Q: Simple harmonic motion (SHM): Relate periodic time and frequency. Fundamental relation between frequency (f) and time period (T) of SHM. Choose the correct value of (T × f).

Concept By definition, frequency f = 1/T. Therefore, T × f = 1. Final Answer One

Q: Vibration isolation: Frequency ratio for unity transmissibility (general damping). In a base-excited isolation system, transmissibility (force or motion) equals unity at a specific frequency ratio r = ω/ω n , independent of damping. Deter…

Concept/Approach Transmissibility T for a damped single-degree system is T = √(1 + (2ζr)²) / √((1 − r²)² + (2ζr)²).Setting T = 1 and simplifying gives (1 + (2ζr)²) = (1 − r²)² + (2ζr)² ⇒ 1 = (1 − r²)² ⇒ r² = 2 (taking r > 0). Result Thus, independent of damping (ζ), T = 1 at r = √2. Final Answer r = √2 (≈ 1.414)

Question 1

Damped vibrations: Compute the logarithmic decrement from the damping ratio.  Given: actual damping coefficient c = 40 N·s/m; critical damping coefficient c_c = 420 N·s/m.  Calculate the logarithmic decrement (δ) for the underdamped system.

Accepted Answer

Given data c = 40 N·s/mcc = 420 N·s/m Concept/Approach For underdamped free vibration, damping ratio ζ = c / cc.Logarithmic decrement: δ = (2πζ)/√(1 − ζ²). Step-by-Step calculation ζ = 40 / 420 = 2/21 ≈ 0.095238√(1 − ζ²) = √(1 − 0.00907) ≈ √(0.99093) ≈ 0.99546δ = (2π × 0.095238) / 0.99546 ≈ 0.598 ≈ 0.6 Verification Small ζ (≈0.095) should give a small δ (< 1), so ≈0.6 is reasonable. Final Answer δ ≈ 0.6

Question 2

Simple harmonic motion (SHM): Relate periodic time and frequency.  Fundamental relation between frequency (f) and time period (T) of SHM.  Choose the correct value of (T × f).

Accepted Answer

Concept By definition, frequency f = 1/T. Therefore, T × f = 1. Final Answer One

Question 3

Vibration isolation: Frequency ratio for unity transmissibility (general damping). In a base-excited isolation system, transmissibility (force or motion) equals unity at a specific frequency ratio r = ω/ω_n, independent of damping. Deter…

Accepted Answer

Concept/Approach Transmissibility T for a damped single-degree system is T = √(1 + (2ζr)²) / √((1 − r²)² + (2ζr)²).Setting T = 1 and simplifying gives (1 + (2ζr)²) = (1 − r²)² + (2ζr)² ⇒ 1 = (1 − r²)² ⇒ r² = 2 (taking r > 0). Result Thus, independent of damping (ζ), T = 1 at r = √2. Final Answer r = √2 (≈ 1.414)

Question 4

Cam design for high-speed engines: Choose suitable follower motion law.  Objective: Minimise shock/jerk and ensure smooth acceleration at high speeds in cam-follower systems.  Select the most appropriate follower motion.

Accepted Answer

Concept Cycloidal motion provides zero jerk at the start and end of stroke (continuous acceleration profile), making it ideal for high-speed cams where dynamic forces must be minimised. Why not others? Uniform velocity and SHM introduce higher jerk; uniform acceleration/retardation has jerk discontinuities at segment junctions. Final Answer Cycloidal motion

Question 5

Gyroscopic effect on a ship: Determine bow/stern reaction for given rotation/turn.  Given: Rotor spins clockwise when viewed from the stern. The ship turns right (starboard) while moving forward.  State the resulting pitching effect due to the gyros…

Accepted Answer

Concept/Approach Angular momentum (H) is along the shaft. Clockwise rotation seen from the stern puts H toward the bow. A right turn corresponds to a precession vector downward (−z). The gyroscopic couple is C = Ω × H, giving a couple about the transverse axis that pitches the ship. Vector result Ω (down) × H (forward) acts about the negative transverse axis, tending to lower the bow and raise the stern. Final Answer Raise the stern and lower the bow

Question 6

Twin-cylinder V-engine balance: Identify the V-angle from given reverse/direct crank positions.  A diagram (not included here) shows the crank positions for primary reverse cranks and secondary direct cranks for a twin-cylinder V-engine.  Choose the…

Accepted Answer

60° V-engine

Question 7

Steering of a ship: Choose the correct definition of the motion.  Marine dynamics terminology: Identify what is meant by ‘‘steering’’ in ship motions (pitch, yaw/turn, roll).  Select the correct description.

