d^2x/dt^2 + ω^2 x = 0. Rearrange to acceleration form: a = d^2x/dt^2 = -ω^2 x. Identify proportionality: |a| ∝ |x| with a restoring (negative) sense. Therefore, the statement is true, acknowledging the restoring sign. Verification / Alternative check: From x(t) = A cos(ω t + φ), differentiate twice: v = -A ω sin(ω t + φ), a = -A ω^2 cos(ω t + φ) = -ω^2 x. The relationship holds for all t. Why Other Options Are Wrong: False: contradicts the defining equation of SHM. “Only for large amplitudes”: SHM model strictly assumes small amplitudes where linear restoring force applies. Damping or circular motion conditions are not required for the basic SHM proportionality. Common Pitfalls: Forgetting the negative sign and calling it “proportional in the same direction”. Applying SHM to systems with non-linear restoring forces where proportionality fails. Final Answer: True

Question 5

Angular Velocity – Valid Units in Engineering Practice Which of the following are valid units for angular velocity (rate of rotation)?

Accepted Answer

Introduction / Context: Angular velocity measures how fast an object rotates about an axis. In mechanical design (shafts, gears, flywheels), expressing angular speed in the correct units is essential for power and kinematics calculations. Given Data / Assumptions: Angular displacement is measured in radians (SI) or revolutions (engineering usage). Time is measured in seconds or minutes. Concept / Approach: Angular velocity ω is change of angular position per unit time. The SI coherent unit is rad/s. In practice, rotating machinery often uses rpm (rev/min). Both are acceptable, with conversion 1 rev = 2π rad, so 1 rpm = 2π/60 rad/s. Step-by-Step Solution: Definition: ω = Δθ / Δt. SI unit: rad/s (radian is dimensionless but treated as a unit of angle). Common industry unit: rev/min (rpm). Therefore, both rad/s and rev/min are valid units of angular velocity. Verification / Alternative check: For a motor rated 1500 rpm: ω = 1500 * 2π/60 ≈ 157.08 rad/s. The conversion shows consistency between the two units. Why Other Options Are Wrong: m/min: linear speed, not angular. N·m/s: power unit (watt) equivalence, not angular speed. Common Pitfalls: Confusing linear speed v with angular speed ω; they relate by v = ω r. Mixing rpm and rad/s without proper conversion. Final Answer: both rad/s and rev/min

Question 6

Dry Friction – Dependence of the Coefficient of Friction On what factor(s) does the coefficient of (Coulomb) friction primarily depend under standard dry contact conditions?

Accepted Answer

Introduction / Context: In machine design and statics, the coefficient of friction μ is used to estimate the maximum friction force F_max = μ N. Misunderstanding what μ depends on can lead to large errors in predicting motion or required effort in mechanisms. Given Data / Assumptions: Classical (Coulomb) dry friction regime. Rigid bodies with typical engineering surface finishes. No lubrication films that change the regime. Concept / Approach: Coulomb’s law models friction as F_max = μ N, where μ is mainly a property of the pair of surfaces and their condition (roughness, cleanliness, presence of oxides). For a given normal load N, μ does not explicitly depend on apparent area of contact in the idealized model because true microscopic contact area adjusts with load. Step-by-Step Solution: Identify regime: dry, unlubricated contact. Use Coulomb model: F_max = μ N. Note: μ is characteristic of surface pair and condition; not of nominal area. Hence, dependence is on the nature/condition of the surfaces, not on the apparent area alone. Verification / Alternative check: Experiments show friction force roughly scales with normal load and only weakly (if at all) with apparent area. Different materials (steel–steel, wood–steel) and surface finishes exhibit different μ values, confirming surface dependence. Why Other Options Are Wrong: Area of contact only: contradicts Coulomb model. Both equally / surface area and load product: not supported by the classical model. None of these: incorrect since surface pair clearly matters. Common Pitfalls: Assuming larger area always gives more friction; in dry friction models, it does not. Ignoring surface contamination or oxidation which alters μ. Final Answer: Nature and condition of the contacting surfaces only

Question 7

Definition Check – Friction Force and Direction of Motion A force acting opposite to the direction of motion (or impending motion) of a body is called the force of friction. Do you agree?

