Q: Critical damping coefficient — a 1 kg mass is attached to a spring of stiffness 0.7 N/mm. What is the critical damping coefficient c c for this system?

Introduction / Context: The critical damping coefficient defines the threshold between oscillatory (underdamped) and non-oscillatory (overdamped) responses in a mass–spring–damper system. It is a key design target when fast return without overshoot is desired (e.g., measuring instruments). Given Data / Assumptions: Mass m = 1 kg. Stiffness k = 0.7 N/mm = 700 N/m. Viscous damping model; find cc. Concept / Approach: For a single-DOF system, cc=2√(km). Substituting k and m gives the required value. Units must be consistent (N·s/m). Step-by-Step Solution: 1) Convert stiffness: 0.7 N/mm = 0.7×1000 = 700 N/m.2) Compute √(k m) = √(700×1) = √700 ≈ 26.4575.3) Compute cc = 2×26.4575 ≈ 52.915 ≈ 52.92 N·s/m. Verification / Alternative check: Natural frequency ωn=√(k/m)=√700≈26.46 rad/s. Since cc=2mωn=2×1×26.46≈52.92 N·s/m, which matches. Why Other Options Are Wrong: 1.4 N·s/m — far below critical; implies very light damping. 18.52 N·s/m — still underdamped relative to cc. 529.2 N·s/m — ten times the correct value; overdamped by a wide margin. Common Pitfalls: Forgetting to convert N/mm to N/m; using c=2k instead of c=2√(km). Final Answer: 52.92 N-s/m.

Q: Static balancing — a disturbing mass m₁ on a rotating shaft can be balanced by a single mass m₂ in the same plane at radius r₂ if which relation holds (r₁ and r₂ are radii of rotation)?

Introduction / Context: Static (force) balancing in a single plane requires the vector sum of centrifugal forces to be zero. When only one disturbing mass exists, a single balancing mass placed appropriately at the same plane can cancel it. Given Data / Assumptions: Disturbing mass m₁ at radius r₁. Balancing mass m₂ at radius r₂ in the same plane. Both rotate at the same angular speed ω. Concept / Approach: Centrifugal force F = m r ω². For zero net force, m₁ r₁ ω² and m₂ r₂ ω² must be equal in magnitude and 180° out of phase. Hence m₁ r₁ = m₂ r₂. Step-by-Step Solution: 1) Write F₁=m₁r₁ω² and F₂=m₂r₂ω².2) Set F₁=F₂ for balance ⇒ m₁r₁ω²=m₂r₂ω².3) Cancel ω² (same plane and speed): m₁r₁=m₂r₂. Verification / Alternative check: Phasor diagram with two opposite vectors of equal length confirms zero resultant. Why Other Options Are Wrong: m₁r₂=m₂r₁ — inverted; does not follow from force equality. m₁m₂=r₁r₂ — dimensionally inconsistent for balancing. none of these — incorrect because a correct relation exists. Common Pitfalls: Forgetting the 180° angular placement; mixing up radii in the proportion. Final Answer: m1r1 = m2r2.

Q: Primary unbalanced force — the primary inertia force of a reciprocating mass attains its maximum value when the crank angle with the line of stroke is at which positions?

Introduction / Context: The primary unbalanced force in a reciprocating mechanism varies sinusoidally with crank angle. Recognizing where it peaks is important for balancing strategies and bearing load estimation. Given Data / Assumptions: Primary force Fp=m r ω² cos θ (first harmonic). θ is the angle between crank and line of stroke. Constant speed ω. Concept / Approach: The magnitude of Fp is maximum when |cos θ|=1 and minimum when cos θ=0. Therefore, maxima occur at θ=0° and 180° (along the line of stroke), not at quadrature. Step-by-Step Solution: 1) Write Fp=m r ω² cos θ.2) Maximize |cos θ| → |cos θ|=1 at θ=0° and 180°.3) Conclude peak primary shake at inner/outer dead centers. Verification / Alternative check: A plot of Fp against θ shows peaks at 0° and 180° and zeros at 90° and 270°. Why Other Options Are Wrong: 0° and 90° — includes one maximum and one zero. 90° and 180° — includes a zero and a maximum. 180° and 360° — 360°≡0°, duplicates same position but omits the other distinct maximum at 180°. Common Pitfalls: Mixing primary and secondary components; forgetting that the cosine term peaks on the line of stroke. Final Answer: 0° and 180°.