Accepted Answer

Concept Steering is the action of changing heading—i.e., turning (yawing) the ship to port or starboard while progressing forward. Why others are wrong? Pitching is rotation about the transverse axis (bow up/down). Rolling is about the longitudinal axis. Heave is vertical translation. Final Answer Turning the ship on a curved path to right or left while moving ahead (yaw/turning)

Question 8

In mechanical design terminology for standard spur or helical gears, what is the precise radial distance measured from the pitch circle down to the bottom (root) of the tooth space (i.e., the depth below the pitch line)?

Accepted Answer

Introduction / Context In gear engineering, clear definitions of tooth proportions ensure correct meshing, strength, and manufacturability. One commonly confused pair of terms is addendum and dedendum. This question tests recognition of the radial distance below the pitch circle that reaches the root of the tooth space in standard spur or helical gears. Given Data / Assumptions Standard involute spur or helical gear terminology applies. Pitch circle is the reference circle for speed ratios and conjugate action. Root (bottom land) is the lowest point of the tooth space. Concept / Approach The pitch circle divides the tooth into two radial portions: above it, the addendum; below it, the dedendum. The dedendum must be greater than the addendum by a small clearance to prevent tip interference of the mating gear. Step-by-Step Solution 1) Identify the reference: the pitch circle.2) Locate the root (bottom land) of the tooth space.3) The radial distance from the pitch circle to this root is, by definition, the dedendum. Verification / Alternative check Standard proportions (for many systems) use addendum ≈ one module and dedendum ≈ addendum + clearance. Thus, the below-pitch portion is the dedendum. Why Other Options Are Wrong Addendum: distance from pitch circle to tooth tip (outside), not to the root. Clearance: the extra difference between dedendum of one gear and addendum of its mate, not a full radial depth below pitch. Working depth: sum of the two addenda of meshing gears, measured within the engaged region, not a single gear's below-pitch depth. Total depth: addendum + dedendum of the same gear; a larger measure than either alone. Common Pitfalls Mixing up addendum and dedendum by focusing on the wrong side of the pitch circle. Confusing clearance with dedendum due to both being below the pitch circle. Final Answer Dedendum

Question 9

In the classification of gear trains, which type specifically involves gears whose axes move relative to a fixed reference axis (i.e., planet or carrier-mounted gears whose centers revolve about a fixed axis), unlike simple, compound, or reverted tr…

Accepted Answer

Introduction / Context Gear trains are grouped by how gear centers move. In simple, compound, and reverted trains, the gear axes are fixed relative to the frame. In contrast, epicyclic (planetary) trains permit gears to orbit while rotating, enabling high reduction in compact spaces and coaxial input–output designs. Given Data / Assumptions The question highlights axes that move relative to a fixed reference axis. Planet gears are mounted on a carrier that revolves about a fixed axis (typically the sun gear axis). Standard machine design terminology is assumed. Concept / Approach Epicyclic trains contain at least one gear (planet) whose center is not fixed; it revolves around another gear (sun). The entire set may include a ring gear and a carrier that can be held, driven, or used as output. Step-by-Step Solution 1) Identify the distinguishing feature: moving gear axes.2) Recognize that in planetary systems, planet centers follow circular paths about the sun axis via the carrier.3) Conclude that such motion defines an epicyclic (planetary) gear train. Verification / Alternative check Check each class: simple, compound, and reverted have fixed gear centers; only planetary systems have orbiting gear centers. Why Other Options Are Wrong Simple: single gear mesh chains with fixed centers. Compound: shafts carrying two or more gears, yet centers remain fixed. Reverted: input and output shafts are coaxial, but all centers are fixed. Differential gear train: a special function often implemented using planetary elements; the core defining feature asked here is the moving axes of planets, which is the planetary mechanism itself. Common Pitfalls Equating “differential” with the general planetary class rather than its specific function. Assuming coaxiality implies planetary; reverted trains are coaxial without moving axes. Final Answer Epicyclic (planetary) gear train

Question 10

In locomotive dynamics, partial balancing of reciprocating parts (balancing only a fraction of the reciprocating mass) can introduce which effects in a steam locomotive operating along the line of stroke?