Accepted Answer

Introduction / Context: Friction is a resistive contact interaction that opposes relative motion between surfaces. Correctly identifying its direction is crucial in free-body diagrams used for equilibrium and dynamics problems. Given Data / Assumptions: Two solid surfaces in contact. Either actual relative motion (kinetic) or tendency to move (static). No lubrication effects altering the regime. Concept / Approach: Static friction acts to oppose the impending relative motion between surfaces. Kinetic friction acts to oppose the actual relative sliding velocity. Therefore friction acts opposite to the direction of relative motion (or its tendency) at the contact interface. Step-by-Step Solution: If the body slides to the right on a rough surface, kinetic friction acts to the left. If a force attempts to move a block rightward but it has not moved yet, static friction acts leftward to resist. Thus, in either case, friction opposes motion or impending motion. Verification / Alternative check: For rolling without slipping, friction direction depends on external torques but still acts to oppose the relative motion at the point of contact (which is instantaneously at rest for pure rolling). The concept remains consistent. Why Other Options Are Wrong: Disagree or restricted cases (only rolling/fluids/zero speed) misstate the general definition for dry contact friction. Common Pitfalls: Confusing the direction of friction in rolling problems; analyze relative motion at the contact point. Assuming friction always points opposite to the body’s center-of-mass velocity; the correct reference is relative motion at the interface. Final Answer: Agree

Question 8

Statics – Full Equilibrium Conditions for Coplanar Force Systems For any system of coplanar forces acting on a rigid body, which conditions must be satisfied simultaneously for equilibrium?

Accepted Answer

Introduction / Context: Equilibrium analysis is the cornerstone of statics. For planar (coplanar) problems, the correct set of independent equations is needed to solve reactions, internal forces, and support conditions for structures and mechanisms. Given Data / Assumptions: Rigid body assumption. All forces lie in one plane (2D problem). Moments are taken about any convenient point in the plane. Concept / Approach: For a rigid body in the plane, there are three independent equilibrium conditions: the sums of force components in two orthogonal directions must be zero, and the sum of moments about any point must be zero. Using all three ensures no net translation or rotation. Step-by-Step Solution: Write ΣFx = 0 to prevent acceleration along x. Write ΣFy = 0 to prevent acceleration along y. Write ΣM_O = 0 (about any point O) to prevent angular acceleration. These three equations fully describe planar equilibrium. Verification / Alternative check: Taking moments about different points yields equivalent conditions when ΣFx = 0 and ΣFy = 0 hold. If any condition is omitted, rotational or translational equilibrium may be violated. Why Other Options Are Wrong: Using only ΣM = 0 ignores possible net forces; the body could translate without rotation. Selecting only ΣFx = 0 or ΣFy = 0 is incomplete. Common Pitfalls: Forgetting to include couple moments or distributed loads as equivalent forces/moments. Choosing an inconvenient moment center leading to algebraic complexity. Final Answer: all of the above

Question 9

Machines – Improving Velocity Ratio of a Single-Purchase Crab Winch Which action increases the velocity ratio (VR) of a single-purchase crab winch?

Accepted Answer

Introduction / Context: In hoisting machines like a single-purchase crab winch, the velocity ratio (VR) determines how much movement of effort produces how much lift of the load. Understanding how design parameters affect VR helps in selecting easier hand effort or faster lift. Given Data / Assumptions: Single-purchase arrangement: one gear and one pinion stage. Let L = handle length, r = load drum radius, Zg = gear teeth, Zp = pinion teeth. Ideal kinematics (no slip). Concept / Approach: A typical VR expression is VR = (Zg / Zp) * (L / r). Thus, VR increases if (Zg / Zp) increases, if L increases, or if r decreases. The question asks which listed action increases VR. Step-by-Step Solution: Check handle length: increasing L increases VR directly. Check drum radius: increasing r decreases VR (opposite effect). Check pinion teeth: increasing Zp reduces (Zg / Zp), thereby reducing VR. Therefore, only increasing L (option A) increases VR among the choices. Verification / Alternative check: Consider proportional changes: doubling L doubles VR; doubling r halves VR; increasing Zp by 20% lowers gear ratio and thus VR by 20% if Zg fixed. Why Other Options Are Wrong: Increasing drum radius reduces VR, making lifting harder. Increasing pinion teeth lowers gear ratio and VR. “All of the above” contradicts the correct trends. Reducing gear ratio reduces VR (opposite of desired). Common Pitfalls: Confusing mechanical advantage with VR and assuming any “increase” helps lifting. Ignoring that VR trades speed for force. Final Answer: Increasing the length of the handle

Question 10

Rocket Propulsion – Law of Action and Reaction A teacher states that Newton’s third law of motion is directly involved when explaining how rockets move in space. Is this statement justified?