Q: Whirling of shafts — a disc fixed at mid-span has CG eccentricity e = 2 mm relative to the shaft axis. The lateral natural frequency is 10 rad/s. If the shaft rotates at 300 r.p.m., the steady whirl radius will be approximately:

Introduction / Context: An unbalanced rotor (disc with eccentric center of gravity) mounted on a flexible shaft will whirl at a radius that depends on running speed relative to the lateral natural frequency. This is the classical Jeffcott (Laval) rotor model. Given Data / Assumptions: Eccentricity e = 2 mm = 0.002 m. Undamped lateral natural frequency ωn = 10 rad/s. Running speed 300 r.p.m. ⇒ ω = 2π×300/60 = 10π ≈ 31.416 rad/s. Neglect damping for the amplitude estimate. Concept / Approach: For an undamped Jeffcott rotor, steady whirl amplitude X equals X = e·ω²/|ωn² − ω²|. Since ω > ωn here, use X = e·ω²/(ω² − ωn²). Step-by-Step Solution: 1) Compute ω = 31.416 rad/s; ω² ≈ 985.96.2) Compute ω² − ωn² ≈ 985.96 − 100 = 885.96.3) Evaluate X = 0.002 × 985.96 / 885.96 ≈ 0.002225 m = 2.225 mm.4) Round suitably → 2.22 mm. Verification / Alternative check: Limiting cases: as ω→ωn, amplitude grows large (resonance); as ω ≫ ωn, X→e (approaches the geometric eccentricity), which aligns with the computed value being slightly above 2 mm. Why Other Options Are Wrong: 2 mm — would be the asymptote as ω ≫ ωn, but ω is only ~3.14× ωn. 2.50 mm and 3.0 mm — overestimate the dynamic amplification at this speed ratio. Common Pitfalls: Using (ωn² − ω²) in the denominator for ω > ωn; forgetting to convert r.p.m. to rad/s. Final Answer: 2.22 mm.

Question 1

Multi-collar thrust bearing — the frictional torque transmitted by a multi-collared shaft compared with a single-collared shaft (for the same total axial load and mean radius) is:

Accepted Answer

Introduction / Context: Thrust bearings transmit axial loads through collar faces. Designers sometimes split a single collar into multiple collars to reduce pressure and heat. A common exam point asks whether total frictional torque changes under the same overall axial load. Given Data / Assumptions: Total axial load on the shaft = W (fixed). Uniform pressure or uniform wear theory may be used for mean radius Rm. Coulomb friction with coefficient μ, and n identical collars share load. Concept / Approach: For one collar, torque T = μ W Rm. For n collars sharing the same total load W, each collar carries W/n. The torque per collar is μ (W/n) Rm, and the total torque becomes n × μ (W/n) Rm = μ W Rm — the same as the single-collar case, provided Rm is unchanged. Step-by-Step Solution: 1) Single collar: T₁ = μ W Rm.2) n collars: each takes W/n ⇒ torque per collar Ti= μ (W/n) Rm.3) Sum torques: ΣTi= n×μ (W/n) Rm= μ W Rm. Verification / Alternative check: If instead each collar saw the full W (not load-sharing), torque would scale with n, but that is not how multi-collar bearings are designed. Why Other Options Are Wrong: Wrong — suggests torque changes despite equal total load and mean radius, which contradicts the load-sharing derivation. Common Pitfalls: Forgetting that the stated W is the total axial load; confusing uniform pressure vs wear (which affects Rm slightly but not the equality under equal Rm). Final Answer: Right.

Question 2

Underdamped vibrations — in a single-degree-of-freedom viscously damped system, how does the amplitude of free vibration vary with time when the system is underdamped?

Accepted Answer

Introduction / Context: Real mechanical systems usually possess some damping. When damping is light (underdamped), oscillations persist but their envelope decays. Recognizing the correct decay law is essential for estimating settling time and vibration severity. Given Data / Assumptions: Single mass–spring–damper with damping ratio 0 < ζ < 1. No external forcing (free vibration). Undamped natural frequency ωn=√(k/m). Concept / Approach: The displacement solution is x(t)=X0e−ζωntcos(ωdt+φ), where ωd=ωn√(1−ζ²). The amplitude envelope X(t)=X0e−ζωnt decays exponentially with time. Step-by-Step Solution: 1) Write characteristic equation for mẍ + cẋ + kx=0.2) For 0<ζ<1, complex roots yield oscillatory motion.3) Envelope is exponential: e−ζωnt → amplitude decreases exponentially. Verification / Alternative check: Log-decrement δ=ln(xi/xi+1)=2πζ/√(1−ζ²) confirms an exponential decay trend across cycles. Why Other Options Are Wrong: Linear increase/decrease — decay is not linear in time. Increases exponentially — contradicts dissipative energy loss. Common Pitfalls: Confusing forced resonance amplitude growth with free-decay behavior; mixing underdamped decay with critically damped non-oscillatory return. Final Answer: decreases exponentially with time.