Accepted Answer

Introduction / Context Steam locomotives have significant reciprocating masses (pistons, piston rods, parts of connecting rods). Full balancing of these masses is impractical because it would demand counterweights that create excessive vertical and lateral dynamic loads. Designers therefore use partial balancing, which reduces some out-of-balance forces but inevitably introduces side effects. This question explores the resulting dynamic phenomena. Given Data / Assumptions Only a fraction of reciprocating mass is balanced. Wheel counterweights are used for balancing. Operation is typical mainline with varying speeds. Concept / Approach Partial balancing trades one form of unbalance for another. The rotating counterbalance cannot perfectly cancel alternating inertia forces. This creates periodic vertical loads (hammer blow), lateral couples (swaying), and fluctuations in effective tractive effort due to changing inertia force components along the line of stroke. Step-by-Step Solution 1) Reciprocating inertia produces horizontal unbalanced force F = m·r·ω²·cosθ and its derivatives.2) Adding rotating counterweights cancels a component but introduces rotating unbalance components at right angles.3) The vertical component on rails varies cyclically, yielding hammer blow.4) Unequal forces on left and right sides create a swaying couple about the locomotive centerline.5) The resultant horizontal component along the line of stroke varies with crank angle, causing variation in tractive force. Verification / Alternative check Historical practice documents the triad of issues for partially balanced reciprocating locomotives: hammer blow, swaying couple, and tractive effort fluctuation, all speed dependent. Why Other Options Are Wrong Hammer blow only: ignores swaying and tractive variation. Swaying couple only: incomplete, as vertical and longitudinal effects also arise. Variation in tractive force only: similarly incomplete. None of these: contradicts well-known locomotive balancing outcomes. Common Pitfalls Assuming counterweights solve all balance problems; they shift unbalance rather than eliminate it. Neglecting speed dependence; dynamic effects grow with ω². Final Answer All of the above

Question 11

For selecting a cam–follower system in mainstream automobile valve gear, which follower form is generally adopted to reduce wear at high speed while maintaining stable contact and low sliding friction at the cam–tappet interface?

Accepted Answer

Introduction / Context Automotive valve trains demand high durability, controlled contact stress, and stable kinematics at high engine speeds. The choice of follower geometry materially affects wear, lubrication behavior, contact stress distribution, and manufacturing simplicity. Given Data / Assumptions Typical mass-production overhead cam or tappet-driven automotive engines. Priority on long life, low wear, and reliable lubrication at high speed. Cam profiles designed for smooth acceleration (limited jerk) and avoidance of edge loading. Concept / Approach Flat-faced (mushroom or bucket) followers present a relatively large, flat contact area that spreads load, reduces Hertzian pressure, and simplifies hydrodynamic or boundary lubrication films. The flat face allows the use of slight crowning and engineered spin to distribute wear uniformly. Manufacturing and packaging are favorable for high-volume engines. Step-by-Step Solution 1) Compare contact types: line/point contacts concentrate stress, while area contacts distribute it.2) Knife-edge produces extreme stress and rapid wear; unsuitable for automotive durability.3) Roller followers reduce sliding but add complexity, mass, and bearing wear modes; widely used in some performance or pushrod applications but less universal in mass-production overhead designs.4) Spherical-faced followers reduce edge effects but still localize contact more than a broad flat, and control of spin may be less predictable.5) Thus, flat-faced (mushroom/bucket) followers are generally adopted for robust, economical high-speed use. Verification / Alternative check Production engines frequently use bucket tappets or flat tappets with surface treatments (nitriding, DLC) and oil-control features to sustain high cycles. Why Other Options Are Wrong Knife-edge: severe wear and stress concentration. Spherical-faced: smaller contact area and potential edge loading relative to a broad flat. Roller follower: reduced sliding but higher complexity and mass; not the general default across mainstream designs. Offset flat: a layout variation, not the core wear-minimizing geometry choice. Common Pitfalls Assuming rollers are always superior; system-level trade-offs matter. Ignoring lubrication film behavior and spin-induced wear distribution. Final Answer Flat-faced (mushroom) follower

