Q: Definition check – what is energy? Select the most accurate definition used in engineering mechanics for “energy”.

Introduction / Context: Clear verbal definitions help prevent mix-ups between fundamental quantities: work, energy, and power. Energy underpins analyses of machines, structures, and thermodynamic systems, representing the budget from which mechanical, thermal, or electrical outputs can be produced. Given Data / Assumptions: Classical mechanics conventions for work, energy, and power. Scalar quantities unless otherwise stated (e.g., potential, kinetic energies). Steady units: Joule for work/energy; Watt for power. Concept / Approach: Energy is the capacity for doing work. Work is the energy transfer via a force acting through a distance. Power is the time rate of doing work. The accepted textbook phrasing distinguishes these three succinctly and avoids ambiguity in problem solving and unit checks. Step-by-Step Solution: Recall: Work W = ∫ F · ds, unit Joule.Energy E (e.g., kinetic 1/2 m v^2, potential m g h) represents the stored capacity to perform work, unit Joule.Power P = dW/dt, unit Watt = Joule/second.Therefore the correct definition is “capacity for doing work”. Verification / Alternative check: Dimensionally, energy and work share the same unit; power differs by per-time. Common engineering references adopt precisely this wording. Why Other Options Are Wrong: (a) “power of doing work” is vague and conflates with power. (c) and (e) explicitly define power, not energy. (d) cannot be correct because (a), (c), and (e) are not definitions of energy. Common Pitfalls: Using “power” and “energy” interchangeably; forgetting that energy is a state property whereas power is a rate. Final Answer: capacity for doing work

Q: Angle of friction from limiting horizontal push on a horizontal surface A 50 kgf block rests on a horizontal plane. A horizontal force of 18 kgf just causes impending motion. Determine the angle of friction φ (where tan φ = μ).

Introduction / Context: At impending motion on a horizontal plane, the frictional force reaches its limiting value μN. Converting the limiting push to the friction coefficient and then to the angle of friction is common in design of wedges, clamps, and handling equipment where stick–slip must be controlled. Given Data / Assumptions: Normal reaction N equals the weight W on a horizontal plane. Limiting horizontal force P at impending motion equals friction μN. Given W = 50 kgf, P = 18 kgf; use kgf consistently (force units). Concept / Approach: At the threshold of motion: P = μ * N. For a horizontal plane, N = W. Thus μ = P / W. The angle of friction φ satisfies tan φ = μ. Compute μ and take inverse tangent to obtain φ in degrees and minutes. Step-by-Step Solution: N = W = 50 kgf.μ = P / W = 18 / 50 = 0.36.φ = arctan(μ) = arctan(0.36) ≈ 19.798°.Convert decimal degrees to degrees–minutes: 0.798° * 60 ≈ 47.9′ ≈ 48′ → φ ≈ 19° 48′. Verification / Alternative check: Check by resolving along motion: Required horizontal push at limiting state equals μW. Using μ = tan φ ensures consistency with the angle interpretation from the friction cone. Why Other Options Are Wrong: Other listed angles correspond to μ values not equal to 0.36. For example, 17°48′ → μ ≈ 0.32; 21°48′ → μ ≈ 0.40, which do not match P/W. Common Pitfalls: Confusing kg (mass) with kgf (force); rounding too early; using sin or cos instead of tan for the friction angle. Final Answer: 19° 48′

Q: Distance of the centroid of the leftover lamina from the original center A circular plate of diameter 6 cm (R = 3 cm) has a circular disc removed whose diameter equals the original radius (3 cm, so r = 1.5 cm). Find the distance of the centroid of t…

Introduction / Context: Designers often remove material from plates to save weight while controlling balance. The centroid of the remaining area shifts toward the heavier side. Computing that shift ensures proper placement of supports, bearings, or counterweights. Given Data / Assumptions: Original circle radius R = 3 cm (area A1 = 9π). Removed circle radius r = 1.5 cm (area A2 = 2.25π). The small circle is cut at the rim, internally tangent to the outer boundary → center spacing d = R − r = 1.5 cm. Uniform thickness and density (area centroid problem). Concept / Approach: Use the negative-area method along the line of centers. The centroid shift x from the original center equals the first moment of removed area divided by the net remaining area. Step-by-Step Solution: Compute A1 − A2 = 9π − 2.25π = 6.75π.First moment of removed area about original center = A2 * d = 2.25π * 1.5 = 3.375π.Centroid shift x = (A2 * d) / (A1 − A2) = 3.375π / 6.75π = 0.5 cm.Direction: along the radius away from the cut-out toward the remaining material mass. Verification / Alternative check: By symmetry, the centroid must lie on the line of centers. Magnitude is less than 1.5 cm and proportional to area ratio A2/(A1 − A2), confirming 0.5 cm. Why Other Options Are Wrong: 2.0 and 1.5 cm exceed d; 1.0 cm overestimates; 0.25 cm underestimates given the removed area fraction 2.25/6.75 = 1/3. Common Pitfalls: Confusing diameter and radius; forgetting to subtract the removed area in the denominator; placing the small circle incorrectly. Final Answer: 0.5 cm

Q: How to double the time period of a simple pendulum For a simple pendulum operating at small amplitude, what change is required to double its period of oscillation?

Introduction / Context: The period of a simple pendulum guides timing devices and dynamic isolation systems. Understanding parameter sensitivity helps engineers tune response without trial-and-error changes in mass or structure. Given Data / Assumptions: Small-angle oscillations so that T = 2π * sqrt(l / g) applies. Mass of bob is concentrated; string is light and inextensible. Local gravity g is constant. Concept / Approach: Period T depends on length l and gravity g only. Mass does not appear in T. To change T by a factor, scale l accordingly using the square-root relationship. Step-by-Step Solution: T ∝ sqrt(l). To double T, set T_new / T_old = 2 = sqrt(l_new / l_old).Square both sides: 4 = l_new / l_old ⇒ l_new = 4 * l_old.Therefore, quadrupling the length doubles the period. Verification / Alternative check: Plug into T = 2π * sqrt(l / g). If l → 4l, then T → 2π * sqrt(4l/g) = 2 * 2π * sqrt(l/g) = 2T, confirming the requirement. Why Other Options Are Wrong: (a) and (b) change mass, which does not affect T in the simple model. (d) doubling length increases T by sqrt(2), not 2. (e) while halving g would increase T by sqrt(2); it does not double T unless g is reduced by a factor of 4, which is unrealistic. Common Pitfalls: Believing mass influences period; forgetting the square-root dependence; applying the formula outside small-angle limits. Final Answer: quadruple the length of the pendulum

Q: Simple harmonic motion (SHM) energy at the mean position: A particle executes simple harmonic motion. While passing through the mean (equilibrium) position, which form(s) of mechanical energy does the particle possess at a maximum or minimum level?