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Introduction / Context: Rockets accelerate by expelling mass (exhaust gases) at high speed. Understanding which fundamental law explains this motion is central to basic mechanics and aerospace engineering. Given Data / Assumptions: Rocket ejects exhaust mass backward at high velocity. System is isolated over short intervals (neglecting external forces). Motion occurs even in vacuum (no need for air to “push against”). Concept / Approach: Newton’s third law: for every action, there is an equal and opposite reaction. The rocket exerts a force on the exhaust gases backward; the gases exert an equal and opposite force on the rocket forward. This interaction creates thrust and accelerates the rocket. Step-by-Step Solution: Exhaust mass flow rate: ṁ. Relative exhaust speed: u (backward relative to rocket). Thrust magnitude: T ≈ ṁ * u (neglecting pressure terms for simplicity). Direction: forward on rocket, backward on exhaust—an action–reaction pair. Verification / Alternative check: Conservation of momentum provides an equivalent explanation: as the rocket imparts backward momentum to exhaust, the rocket gains equal forward momentum. This is consistent with Newton’s third law. Why Other Options Are Wrong: “Not justified” or “only near Earth’s surface”: incorrect because rockets operate using internal reaction forces, not relying on air or ground. “Only with aerodynamic lift” or “requires gravity”: thrust arises from mass ejection, independent of atmosphere or gravity. Common Pitfalls: Thinking rockets push against air; in vacuum they still accelerate by ejecting mass. Final Answer: justified

Question 11

Vibrations – Definition of Amplitude in Oscillatory Motion The maximum displacement of a vibrating body from its mean (equilibrium) position is called the amplitude. Is this definition correct?

Accepted Answer

Introduction / Context: When describing oscillations—whether mechanical (springs), structural (beams), or electrical (AC signals)—we routinely use the term amplitude. A precise definition helps prevent confusion in analysis and communication. Given Data / Assumptions: Periodic motion about a mean (equilibrium) position. Displacement measured along a defined axis. No restriction on the nature of oscillation other than being measurable relative to an equilibrium. Concept / Approach: Amplitude A is defined as the maximum absolute displacement from the mean position achieved during oscillation. For SHM, x(t) = A cos(ω t + φ), the amplitude is the constant A, representing the peak magnitude. This definition does not require damping, circular motion, or large angles. Step-by-Step Solution: Identify equilibrium position where net restoring force is zero. Track displacement x(t) over time. Determine the maximum absolute value reached: A = max|x(t)|. This maximum is the amplitude by definition. Verification / Alternative check: For SHM, differentiate x(t) to find turning points where velocity v = 0; the corresponding x values are ±A. These are the amplitude limits irrespective of damping presence (though for damping the amplitude decays over time). Why Other Options Are Wrong: Restricting amplitude to special cases (circular motion, large oscillations, damping) is unnecessary; the definition is general. “Incorrect” contradicts widely accepted terminology in physics and engineering. Common Pitfalls: Confusing amplitude with instantaneous displacement at a given time. In damped systems, forgetting that amplitude can be time-varying (envelope), but the definition per cycle remains maximum absolute displacement. Final Answer: Correct

Question 12

Area Properties — Square Section (Centroidal Axis) What is the area moment of inertia of a square plate of side length a about an axis through its centre of gravity (centroid) and perpendicular to the plane of the square?

Accepted Answer

Introduction / Context: Engineers frequently need the area moment of inertia (also called the second moment of area) of basic shapes to evaluate bending stress and deflection. For a square plate, the centroidal moments about in-plane axes are standard results used in beam and plate calculations. Given Data / Assumptions: Shape: square, side length a. Axis: through the centroid, perpendicular to the plate for polar, but here we need the in-plane centroidal axis result commonly quoted as I_x or I_y for a square (either is the same by symmetry). Material distribution is uniform; this is an area property, not a mass property. Concept / Approach: The area moment of inertia quantifies how area is distributed relative to an axis. For a rectangle of breadth b and depth d, I_centroid about the axis parallel to b is (b d^3) / 12. Setting b = d = a for a square gives I = a^4 / 12 about either centroidal in-plane axis. The polar area moment about the perpendicular centroidal axis would be J = I_x + I_y = a^4 / 6, but that is not asked here. Step-by-Step Solution: Start with rectangle formula: I = (b d^3) / 12. For a square: b = a and d = a ⇒ I = (a * a^3) / 12 = a^4 / 12. By symmetry, I_x = I_y for a square; result holds for either centroidal in-plane axis. Verification / Alternative check: Compare with polar relation: J = I_x + I_y = 2 * (a^4 / 12) = a^4 / 6. This is a known closed-form identity for a square, confirming I_x = a^4 / 12. Why Other Options Are Wrong: a^4 / 4 and a^4 / 8 are too large and would overestimate stiffness; a^4 / 36 is the centroidal inertia of a triangle-like scaling, not applicable to a square; the correct rectangular/square formula yields a^4 / 12. Common Pitfalls: Confusing area moment (used in bending) with mass moment (used in dynamics); mixing the polar moment with the in-plane centroidal moment. Final Answer: a^4 / 12.