Question 3

Critical damping coefficient — a 1 kg mass is attached to a spring of stiffness 0.7 N/mm. What is the critical damping coefficient c_c for this system?

Accepted Answer

Introduction / Context: The critical damping coefficient defines the threshold between oscillatory (underdamped) and non-oscillatory (overdamped) responses in a mass–spring–damper system. It is a key design target when fast return without overshoot is desired (e.g., measuring instruments). Given Data / Assumptions: Mass m = 1 kg. Stiffness k = 0.7 N/mm = 700 N/m. Viscous damping model; find cc. Concept / Approach: For a single-DOF system, cc=2√(km). Substituting k and m gives the required value. Units must be consistent (N·s/m). Step-by-Step Solution: 1) Convert stiffness: 0.7 N/mm = 0.7×1000 = 700 N/m.2) Compute √(k m) = √(700×1) = √700 ≈ 26.4575.3) Compute cc = 2×26.4575 ≈ 52.915 ≈ 52.92 N·s/m. Verification / Alternative check: Natural frequency ωn=√(k/m)=√700≈26.46 rad/s. Since cc=2mωn=2×1×26.46≈52.92 N·s/m, which matches. Why Other Options Are Wrong: 1.4 N·s/m — far below critical; implies very light damping. 18.52 N·s/m — still underdamped relative to cc. 529.2 N·s/m — ten times the correct value; overdamped by a wide margin. Common Pitfalls: Forgetting to convert N/mm to N/m; using c=2k instead of c=2√(km). Final Answer: 52.92 N-s/m.

Question 4

Static balancing — a disturbing mass m₁ on a rotating shaft can be balanced by a single mass m₂ in the same plane at radius r₂ if which relation holds (r₁ and r₂ are radii of rotation)?

Accepted Answer

Introduction / Context: Static (force) balancing in a single plane requires the vector sum of centrifugal forces to be zero. When only one disturbing mass exists, a single balancing mass placed appropriately at the same plane can cancel it. Given Data / Assumptions: Disturbing mass m₁ at radius r₁. Balancing mass m₂ at radius r₂ in the same plane. Both rotate at the same angular speed ω. Concept / Approach: Centrifugal force F = m r ω². For zero net force, m₁ r₁ ω² and m₂ r₂ ω² must be equal in magnitude and 180° out of phase. Hence m₁ r₁ = m₂ r₂. Step-by-Step Solution: 1) Write F₁=m₁r₁ω² and F₂=m₂r₂ω².2) Set F₁=F₂ for balance ⇒ m₁r₁ω²=m₂r₂ω².3) Cancel ω² (same plane and speed): m₁r₁=m₂r₂. Verification / Alternative check: Phasor diagram with two opposite vectors of equal length confirms zero resultant. Why Other Options Are Wrong: m₁r₂=m₂r₁ — inverted; does not follow from force equality. m₁m₂=r₁r₂ — dimensionally inconsistent for balancing. none of these — incorrect because a correct relation exists. Common Pitfalls: Forgetting the 180° angular placement; mixing up radii in the proportion. Final Answer: m1r1 = m2r2.

Question 5

Primary unbalanced force — the primary inertia force of a reciprocating mass attains its maximum value when the crank angle with the line of stroke is at which positions?

Accepted Answer

Introduction / Context: The primary unbalanced force in a reciprocating mechanism varies sinusoidally with crank angle. Recognizing where it peaks is important for balancing strategies and bearing load estimation. Given Data / Assumptions: Primary force Fp=m r ω² cos θ (first harmonic). θ is the angle between crank and line of stroke. Constant speed ω. Concept / Approach: The magnitude of Fp is maximum when |cos θ|=1 and minimum when cos θ=0. Therefore, maxima occur at θ=0° and 180° (along the line of stroke), not at quadrature. Step-by-Step Solution: 1) Write Fp=m r ω² cos θ.2) Maximize |cos θ| → |cos θ|=1 at θ=0° and 180°.3) Conclude peak primary shake at inner/outer dead centers. Verification / Alternative check: A plot of Fp against θ shows peaks at 0° and 180° and zeros at 90° and 270°. Why Other Options Are Wrong: 0° and 90° — includes one maximum and one zero. 90° and 180° — includes a zero and a maximum. 180° and 360° — 360°≡0°, duplicates same position but omits the other distinct maximum at 180°. Common Pitfalls: Mixing primary and secondary components; forgetting that the cosine term peaks on the line of stroke. Final Answer: 0° and 180°.