Introduction / Context: In simple harmonic motion (SHM), such as a mass–spring system or a small-angle pendulum, energy oscillates between kinetic and potential forms. Understanding how kinetic energy (KE) and potential energy (PE) vary with displacement and velocity is fundamental in vibration analysis and engineering dynamics. Given Data / Assumptions: Motion is ideal SHM with no damping. Mean position is the equilibrium point where displacement is zero. Total mechanical energy remains constant (no losses). Concept / Approach: For SHM with displacement x(t) and amplitude A, potential energy is proportional to displacement squared (for a spring, PE = 1/2 k x^2). Kinetic energy depends on velocity squared (KE = 1/2 m v^2). At the mean position, x = 0 → PE is minimum (zero for a spring), and velocity is maximum → KE is maximum. At extreme positions, velocity is zero → KE is zero and PE is maximum. Step-by-Step Solution: At mean position: x = 0 → PE = 1/2 k (0)^2 = 0 (minimum).Velocity magnitude is maximum since v = ω√(A^2 − x^2) → v = ωA.Therefore, KE = 1/2 m (ωA)^2 is maximum at the mean position. Verification / Alternative check: Energy conservation: total energy E = 1/2 k A^2 is constant. At x = 0, all of E is kinetic; at x = ±A, all of E is potential—consistent with the conclusion. Why Other Options Are Wrong: (b) PE cannot be maximum at the mean position; (c) KE is not minimum at x = 0; (d) both minimum is impossible because total energy is constant and nonzero for A ≠ 0; (e) is incorrect because one option exactly describes the SHM energy distribution. Common Pitfalls: Confusing displacement and velocity extrema; assuming KE and PE are equal at the mean position (they are equal only at intermediate displacements, not at x = 0). Final Answer: maximum kinetic energy and minimum potential energy

Q: Compound pendulum — small-angle SHM relation for frequency: If a compound pendulum undergoes small oscillations with angular displacement θ and angular acceleration α proportional to θ (i.e., α = −(g/ℓ) θ), what is the frequency f of its motion expr…

Introduction / Context: The compound pendulum (physical pendulum) exhibits simple harmonic motion for small angular displacements. The relationship between the restoring angular acceleration and angular displacement determines the natural frequency. Expressing frequency directly in terms of the proportionality α/θ reinforces the SHM analogy to linear systems. Given Data / Assumptions: Small-angle oscillations so that sin θ ≈ θ (in radians). Angular acceleration α is proportional to −θ: α = −(g/ℓ) θ. g is acceleration due to gravity; ℓ is the equivalent pendulum length. Concept / Approach: For SHM, the governing rotational equation is θ¨ + (g/ℓ) θ = 0. Angular frequency ω satisfies ω^2 = g/ℓ. Because α/θ = −g/ℓ (magnitude g/ℓ), we can write ω = sqrt(α/θ) using magnitudes. The cyclic frequency f relates to angular frequency by f = ω / (2π). Step-by-Step Solution: Start with α = θ¨ = −(g/ℓ) θ → |α/θ| = g/ℓ.Hence ω^2 = g/ℓ = |α/θ|.Therefore ω = sqrt(α/θ) (using positive magnitudes).Frequency f = ω / (2π) = (1 / 2π) * sqrt(α / θ). Verification / Alternative check: From standard formula for a compound pendulum, f = (1 / 2π) * sqrt(g / ℓ). Substituting g/ℓ = α/θ reproduces the same expression. Why Other Options Are Wrong: (b) and (c) have incorrect dimensions (frequency cannot be linear in α/θ without a square root). (d) is not dimensionally consistent. (e) inverts the needed relationship and lacks the square root. Common Pitfalls: Forgetting the small-angle assumption; ignoring sign (use magnitudes for frequency); confusing angular frequency ω with cyclic frequency f. Final Answer: (1 / 2π) * sqrt(α / θ)

Q: Uniform circular motion — direction of acceleration: A particle moves along a circle with uniform speed. At any instant, the direction of its acceleration vector is which of the following?

Introduction / Context: Uniform circular motion (UCM) features constant speed but continuously changing velocity direction. This change in direction implies a nonzero acceleration known as centripetal acceleration. Recognizing its direction is vital in dynamics of rotating machinery, vehicle cornering, and orbital motion. Given Data / Assumptions: Particle moves with constant speed on a circular path of radius r. Angular speed may be denoted ω; linear speed v = ω r. No tangential acceleration component (speed constant). Concept / Approach: The acceleration responsible for changing the velocity direction points toward the centre of curvature. Its magnitude is a_c = v^2 / r = ω^2 r. Because speed is constant, any tangential component would imply a change in speed, which is absent in UCM; hence acceleration must be purely radial and inward. Step-by-Step Solution: Identify that only direction changes, not speed → tangential acceleration = 0.Use a_c = v^2 / r directed toward the centre (centripetal).Therefore, the acceleration vector is radially inward at all times. Verification / Alternative check: Differentiating the velocity vector in polar coordinates shows a radial component −ω^2 r pointing toward the centre, with zero tangential component for constant ω. Why Other Options Are Wrong: (a) and (c) incorrectly suggest outward direction; (b) implies a tangential component which would change speed; (e) is arbitrary and not supported by UCM kinematics. Common Pitfalls: Confusing centripetal (inward) with centrifugal (a pseudo force in a rotating frame) which appears outward; assuming any tangential component exists when speed is constant. Final Answer: towards the centre (radially inward)