Question 13

Friction Fundamentals — Static vs Dynamic What is the typical ratio of the coefficient of static friction to the coefficient of dynamic (kinetic) friction for the same pair of materials?

Accepted Answer

Introduction / Context: In contact mechanics, the force required to initiate motion is generally higher than the force required to maintain motion. This behavioral difference is captured by coefficients of static and dynamic friction and is critical in design of brakes, clutches, and handling equipment. Given Data / Assumptions: Same two surfaces in contact under similar conditions (normal force, surface preparation). Static regime: surfaces at rest relative to each other. Dynamic regime: surfaces sliding relative to each other. Concept / Approach: Static friction adapts up to a maximum (limiting) value F_s,max = μ_s * N before sliding begins. Once motion starts, the resisting force is F_k = μ_k * N, where usually μ_k < μ_s. Therefore μ_s / μ_k > 1 in typical engineering situations. Step-by-Step Solution: Define μ_s: coefficient at the verge of motion. Define μ_k: coefficient during steady sliding. Observation from experiments: μ_s ≥ μ_k ⇒ μ_s / μ_k ≥ 1 and typically > 1. Verification / Alternative check: Empirical charts for common pairs (steel–steel, rubber–concrete, wood–wood) consistently show μ_s exceeding μ_k under comparable conditions, confirming the ratio > 1. Why Other Options Are Wrong: “Equal to one” would imply identical onset and sliding resistance, rarely true; “Less than one” contradicts experimental evidence; “Depends only on speed” ignores strong material and surface dependence. Common Pitfalls: Using μ_s for sliding calculations; ignoring that lubrication or contamination can modify both coefficients, though μ_s usually remains ≥ μ_k. Final Answer: Greater than one.

Question 14

Routh’s Rule for Bodies Symmetrical About a Perpendicular Axis For a spherical (or axisymmetric) body of mass M, let S be the sum of squares of the two semi-axes perpendicular to the axis considered. According to Routh’s rule, what is the mass momen…

Accepted Answer

Introduction / Context: Routh’s rule provides compact expressions for the mass moment of inertia of ellipsoids or bodies symmetrical about a perpendicular axis. For such a body, the inertia about one principal axis depends on the squares of the other two semi-axes. Given Data / Assumptions: Mass M, centroidal principal axes. Axisymmetry about a perpendicular axis; S is the sum of squares of the two semi-axes perpendicular to the axis considered. Uniform mass distribution. Concept / Approach: For an ellipsoid with semi-axes a, b, c, the mass moments are I_x = (M/5)(b^2 + c^2), I_y = (M/5)(c^2 + a^2), I_z = (M/5)(a^2 + b^2). Writing S as the sum of the squares of the two semi-axes perpendicular to the chosen axis leads directly to I = (M/5) S. Step-by-Step Solution: Identify the axis and the other two semi-axes. Compute S = (semi-axis_1)^2 + (semi-axis_2)^2. Apply Routh’s rule: I_axis = (M / 5) * S. Verification / Alternative check: For a solid sphere (a = b = c = r), S = r^2 + r^2 = 2 r^2 ⇒ I = (M/5) * 2 r^2 = (2/5) M r^2, the well-known result for a solid sphere about any diameter. Why Other Options Are Wrong: M S / 3 and M S / 4 would overpredict inertia and do not match standard ellipsoid formulas; “None of these” is incorrect because M S / 5 is the exact expression. Common Pitfalls: Confusing area moments with mass moments; forgetting that S excludes the semi-axis along the chosen axis. Final Answer: M S / 5.