Question 6

Whirling of shafts — a disc fixed at mid-span has CG eccentricity e = 2 mm relative to the shaft axis. The lateral natural frequency is 10 rad/s. If the shaft rotates at 300 r.p.m., the steady whirl radius will be approximately:

Accepted Answer

Introduction / Context: An unbalanced rotor (disc with eccentric center of gravity) mounted on a flexible shaft will whirl at a radius that depends on running speed relative to the lateral natural frequency. This is the classical Jeffcott (Laval) rotor model. Given Data / Assumptions: Eccentricity e = 2 mm = 0.002 m. Undamped lateral natural frequency ωn = 10 rad/s. Running speed 300 r.p.m. ⇒ ω = 2π×300/60 = 10π ≈ 31.416 rad/s. Neglect damping for the amplitude estimate. Concept / Approach: For an undamped Jeffcott rotor, steady whirl amplitude X equals X = e·ω²/|ωn² − ω²|. Since ω > ωn here, use X = e·ω²/(ω² − ωn²). Step-by-Step Solution: 1) Compute ω = 31.416 rad/s; ω² ≈ 985.96.2) Compute ω² − ωn² ≈ 985.96 − 100 = 885.96.3) Evaluate X = 0.002 × 985.96 / 885.96 ≈ 0.002225 m = 2.225 mm.4) Round suitably → 2.22 mm. Verification / Alternative check: Limiting cases: as ω→ωn, amplitude grows large (resonance); as ω ≫ ωn, X→e (approaches the geometric eccentricity), which aligns with the computed value being slightly above 2 mm. Why Other Options Are Wrong: 2 mm — would be the asymptote as ω ≫ ωn, but ω is only ~3.14× ωn. 2.50 mm and 3.0 mm — overestimate the dynamic amplification at this speed ratio. Common Pitfalls: Using (ωn² − ω²) in the denominator for ω > ωn; forgetting to convert r.p.m. to rad/s. Final Answer: 2.22 mm.

Question 7

Cam terminology — the pressure angle of a cam–follower system is the angle between the direction of follower motion and a normal drawn to which reference curve?

Accepted Answer

Introduction / Context: Pressure angle is a crucial parameter in cam design because it controls side thrust and potential jamming in the follower guide. Designers aim to limit it for smooth motion and lower contact stresses. Given Data / Assumptions: Translating or oscillating follower with a defined motion law. Pitch curve is the trajectory of the follower center (roller center) relative to the cam. Normal is taken at the point of contact along this curve. Concept / Approach: By definition, the pressure angle is the angle between the follower’s line of motion and the common normal to the pitch curve at the contact point. Using the pitch curve (not merely the base or pitch circle) accounts for the actual geometry that drives the follower. Step-by-Step Solution: 1) Identify follower motion direction (e.g., line of stroke for a translating follower).2) At the contact point, construct the normal to the pitch curve.3) Measure the angle between these two directions; that is the pressure angle. Verification / Alternative check: For roller followers, the pitch curve is the cam profile offset by the roller radius; the definition still references the pitch curve normal. Why Other Options Are Wrong: pitch circle/base circle/prime circle — circles define basic sizes but not the instantaneous geometry of contact and force transmission. Common Pitfalls: Confusing the base circle (used to generate profiles) with the functional pitch curve; ignoring that excessive pressure angle increases side load on guides. Final Answer: pitch curve.

Question 8

In classical mechanism theory for manufacturing and machine design, is a pair of friction discs (as in a plate clutch or friction drive) correctly classified as a rolling pair, or does it belong to a different kinematic contact category?