Q: Apparent weight in a moving lift (elevator): A body that truly weighs 14 g (measured at rest) appears to weigh 13 g when measured by a spring balance in a moving lift. Determine the magnitude of the lift’s acceleration at that instant (state the clo…

Introduction / Context: Apparent weight changes in an accelerating frame such as a lift (elevator). A spring balance reads the normal reaction on the body, which differs from the true weight when the frame accelerates. This classic dynamics problem reinforces Newton’s second law in non-inertial scenarios treated from an inertial frame. Given Data / Assumptions: True weight corresponds to mass m such that W = m g; the balance reads 14 g at rest. Apparent weight during motion is equivalent to 13 g on the same balance. Take g ≈ 9.8 m/s^2 (ratios will cancel). Motion is vertical; friction and air resistance neglected. Concept / Approach: When the lift accelerates downward with acceleration a, the apparent weight N satisfies N = m (g − a). When it accelerates upward, N = m (g + a). A reading lower than the true weight indicates downward acceleration (or upward acceleration with less likely opposing sign choice). Use the ratio of apparent to true weight to solve for a/g. Step-by-Step Solution: True weight W_true = m g corresponds to “14 g units” on the balance.Apparent weight W_app = m (g − a) corresponds to “13 g units”.Form the ratio: W_app / W_true = (g − a)/g = 13/14.Solve for a: g − a = (13/14) g → a = g (1 − 13/14) = g / 14.Compute: a ≈ 9.8 / 14 ≈ 0.7 m/s^2 (downward). Verification / Alternative check: If the lift were accelerating upward, apparent weight would exceed true weight (contrary to the observation). Thus the direction is downward with magnitude about 0.7 m/s^2, consistent with the sign logic. Why Other Options Are Wrong: 0.5 and 1.0 m/s^2 do not satisfy the precise ratio 13/14; 0.01 m/s^2 is negligible; 1.5 m/s^2 is too large for the given readings. Common Pitfalls: Confusing mass units with force units; using W = m a incorrectly; forgetting that a reduced reading implies downward acceleration. Final Answer: 0.7 m/s^2

Question 1

Effect of a retarding force – what does it not do? A retarding (opposing) force acts on a moving body. Identify what a retarding force does not cause.

Accepted Answer

Introduction / Context: Forces are classified by the effect they produce on motion. A retarding force specifically opposes the current direction of motion, such as kinetic friction or aerodynamic drag, and thus reduces speed or kinetic energy. Recognizing what a retarding force cannot do sharpens physical intuition for dynamics problems and design scenarios. Given Data / Assumptions: The body is subject to an opposing (retarding) force. Other forces may or may not be present, but the question focuses on the inherent effect of the retarding force itself. Classical Newtonian framework applies. Concept / Approach: A retarding force does negative work on a moving body, reducing its kinetic energy. If no additional driving force exists, the speed decreases and may eventually reach zero. By definition, a retarding force cannot by itself initiate motion of a body initially at rest, because it acts opposite to any would-be motion and cannot supply positive work to start movement. Step-by-Step Solution: Consider a block sliding on a surface with kinetic friction F_k opposing motion.Work by friction W = −F_k * s is negative, lowering kinetic energy KE.If the block is initially at rest and only friction acts, net force is zero or opposing any attempted motion; no acceleration starts the block moving.Therefore, a retarding force cannot introduce motion; it can only resist or reduce it. Verification / Alternative check: Newton’s second law: ΣF = m * a. With only a retarding force present and the body at rest, the net force does not favor motion initiation in the positive direction; hence a = 0 or negative relative to that direction. Why Other Options Are Wrong: (a) and (b) are true descriptions of a retarding force; it changes motion by reducing speed. (d) “none of these” is wrong because one option is correct. (e) Reversing direction would require additional dynamics; a retarding force alone cannot reverse unless other conditions change. Common Pitfalls: Confusing cause (force) with outcome under multiple forces; ignoring sign conventions for work and acceleration. Final Answer: introduce (initiate) the motion of a body at rest

Question 2

Units check – mass moment of inertia Choose the correct SI-style units for the mass moment of inertia of a rigid body about an axis.

Accepted Answer

Introduction / Context: Mass moment of inertia quantifies a body’s resistance to angular acceleration about a given axis. Correct units matter for validating formulas, checking finite element outputs, and converting catalog data for rotors, gears, and flywheels. Given Data / Assumptions: Using SI base units: mass in kg, length in m, time in s. Mass moment of inertia I relates to torque and angular acceleration via τ = I * α. We distinguish mass moment of inertia from area moment of inertia used in beam bending. Concept / Approach: From τ = I * α and dimensional analysis: [τ] = N·m = kg·m^2/s^2 and [α] = 1/s^2. Therefore [I] = [τ]/[α] = (kg·m^2/s^2) / (1/s^2) = kg·m^2. This matches the integral definition I = ∫ r^2 dm, where r has units of meters and dm has units of kilograms, producing kg·m^2. Step-by-Step Solution: Start with τ = I * α.Insert units: N·m = kg·m^2/s^2; α = 1/s^2.Compute: I units = (kg·m^2/s^2) / (1/s^2) = kg·m^2. Verification / Alternative check: Integral form: I = ∫ r^2 dm → units r^2 (m^2) times dm (kg) gives kg·m^2 directly. Why Other Options Are Wrong: (a) and (b) involve division by meters; incorrect. (c) m^4 is the area moment of inertia unit, not mass moment. (d) m^3 is a volume unit. Common Pitfalls: Confusing mass moment of inertia (kg·m^2) with second moment of area (m^4); mixing torque with energy units. Final Answer: kg·m^2

Question 3

Kinematics of mechanisms – total number of instantaneous centers For a mechanism (kinematic chain) having n links, what is the total number of instantaneous centers of rotation that can be identified (Kennedy’s theorem framework)?