Question 15

Ladder Equilibrium — Direction of Friction at the Wall A ladder rests on smooth ground and leans against a rough vertical wall. At the upper end (wall contact), the direction of the frictional force will be:

Accepted Answer

Introduction / Context: Ladder problems test understanding of friction directions. When one contact is smooth (no friction) and the other is rough (with friction), the frictional force must oppose the impending motion at the rough contact surface. Given Data / Assumptions: Ground is smooth ⇒ no friction at the foot; reaction at ground is normal (vertical). Wall is rough ⇒ friction is available along the wall at the top contact. Ladder tends to slip due to its weight acting through its center of gravity. Concept / Approach: With no horizontal reaction at the smooth ground, the wall must supply the horizontal reaction. The ladder’s upper end tends to slide downward along the wall under weight and geometry. Friction acts opposite to the direction of relative motion (or impending motion) along the surface of contact; hence it must act upward along the wall at the upper end to resist sliding down. Step-by-Step Reasoning: Identify possible motion: upper end wants to move downward along the wall. Friction opposes motion along contact: therefore friction acts upward on the ladder at the wall. Equilibrium then involves: upward friction at wall, horizontal wall reaction, vertical ground reaction, and weight. Verification / Alternative check: Remove friction at wall hypothetically: the upper end would certainly slide downward. Thus actual friction must be upward to prevent that motion. Why Other Options Are Wrong: “Towards/away from the wall” describe horizontal directions, but friction acts tangential to the wall (vertical). “Downward” would aid the slip, not oppose it. Common Pitfalls: Reversing friction direction; assuming friction acts at the smooth contact, which it does not. Final Answer: Upward at the upper end.

Question 16

Kinematics of Rolling — Wheel of a Car What best describes the motion of a car’s wheel (assuming rolling without slipping)?

Accepted Answer

Introduction / Context: Rolling without slipping is a fundamental kinematic model in vehicle dynamics. Each wheel simultaneously rotates about its own axis while the wheel’s center translates along the road. Given Data / Assumptions: No slip condition at the contact (ideal rolling). Rigid wheel model for kinematics. Concept / Approach: For rolling without slipping, the instantaneous velocity of the contact point relative to the ground is zero. The motion can be decomposed into translation of the center of mass plus rotation about the center with angular speed ω such that v_center = ω * r. Step-by-Step Explanation: Let r be wheel radius; center translates at speed v. Wheel rotates with ω satisfying v = ω * r (no slip). Therefore overall motion = translation (of center) + rotation (about center). Verification / Alternative check: Observe tire tread: points above the center move faster than the vehicle speed, while the contact point is instantaneously at rest. This only occurs with combined translation and rotation. Why Other Options Are Wrong: Pure translation would imply no wheel spin; pure rotation would imply spinning in place; “neither” is not physical. Common Pitfalls: Confusing slip conditions; forgetting the v = ω r condition for ideal rolling. Final Answer: Combined translation and rotation (rolling motion).

Question 17

Highway Engineering — Superelevation Terminology The slope provided on a road surface at horizontal curves to counteract centrifugal effects is called the:

Accepted Answer

Introduction / Context: Vehicles negotiating a horizontal curve experience lateral acceleration. Roads are raised on the outer edge (banked) to develop a component of normal reaction that supplies the necessary centripetal force, improving safety and comfort. Given Data / Assumptions: Highway curve with designed speed and radius. Uniform cross slope (superelevation) on the curve. Concept / Approach: Banking (superelevation) tilts the pavement at an angle so that part of the weight’s normal reaction provides inward radial force. The “angle of banking” or the corresponding superelevation rate e is set by speed, radius, and side friction allowance. Step-by-Step Reasoning: Identify purpose: balance lateral acceleration a = v^2 / r. Provide slope e so that N’s components contribute to centripetal requirement. Terminology: the slope is called the angle of banking (superelevation). Verification / Alternative check: Standards specify e = v^2 / (g r) (basic form without side friction) for design checks, confirming the role of banking. Why Other Options Are Wrong: Angle of friction is related to μ and impending slip; angle of repose is for granular piles and sliding initiation, not road design; “none” is incorrect. Common Pitfalls: Confusing banking with camber (for drainage); mixing angle units with the superelevation rate e. Final Answer: Angle of banking (superelevation).