Accepted Answer

Introduction / Context: In kinematics of machines, pairs are classified by the nature of contact between mating elements: surface (lower pairs) versus line/point (higher pairs), and by the dominant relative motion—sliding, rolling, screw (helical), etc. A pair of friction discs, such as those used in plate clutches and friction drives, is often confused with rolling contact. This item clarifies the correct classification. Given Data / Assumptions: Two flat or conical discs pressed together to transmit torque by friction. Relative motion at the interface can be stick (no gross slip) or involve micro-slip; pure rolling is not the intended mode. Standard machine-design terminology for kinematic pairs is used. Concept / Approach: A rolling pair requires predominantly rolling contact with line/point contact (e.g., gear tooth idealization, pure rolling cylinders). Friction discs transmit torque through surface contact with primarily sliding (or impending sliding) motion at the interface when relative angular velocity exists. Hence, the pair is a lower pair characterized by surface contact and sliding—not a rolling pair. Step-by-Step Solution: 1) Identify contact type: friction discs have surface-to-surface contact.2) Identify relative motion: when engaged with speed difference, motion is sliding (or mixed stick–slip), not pure rolling.3) Conclude: classification is a sliding lower pair; therefore, the statement that it is a rolling pair is false. Verification / Alternative check: Clutches are analyzed with Coulomb friction models; energy is dissipated as heat due to sliding. Rolling contact would not yield such heat generation nor the wear patterns typical of clutches. Why Other Options Are Wrong: True/True only at very low slip speeds/True if lubrication is perfectly hydrodynamic/True only when discs are crowned: none convert the fundamental sliding-dominated, surface-contact pair into a rolling pair. Common Pitfalls: Equating ‘‘no gross slip’’ during torque transmission with pure rolling; at the interface, shear and micro-slip persist. Confusing gear rolling (higher pair idealization) with friction-disc action. Final Answer: False

Question 9

Within the taxonomy of kinematic pairs, do sliding (prismatic) pairs, turning (revolute) pairs, and screw (helical) pairs all belong to the class of lower pairs characterized by surface contact?

Accepted Answer

Introduction / Context: Kinematic pairs are divided into lower and higher categories based on contact type. Lower pairs have surface contact, while higher pairs have line/point contact. This question checks whether three common joints—sliding, turning, and screw—are correctly grouped as lower pairs. Given Data / Assumptions: Standard planar/3D mechanism definitions. Ideal joints with rigid elements. Focus on contact geometry, not on friction model or lubrication. Concept / Approach: Sliding (prismatic) pairs maintain surface contact between a slider and guide; turning (revolute) pairs maintain surface contact between pin and bearing; screw (helical) pairs maintain mating surface contact along helical threads. Thus, all three are lower pairs. Higher pairs include cam–follower and gear tooth idealizations (line/point contact). Step-by-Step Solution: 1) Sliding pair → surface-to-surface guide contact → lower pair.2) Turning (revolute) pair → pin-in-hole surface contact → lower pair.3) Screw pair → nut and bolt thread flanks in surface contact → lower pair. Verification / Alternative check: Textbook classifications consistently list these three as lower pairs; only pairs like cams, gears, and pure rolling contacts are higher pairs. Why Other Options Are Wrong: False and the two mixed options: contradict accepted definitions. Depends on lubrication: lubrication affects friction/wear, not the geometric contact classification. Common Pitfalls: Confusing ‘‘higher’’ with ‘‘more sophisticated’’—it strictly refers to contact geometry. Final Answer: True

Question 10

In mechanism terminology, the two elements of a pair are said to form a ________ when they permit suitably constrained relative motion between them (i.e., they constitute a recognized joint in kinematics).

Accepted Answer

Introduction / Context: Mechanisms are built from links connected by pairs (joints). The fundamental term for two elements that permit constrained relative motion is crucial for classification and mobility analysis. Given Data / Assumptions: Standard kinematics definitions. Relative motion is constrained (not arbitrary) and repeatable. Contact may be surface or line/point depending on pair type. Concept / Approach: A kinematic pair is formed when two elements are joined in such a way that their relative motion is constrained to a definite direction(s)/mode(s). Terms like open pair, force-closed pair, and self-closed pair describe how contact is maintained, not whether the joint is kinematic in the first place. Step-by-Step Solution: 1) Identify the key requirement: constrained relative motion → definition of a kinematic pair.2) Recognize that ‘‘open pair’’ refers to contact maintained by external force (e.g., spring), not the existence of a kinematic joint.3) Conclude the blank is ‘‘kinematic pair.’’ Verification / Alternative check: Mobility (Gruebler/Kutzbach) analyses count joints as kinematic pairs; closure type is secondary. Why Other Options Are Wrong: open pair/force-closed pair/self-closed pair: refer to maintenance of contact, not the core definition. rolling pair: a subtype of kinematic pair, not the general term. Common Pitfalls: Mixing classification axes: contact geometry vs. closure vs. the fact of being a kinematic pair. Final Answer: kinematic pair

Theory of machines Questions