Accepted Answer

Introduction / Context: Instantaneous centers (ICs) are pivotal in graphical kinematics to determine velocities without calculus. In a mechanism with multiple links, every pair of links has one IC, enabling systematic velocity mapping via Kennedy’s theorem of three centers in relative motion. Given Data / Assumptions: Planar mechanism with n rigid links connected by pairs. All pairs of links possess a unique instantaneous center. Links are numbered and considered in pairs. Concept / Approach: The count of ICs equals the number of unique unordered pairs of links, which is a combination count C(n, 2) = n(n − 1)/2. This is independent of the particular mechanism layout; location of ICs depends on geometry, but the count is purely combinatorial. Step-by-Step Solution: Let the set of links be {1, 2, …, n}.Each pair (i, j) with i < j defines one IC, giving C(n, 2) pairs.Therefore, total ICs = n(n − 1)/2. Verification / Alternative check: Check small cases: n = 2 → 1 IC; n = 3 → 3 ICs (a triangle); n = 4 → 6 ICs, matching known results for four-bar linkages. Why Other Options Are Wrong: (a) and (b) undercount; (d) n/2 is not combinatorial. (e) 2n − 3 is a formula seen for joints in trees, not for ICs. Common Pitfalls: Confusing number of joints with number of ICs; counting ordered pairs rather than unordered; forgetting that even non-adjacent links have an IC (possibly at infinity). Final Answer: n(n − 1)/2

Question 4

Definition check – what is energy? Select the most accurate definition used in engineering mechanics for “energy”.

Accepted Answer

Introduction / Context: Clear verbal definitions help prevent mix-ups between fundamental quantities: work, energy, and power. Energy underpins analyses of machines, structures, and thermodynamic systems, representing the budget from which mechanical, thermal, or electrical outputs can be produced. Given Data / Assumptions: Classical mechanics conventions for work, energy, and power. Scalar quantities unless otherwise stated (e.g., potential, kinetic energies). Steady units: Joule for work/energy; Watt for power. Concept / Approach: Energy is the capacity for doing work. Work is the energy transfer via a force acting through a distance. Power is the time rate of doing work. The accepted textbook phrasing distinguishes these three succinctly and avoids ambiguity in problem solving and unit checks. Step-by-Step Solution: Recall: Work W = ∫ F · ds, unit Joule.Energy E (e.g., kinetic 1/2 m v^2, potential m g h) represents the stored capacity to perform work, unit Joule.Power P = dW/dt, unit Watt = Joule/second.Therefore the correct definition is “capacity for doing work”. Verification / Alternative check: Dimensionally, energy and work share the same unit; power differs by per-time. Common engineering references adopt precisely this wording. Why Other Options Are Wrong: (a) “power of doing work” is vague and conflates with power. (c) and (e) explicitly define power, not energy. (d) cannot be correct because (a), (c), and (e) are not definitions of energy. Common Pitfalls: Using “power” and “energy” interchangeably; forgetting that energy is a state property whereas power is a rate. Final Answer: capacity for doing work

Question 5

Angle of friction from limiting horizontal push on a horizontal surface A 50 kgf block rests on a horizontal plane. A horizontal force of 18 kgf just causes impending motion. Determine the angle of friction φ (where tan φ = μ).

Accepted Answer

Introduction / Context: At impending motion on a horizontal plane, the frictional force reaches its limiting value μN. Converting the limiting push to the friction coefficient and then to the angle of friction is common in design of wedges, clamps, and handling equipment where stick–slip must be controlled. Given Data / Assumptions: Normal reaction N equals the weight W on a horizontal plane. Limiting horizontal force P at impending motion equals friction μN. Given W = 50 kgf, P = 18 kgf; use kgf consistently (force units). Concept / Approach: At the threshold of motion: P = μ * N. For a horizontal plane, N = W. Thus μ = P / W. The angle of friction φ satisfies tan φ = μ. Compute μ and take inverse tangent to obtain φ in degrees and minutes. Step-by-Step Solution: N = W = 50 kgf.μ = P / W = 18 / 50 = 0.36.φ = arctan(μ) = arctan(0.36) ≈ 19.798°.Convert decimal degrees to degrees–minutes: 0.798° * 60 ≈ 47.9′ ≈ 48′ → φ ≈ 19° 48′. Verification / Alternative check: Check by resolving along motion: Required horizontal push at limiting state equals μW. Using μ = tan φ ensures consistency with the angle interpretation from the friction cone. Why Other Options Are Wrong: Other listed angles correspond to μ values not equal to 0.36. For example, 17°48′ → μ ≈ 0.32; 21°48′ → μ ≈ 0.40, which do not match P/W. Common Pitfalls: Confusing kg (mass) with kgf (force); rounding too early; using sin or cos instead of tan for the friction angle. Final Answer: 19° 48′

Question 6

Distance of the centroid of the leftover lamina from the original center A circular plate of diameter 6 cm (R = 3 cm) has a circular disc removed whose diameter equals the original radius (3 cm, so r = 1.5 cm). Find the distance of the centroid of t…

Accepted Answer

Introduction / Context: Designers often remove material from plates to save weight while controlling balance. The centroid of the remaining area shifts toward the heavier side. Computing that shift ensures proper placement of supports, bearings, or counterweights. Given Data / Assumptions: Original circle radius R = 3 cm (area A1 = 9π). Removed circle radius r = 1.5 cm (area A2 = 2.25π). The small circle is cut at the rim, internally tangent to the outer boundary → center spacing d = R − r = 1.5 cm. Uniform thickness and density (area centroid problem). Concept / Approach: Use the negative-area method along the line of centers. The centroid shift x from the original center equals the first moment of removed area divided by the net remaining area. Step-by-Step Solution: Compute A1 − A2 = 9π − 2.25π = 6.75π.First moment of removed area about original center = A2 * d = 2.25π * 1.5 = 3.375π.Centroid shift x = (A2 * d) / (A1 − A2) = 3.375π / 6.75π = 0.5 cm.Direction: along the radius away from the cut-out toward the remaining material mass. Verification / Alternative check: By symmetry, the centroid must lie on the line of centers. Magnitude is less than 1.5 cm and proportional to area ratio A2/(A1 − A2), confirming 0.5 cm. Why Other Options Are Wrong: 2.0 and 1.5 cm exceed d; 1.0 cm overestimates; 0.25 cm underestimates given the removed area fraction 2.25/6.75 = 1/3. Common Pitfalls: Confusing diameter and radius; forgetting to subtract the removed area in the denominator; placing the small circle incorrectly. Final Answer: 0.5 cm

Question 7

How to double the time period of a simple pendulum For a simple pendulum operating at small amplitude, what change is required to double its period of oscillation?