Question 18

Simple Harmonic Motion — Direction of Acceleration A body is said to execute simple harmonic motion (SHM) if its acceleration is directed toward the mean (equilibrium) position and proportional to displacement from it. Is the statement (direction to…

Accepted Answer

Introduction / Context: SHM is a foundational motion model for oscillators (springs, pendulums for small angles, masses). It underpins vibration theory, acoustics, and many control-system approximations near equilibrium. Given Data / Assumptions: Displacement x measured from equilibrium. Restoring mechanism (spring force, gravity component) linear near equilibrium. Concept / Approach: The defining differential equation is a = d^2x/dt^2 = −ω^2 x, showing acceleration proportional to displacement and directed toward x = 0 (negative sign). Thus, acceleration always points toward the mean position, reversing direction as the particle passes through equilibrium. Step-by-Step Reasoning: Assume linear restoring force: F = −k x. Newton’s law: m a = −k x ⇒ a = −(k/m) x. Let ω^2 = k/m ⇒ a = −ω^2 x (toward equilibrium). Verification / Alternative check: Energy view: total energy E = (1/2)k x^2 + (1/2)m v^2 remains constant; slope of potential pulls the mass back toward x = 0, consistent with acceleration direction. Why Other Options Are Wrong: Saying “False” would contradict the SHM definition; without the “proportional to displacement” part, the motion may not be strictly SHM, but the directionality statement is still correct for SHM. Common Pitfalls: Forgetting proportionality as well as direction; assuming large-angle pendulum motion is perfectly SHM (it is only approximate). Final Answer: True.

Question 19

Mass Moment — Thin Rod About Its Centre For a thin, uniform rod of mass m and length l, what is the mass moment of inertia about an axis through its centre and perpendicular to its length?

Accepted Answer

Introduction / Context: Standard mass moment formulas for slender members are used in vibration and rotational dynamics. The rod-about-centre result is one of the most cited identities in mechanical design and analysis. Given Data / Assumptions: Uniform rod, length l, mass m. Axis through the midpoint, perpendicular to the rod’s length. Concept / Approach: For a slender rod, distributing mass along a line, the integral I = ∫ r^2 dm with symmetric limits gives I_center = (1/12) m l^2. About an end, it is (1/3) m l^2; the two are connected by the parallel-axis theorem. Step-by-Step Solution (outline): Place origin at the center; coordinate x runs from −l/2 to +l/2. Let linear density λ = m / l, and r = x for the perpendicular axis. Compute I = ∫(x^2) dm = ∫(x^2) λ dx = λ ∫_{−l/2}^{+l/2} x^2 dx = (m/l) * (l^3 / 12) = (1/12) m l^2. Verification / Alternative check: Using parallel-axis: I_end = I_center + m (l/2)^2 = (1/12) m l^2 + (1/4) m l^2 = (1/3) m l^2, a known correct endpoint value. Why Other Options Are Wrong: m l^2 / 4 and m l^2 / 6 overpredict; m l^2 / 8 is not a standard value for any principal rod axis. Common Pitfalls: Confusing area versus mass moments; mixing the end-axis (1/3 m l^2) with center-axis (1/12 m l^2). Final Answer: m l^2 / 12.

Question 20

Projectile Motion — Cartesian Trajectory Equation Which of the following correctly gives y as a function of x for a projectile launched with speed u at angle α above the horizontal (neglecting air resistance)?

Accepted Answer

Introduction / Context: The Cartesian equation of a projectile’s trajectory is a parabola. Deriving y(x) helps solve range, height, and position problems without explicitly using time after elimination. Given Data / Assumptions: Initial speed u, projection angle α above horizontal. Gravity g constant; no air resistance. Concept / Approach: Resolve initial velocity: u_x = u cos α, u_y = u sin α. Use horizontal motion x = u cos α * t and vertical motion y = u sin α * t − (1/2) g t^2. Eliminate time t by t = x / (u cos α) to get the Cartesian form. Step-by-Step Derivation: x = u cos α * t ⇒ t = x / (u cos α). y = u sin α * t − (1/2) g t^2. Substitute t: y = x tan α − (g x^2) / (2 u^2 cos^2 α). Verification / Alternative check: At x = 0 ⇒ y = 0; the curvature term is negative (concave down), consistent with gravity. Units also match length on both sides. Why Other Options Are Wrong: The “+” sign before the gravity term would make the path open upward; the other two options are unrelated forms and do not represent y(x) for projectile motion. Common Pitfalls: Forgetting the cos^2 α in the denominator; sign errors on the gravity term; failing to eliminate time correctly. Final Answer: y = x tan α − (g x^2) / (2 u^2 cos^2 α).

Engineering Mechanics Questions