Accepted Answer

Introduction / Context: The period of a simple pendulum guides timing devices and dynamic isolation systems. Understanding parameter sensitivity helps engineers tune response without trial-and-error changes in mass or structure. Given Data / Assumptions: Small-angle oscillations so that T = 2π * sqrt(l / g) applies. Mass of bob is concentrated; string is light and inextensible. Local gravity g is constant. Concept / Approach: Period T depends on length l and gravity g only. Mass does not appear in T. To change T by a factor, scale l accordingly using the square-root relationship. Step-by-Step Solution: T ∝ sqrt(l). To double T, set T_new / T_old = 2 = sqrt(l_new / l_old).Square both sides: 4 = l_new / l_old ⇒ l_new = 4 * l_old.Therefore, quadrupling the length doubles the period. Verification / Alternative check: Plug into T = 2π * sqrt(l / g). If l → 4l, then T → 2π * sqrt(4l/g) = 2 * 2π * sqrt(l/g) = 2T, confirming the requirement. Why Other Options Are Wrong: (a) and (b) change mass, which does not affect T in the simple model. (d) doubling length increases T by sqrt(2), not 2. (e) while halving g would increase T by sqrt(2); it does not double T unless g is reduced by a factor of 4, which is unrealistic. Common Pitfalls: Believing mass influences period; forgetting the square-root dependence; applying the formula outside small-angle limits. Final Answer: quadruple the length of the pendulum

Question 8

Simple harmonic motion (SHM) energy at the mean position: A particle executes simple harmonic motion. While passing through the mean (equilibrium) position, which form(s) of mechanical energy does the particle possess at a maximum or minimum level?

Accepted Answer

Introduction / Context: In simple harmonic motion (SHM), such as a mass–spring system or a small-angle pendulum, energy oscillates between kinetic and potential forms. Understanding how kinetic energy (KE) and potential energy (PE) vary with displacement and velocity is fundamental in vibration analysis and engineering dynamics. Given Data / Assumptions: Motion is ideal SHM with no damping. Mean position is the equilibrium point where displacement is zero. Total mechanical energy remains constant (no losses). Concept / Approach: For SHM with displacement x(t) and amplitude A, potential energy is proportional to displacement squared (for a spring, PE = 1/2 k x^2). Kinetic energy depends on velocity squared (KE = 1/2 m v^2). At the mean position, x = 0 → PE is minimum (zero for a spring), and velocity is maximum → KE is maximum. At extreme positions, velocity is zero → KE is zero and PE is maximum. Step-by-Step Solution: At mean position: x = 0 → PE = 1/2 k (0)^2 = 0 (minimum).Velocity magnitude is maximum since v = ω√(A^2 − x^2) → v = ωA.Therefore, KE = 1/2 m (ωA)^2 is maximum at the mean position. Verification / Alternative check: Energy conservation: total energy E = 1/2 k A^2 is constant. At x = 0, all of E is kinetic; at x = ±A, all of E is potential—consistent with the conclusion. Why Other Options Are Wrong: (b) PE cannot be maximum at the mean position; (c) KE is not minimum at x = 0; (d) both minimum is impossible because total energy is constant and nonzero for A ≠ 0; (e) is incorrect because one option exactly describes the SHM energy distribution. Common Pitfalls: Confusing displacement and velocity extrema; assuming KE and PE are equal at the mean position (they are equal only at intermediate displacements, not at x = 0). Final Answer: maximum kinetic energy and minimum potential energy

Question 9

Compound pendulum — small-angle SHM relation for frequency: If a compound pendulum undergoes small oscillations with angular displacement θ and angular acceleration α proportional to θ (i.e., α = −(g/ℓ) θ), what is the frequency f of its motion expr…

Accepted Answer

Introduction / Context: The compound pendulum (physical pendulum) exhibits simple harmonic motion for small angular displacements. The relationship between the restoring angular acceleration and angular displacement determines the natural frequency. Expressing frequency directly in terms of the proportionality α/θ reinforces the SHM analogy to linear systems. Given Data / Assumptions: Small-angle oscillations so that sin θ ≈ θ (in radians). Angular acceleration α is proportional to −θ: α = −(g/ℓ) θ. g is acceleration due to gravity; ℓ is the equivalent pendulum length. Concept / Approach: For SHM, the governing rotational equation is θ¨ + (g/ℓ) θ = 0. Angular frequency ω satisfies ω^2 = g/ℓ. Because α/θ = −g/ℓ (magnitude g/ℓ), we can write ω = sqrt(α/θ) using magnitudes. The cyclic frequency f relates to angular frequency by f = ω / (2π). Step-by-Step Solution: Start with α = θ¨ = −(g/ℓ) θ → |α/θ| = g/ℓ.Hence ω^2 = g/ℓ = |α/θ|.Therefore ω = sqrt(α/θ) (using positive magnitudes).Frequency f = ω / (2π) = (1 / 2π) * sqrt(α / θ). Verification / Alternative check: From standard formula for a compound pendulum, f = (1 / 2π) * sqrt(g / ℓ). Substituting g/ℓ = α/θ reproduces the same expression. Why Other Options Are Wrong: (b) and (c) have incorrect dimensions (frequency cannot be linear in α/θ without a square root). (d) is not dimensionally consistent. (e) inverts the needed relationship and lacks the square root. Common Pitfalls: Forgetting the small-angle assumption; ignoring sign (use magnitudes for frequency); confusing angular frequency ω with cyclic frequency f. Final Answer: (1 / 2π) * sqrt(α / θ)

Question 10

Uniform circular motion — direction of acceleration: A particle moves along a circle with uniform speed. At any instant, the direction of its acceleration vector is which of the following?

Accepted Answer

Introduction / Context: Uniform circular motion (UCM) features constant speed but continuously changing velocity direction. This change in direction implies a nonzero acceleration known as centripetal acceleration. Recognizing its direction is vital in dynamics of rotating machinery, vehicle cornering, and orbital motion. Given Data / Assumptions: Particle moves with constant speed on a circular path of radius r. Angular speed may be denoted ω; linear speed v = ω r. No tangential acceleration component (speed constant). Concept / Approach: The acceleration responsible for changing the velocity direction points toward the centre of curvature. Its magnitude is a_c = v^2 / r = ω^2 r. Because speed is constant, any tangential component would imply a change in speed, which is absent in UCM; hence acceleration must be purely radial and inward. Step-by-Step Solution: Identify that only direction changes, not speed → tangential acceleration = 0.Use a_c = v^2 / r directed toward the centre (centripetal).Therefore, the acceleration vector is radially inward at all times. Verification / Alternative check: Differentiating the velocity vector in polar coordinates shows a radial component −ω^2 r pointing toward the centre, with zero tangential component for constant ω. Why Other Options Are Wrong: (a) and (c) incorrectly suggest outward direction; (b) implies a tangential component which would change speed; (e) is arbitrary and not supported by UCM kinematics. Common Pitfalls: Confusing centripetal (inward) with centrifugal (a pseudo force in a rotating frame) which appears outward; assuming any tangential component exists when speed is constant. Final Answer: towards the centre (radially inward)

Question 11

Apparent weight in a moving lift (elevator): A body that truly weighs 14 g (measured at rest) appears to weigh 13 g when measured by a spring balance in a moving lift. Determine the magnitude of the lift’s acceleration at that instant (state the clo…

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Introduction / Context: Apparent weight changes in an accelerating frame such as a lift (elevator). A spring balance reads the normal reaction on the body, which differs from the true weight when the frame accelerates. This classic dynamics problem reinforces Newton’s second law in non-inertial scenarios treated from an inertial frame. Given Data / Assumptions: True weight corresponds to mass m such that W = m g; the balance reads 14 g at rest. Apparent weight during motion is equivalent to 13 g on the same balance. Take g ≈ 9.8 m/s^2 (ratios will cancel). Motion is vertical; friction and air resistance neglected. Concept / Approach: When the lift accelerates downward with acceleration a, the apparent weight N satisfies N = m (g − a). When it accelerates upward, N = m (g + a). A reading lower than the true weight indicates downward acceleration (or upward acceleration with less likely opposing sign choice). Use the ratio of apparent to true weight to solve for a/g. Step-by-Step Solution: True weight W_true = m g corresponds to “14 g units” on the balance.Apparent weight W_app = m (g − a) corresponds to “13 g units”.Form the ratio: W_app / W_true = (g − a)/g = 13/14.Solve for a: g − a = (13/14) g → a = g (1 − 13/14) = g / 14.Compute: a ≈ 9.8 / 14 ≈ 0.7 m/s^2 (downward). Verification / Alternative check: If the lift were accelerating upward, apparent weight would exceed true weight (contrary to the observation). Thus the direction is downward with magnitude about 0.7 m/s^2, consistent with the sign logic. Why Other Options Are Wrong: 0.5 and 1.0 m/s^2 do not satisfy the precise ratio 13/14; 0.01 m/s^2 is negligible; 1.5 m/s^2 is too large for the given readings. Common Pitfalls: Confusing mass units with force units; using W = m a incorrectly; forgetting that a reduced reading implies downward acceleration. Final Answer: 0.7 m/s^2

Question 12

Kinematics terminology — precise definition: The rate of change of displacement of a body with respect to its surroundings is known as which of the following quantities?

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Introduction / Context: Clear terminology in kinematics avoids design and analysis mistakes. Displacement, velocity, speed, and acceleration each have specific meanings used in motion studies for mechanisms, vehicles, and structures subjected to dynamic loads. Given Data / Assumptions: Displacement is a vector (magnitude and direction). Time is continuous; differentiability assumed. Reference to “with respect to its surroundings” indicates measurement in a chosen inertial frame. Concept / Approach: Velocity is defined as the time rate of change of displacement: v = ds/dt (vector quantity). Speed is the magnitude of velocity and equals the time rate of change of distance traveled, which is scalar. Acceleration is the rate of change of velocity: a = dv/dt. Jerk is the rate of change of acceleration: j = da/dt. Step-by-Step Solution: Identify the quantity changing: displacement (vector).Apply definition: velocity = time derivative of displacement.Therefore, the precise term is velocity. Verification / Alternative check: In component form, if displacement s(t) has components x(t), y(t), z(t), then velocity vector v(t) = (dx/dt, dy/dt, dz/dt). This matches the definition requested. Why Other Options Are Wrong: Acceleration (b) differentiates velocity, not displacement directly; speed (c) discards direction; “none of these” (d) is false; jerk (e) is a higher derivative not asked for. Common Pitfalls: Interchanging speed and velocity; ignoring vector nature leading to sign/direction errors in analyses. Final Answer: velocity

Question 13

Oscillation frequency on the Moon versus on the Earth: For identical small-oscillation systems (e.g., a simple pendulum), how does the frequency of oscillation on the Moon compare with that on the Earth?

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Introduction / Context: Natural frequency of many gravity-controlled oscillators (like a simple pendulum) depends on the local gravitational acceleration g. Comparing environments with different g values, such as Earth and Moon, illustrates how gravity influences timing devices and dynamic response. Given Data / Assumptions: For a simple pendulum of length L, small-angle frequency f = (1 / 2π) * sqrt(g / L). Approximate lunar gravity g_moon ≈ g_earth / 6. Length L is unchanged; small-angle approximation applies. Concept / Approach: Frequency scales with the square root of gravity: f ∝ sqrt(g). Therefore, the ratio f_moon / f_earth = sqrt(g_moon / g_earth) ≈ sqrt(1/6) ≈ 0.408. Thus, the lunar frequency is about 0.41 times the terrestrial value, meaning it is roughly 2.44 times less (the period is 2.44 times longer). Step-by-Step Solution: Compute ratio: f_moon / f_earth = sqrt( (g/6) / g ) = sqrt(1/6).Evaluate: sqrt(1/6) ≈ 0.408.Hence frequency on Moon ≈ 0.41 of that on Earth → “2.44 times less”. Verification / Alternative check: Period scales inversely with frequency: T = 1/f. Therefore, T_moon / T_earth = 1 / 0.408 ≈ 2.45, matching the “2.44 times less” frequency statement. Why Other Options Are Wrong: (a) and (d) suggest frequency increases on the Moon, which contradicts f ∝ sqrt(g); (c) “3 times less” is an overestimate; (e) ignores the large difference in g. Common Pitfalls: Confusing frequency and period; using a linear instead of square-root dependence on g; rounding errors that change the qualitative conclusion. Final Answer: 2.44 times less

Question 14

Pure rotation about a fixed axis: When a rigid body moves about a fixed axis (with no translation of the axis), which description best fits its motion?

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Introduction / Context: Rigid body kinematics distinguishes between translation, rotation, and general plane motion. Many machine elements (gears, pulleys, flywheels) rotate about a fixed axis, which is a special and very common case in engineering mechanics. Given Data / Assumptions: Axis is fixed in position and orientation in an inertial frame. No motion of the axis itself; points on the body trace circular paths about this axis. Body is rigid (distances between material points remain constant). Concept / Approach: Pure rotation occurs when every point in the body moves in a circle whose centre lies on the fixed axis, with angular position described by a single generalized coordinate (angle). If the axis is fixed, the body has no translational motion of its centroid; translation would imply the axis itself changes position or the body slides bodily, which is not the case. Step-by-Step Solution: Identify the constraint: fixed axis → all points move on circles centred on the axis.Rotation fully characterizes motion: angle θ(t), angular velocity ω(t), angular acceleration α(t).Therefore, the correct description is rotary motion only. Verification / Alternative check: Velocity of a point r from the axis is v = ω × r; acceleration has tangential and normal parts α × r and ω × (ω × r). No net translation of the axis exists. Why Other Options Are Wrong: (b) “a circular motion” describes a point, not the body as a whole; (c) translatory motion contradicts a fixed axis; (d) combines both, which is general plane motion (not applicable here); (e) rectilinear motion is incompatible with circles about an axis. Common Pitfalls: Describing motion of one point (circular) instead of the entire body (rotary); assuming translation of the centroid when the axis is fixed. Final Answer: a rotary motion

Question 15

Outcomes for a body acted upon by coplanar, non-concurrent forces In elementary engineering mechanics, if several forces lie in one plane but their lines of action do not meet at a single point (non-concurrent), what motions or states can the body e…

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Introduction / Context: Coplanar non-concurrent force systems are common in structures and machines. Because the forces do not intersect at a single point, their combined effect can be a pure force (resultant through the center of mass), a pure couple (net moment only), a general force-couple system, or zero (equilibrium). Understanding all possible outcomes is essential for analyzing beams, frames, and rigid bodies. Given Data / Assumptions: Forces lie in a single plane (coplanar). Lines of action are not concurrent. Rigid body assumption; small deformations are neglected. No frictional complications unless implied by the net effect. Concept / Approach: Any system of forces on a rigid body can be reduced to an equivalent resultant force R acting at a chosen reference point, plus a resultant moment M at that point. Special cases arise: R = 0 with M ≠ 0 gives a pure couple (tendency to rotate); M = 0 with R ≠ 0 gives pure translation; both nonzero give general plane motion; both zero give equilibrium. Step-by-Step Solution: Compute vector sum of forces: R = ΣF.Compute resultant moment about a reference point O: M = Σ(r × F).Classify: if R = 0, M ≠ 0 ⇒ pure rotation; if M = 0, R ≠ 0 ⇒ pure translation; if both nonzero ⇒ translation + rotation; if both zero ⇒ rest. Verification / Alternative check: Replace the system by a single force at the center of mass and a couple. Kinematics of rigid bodies confirms the motion classes correspond exactly to the force–moment combinations above. Why Other Options Are Wrong: Each individual option lists only one possible outcome; in reality, all listed outcomes are possible depending on ΣF and ΣM. Hence only “All of the above” is comprehensive and correct. Common Pitfalls: Assuming non-concurrency always causes rotation; forgetting that geometry and magnitudes can cancel to yield equilibrium; ignoring that a pure couple can exist with zero net force. Final Answer: All of the above

Question 16

Definition of a geostationary satellite A geostationary (geosynchronous equatorial) satellite appears fixed over one point on Earth. It orbits with the same angular velocity as which celestial body?

Accepted Answer

Introduction / Context: Communication and weather satellites are often placed in geostationary orbit so they appear fixed above a point on the equator. This condition is achieved only if the satellite’s orbital period matches Earth’s rotation and the orbit is equatorial and circular. Given Data / Assumptions: Satellite is over the equator (zero inclination). Orbit is circular and prograde. Neglect perturbations (idealized definition). Concept / Approach: A geostationary satellite must have the same angular velocity as Earth’s spin so that its orbital period equals 1 sidereal day. When this happens, the satellite’s subpoint on Earth does not move in longitude, and it appears stationary to ground observers. Step-by-Step Solution: Condition for geostationary: ω_sat = ω_earth.Period T_sat = T_earth ≈ 23 h 56 min (sidereal day).Orbit radius adjusts so that gravitational centripetal requirement gives the above period. Verification / Alternative check: If the satellite had the Moon’s angular velocity or any other, it would drift in longitude. Only matching Earth’s rotational angular speed keeps it fixed in the sky. Why Other Options Are Wrong: Moon and Sun have very different angular velocities; a “pole” does not define a rotational rate; therefore only Earth’s own rotation is relevant. Common Pitfalls: Confusing “geosynchronous” with “geostationary”; the former only matches period, while the latter also requires zero inclination and zero eccentricity to remain fixed over one point. Final Answer: The Earth (about its own axis)

Question 17

Kinematics from displacement function A particle moves on a straight line with displacement S(t) = 4 t^3 + 3 t^2 − 10 (S in m, t in s). Starting from rest at t = 0, how long does it take for its velocity to reach 18 m/s?

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Introduction / Context: This problem exercises basic calculus in kinematics: velocity is the time derivative of displacement. Given S(t) explicitly, we compute v(t), apply the initial condition “from rest”, and solve for the time when v reaches a specified value. Given Data / Assumptions: S(t) = 4 t^3 + 3 t^2 − 10 (meters). Velocity v(t) = dS/dt (m/s). Starts from rest at t = 0 (v = 0). Find t when v = 18 m/s, with t ≥ 0. Concept / Approach: Differentiation gives the velocity function. We then set v = 18 and solve the resulting quadratic. Only the physically meaningful positive root is accepted because time is measured forward from the instant the particle is at rest (t = 0). Step-by-Step Solution: Compute velocity: v(t) = dS/dt = 12 t^2 + 6 t.Check initial rest: v(0) = 0 (satisfied).Set target velocity: 12 t^2 + 6 t = 18.Divide by 6: 2 t^2 + t = 3.Rearrange: 2 t^2 + t − 3 = 0.Solve quadratic: t = [−1 ± √(1 + 24)] / 4 = (−1 ± 5) / 4.Physical root: t = 1 s (discard t = −1.5 s). Verification / Alternative check: Substitute t = 1 s into v(t): 12(1)^2 + 6(1) = 18 m/s, confirming the result. Why Other Options Are Wrong: 0.5 s, 1.5 s, 2.0 s give velocities not equal to 18 m/s; “Not attainable” is incorrect because a valid positive solution exists. Common Pitfalls: Forgetting to differentiate S(t); using the negative root; arithmetic errors when solving the quadratic. Final Answer: 1.0 s

Question 18

Identify the incorrect statement for simple harmonic motion (SHM) Which of the following statements about SHM is incorrect?

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Introduction / Context: In simple harmonic motion (SHM), displacement x, velocity v, and acceleration a have well-defined phase relations. Knowing where these kinematic quantities are maximum or zero is essential in vibrations and oscillations analysis. Given Data / Assumptions: Standard SHM with a = −ω^2 x. Motion is about a mean (equilibrium) position. No damping or external forcing. Concept / Approach: From a = −ω^2 x, acceleration magnitude is proportional to |x|. Thus |a| is maximum at extreme displacements and zero at the mean position. Velocity is the time derivative of displacement and reaches its maximum magnitude at the mean position, becoming zero at the ends of stroke. Step-by-Step Solution: Model: a = −ω^2 x; therefore |a| ∝ |x|.At x = 0 (mean), a = 0; |v| is maximum.At extremes (|x| = A), v = 0; |a| = ω^2 A (maximum).Hence the statement “acceleration is minimum at the ends” is false; it is actually maximum there. Verification / Alternative check: Using energy: total energy E = (1/2) k A^2. At ends, all energy is potential ⇒ v = 0; restoring force kx is maximum ⇒ acceleration is maximum, consistent with the differential equation. Why Other Options Are Wrong: (a), (b), and (d) are correct properties of SHM. (e) is also correct since a = −ω^2 x indicates opposite signs of a and x. Common Pitfalls: Confusing velocity and acceleration maxima; thinking acceleration is tied to speed rather than displacement in SHM. Final Answer: Acceleration is minimum at the ends of stroke

Question 19

Direction of centrifugal (inertial) force in circular motion For a mass m moving in a circular path of radius r with speed v (or angular speed ω), the centrifugal force acts in which direction?

Accepted Answer

Introduction / Context: In circular motion analysis, two viewpoints exist: an inertial (laboratory) frame and a rotating (body-attached) frame. The real required force is centripetal (toward the center). In the rotating frame, an equal and opposite inertial (fictitious) force called centrifugal appears to act on the mass. Given Data / Assumptions: Uniform circular motion with radius r and speed v (or angular speed ω). Rotating-frame interpretation for centrifugal force. Rigid, frictionless idealization where applicable. Concept / Approach: Centripetal requirement: a_c = v^2 / r = ω^2 r directed toward the center. In a rotating frame, this is balanced by the inertial (centrifugal) force of magnitude m v^2 / r = m ω^2 r directed radially outward, away from the center. Step-by-Step Solution: Centripetal acceleration: a_c = v^2 / r (toward center).Real force that provides it (e.g., tension, normal reaction) must point inward.In the rotating frame, introduce centrifugal force F_cf = m v^2 / r = m ω^2 r, directed outward. Verification / Alternative check: Free-body diagram in rotating frame: inward real forces are balanced by outward centrifugal force, giving apparent equilibrium for steady rotation. Why Other Options Are Wrong: (a) Tangential direction corresponds to tangential acceleration when speed changes. (b) Describes centripetal, not centrifugal. (d) and (e) are incorrect for nonzero circular motion. Common Pitfalls: Confusing centripetal with centrifugal; forgetting that centrifugal is a fictitious force used only in non-inertial frames. Final Answer: Away from the center of rotation

Question 20

Coefficient of restitution from bounce heights A ball drops from 2.25 m on a smooth floor and rebounds to a height of 1.00 m. What is the coefficient of restitution e between the ball and the floor?

Accepted Answer

Introduction / Context: The coefficient of restitution e quantifies collision elasticity along the line of impact. For vertical bounces under gravity with negligible air resistance, e can be found from pre- and post-impact speeds, which relate to drop and rebound heights via energy considerations. Given Data / Assumptions: Drop height h1 = 2.25 m. Rebound height h2 = 1.00 m. Neglect air resistance; same impact point; smooth floor. Concept / Approach: Before impact, speed v1 just before hitting the floor satisfies v1^2 = 2 g h1. After impact, rebound speed v2 satisfies v2^2 = 2 g h2. By definition, e = v2 / v1 for a vertical bounce on a massive floor. Step-by-Step Solution: v1 = √(2 g h1).v2 = √(2 g h2).e = v2 / v1 = √(h2 / h1) = √(1.00 / 2.25).Compute: h2/h1 = 0.444..., √(0.444...) ≈ 0.6667 ≈ 0.67. Verification / Alternative check: Energy ratios for ideal vertical motion lead to the same relation e^2 = h2/h1, confirming the calculation. Why Other Options Are Wrong: 0.25, 0.33, 0.50, and 0.75 do not match √(1/2.25). Only 0.67 (rounded) is consistent with the measured heights. Common Pitfalls: Using height ratio directly instead of its square root; mixing up drop and rebound heights; rounding too early. Final Answer: 0.67

Applied Mechanics Questions