Question 12

According to standard machine design bands for pitch-line speed, over which peripheral velocity range are gears classified as medium-velocity gears?

Accepted Answer

Introduction / Context Gear classification by pitch-line speed (peripheral velocity at the pitch circle) guides material selection, heat treatment, lubrication strategy, and tooth form accuracy. Typical textbooks and standards group speeds into low, medium, and high ranges for preliminary design decisions. Given Data / Assumptions Standard spur/helical gear practice for industrial machinery. Pitch-line speed v = π·d·n in consistent units. Conventional ranges: low, medium, high, with approximate thresholds. Concept / Approach In many design references, medium-velocity gears operate where hydrodynamic lubrication becomes more reliable and dynamic effects begin to influence accuracy requirements, yet speeds are not so high that special high-speed finishing becomes mandatory. Step-by-Step Solution 1) Identify the common banding used in design guides.2) Low-velocity is typically up to about 3 m/s.3) Medium-velocity is broadly from 3 m/s to about 15 m/s.4) High-velocity regimes exceed roughly 15 m/s, with very high-speed applications going beyond 30–50 m/s. Verification / Alternative check Machine design texts align on 3–15 m/s as the medium band, recognizing variations by author or application without altering the central value. Why Other Options Are Wrong 1 – 3 m/s: characteristic of low-speed gears; boundary lubrication and wear dominate. 15 – 30 m/s and 30 – 50 m/s: these are high-speed domains requiring higher accuracy grades and improved finishing. Above 50 m/s: very high speed; special design and balancing are typical. Common Pitfalls Using exact cutoffs rigidly; these are practical bands, not hard standards. Confusing angular speed with pitch-line speed; diameter matters via v = π·d·n. Final Answer 3 – 15 m/s

Question 13

In the Whitworth quick-return (crank-and-slotted lever) mechanism obtained as an inversion of the slider–crank chain, which link must be kept fixed to produce unequal cutting and return strokes characteristic of quick return?

Accepted Answer

Introduction / Context Quick-return mechanisms allow the tool (or ram) to spend more time on the cutting stroke and less on the return stroke, improving productivity. The Whitworth mechanism is a classic inversion of the slider–crank chain using a crank and a slotted lever. Given Data / Assumptions A slider–crank chain is the parent mechanism. We need the inversion that yields a crank driving a slotted lever about a fixed pivot. Goal: unequal forward and return stroke times. Concept / Approach In the Whitworth arrangement, the slotted lever oscillates about a fixed pivot on the frame, while a rotating crank drives a slider (block) that runs within the slot. This configuration is obtained by fixing the link opposite the slider, i.e., the frame that houses the lever pivot. The geometry causes non-uniform angular motion of the lever for uniform crank rotation, delivering the quick-return effect. Step-by-Step Solution 1) Start with the slider–crank chain and identify links: frame, crank, connecting rod (coupler), and slider.2) To form a slotted lever pivoted to the frame, fix the link opposite the slider so the lever has a stationary center.3) Rotate the crank; the slider (block) in the lever slot converts rotation to oscillation with unequal forward/return timing. Verification / Alternative check Textbook kinematic inversions list the Whitworth mechanism as the inversion where the frame opposite the slider is the fixed link, creating a slotted lever pivot. Why Other Options Are Wrong Coupler fixed: yields other inversions (e.g., oscillating cylinder) but not the Whitworth quick return. Slider fixed: produces different layouts (e.g., crank and guide) without the characteristic slotted lever geometry. Smallest link fixed or longest link fixed: size is not the defining criterion; the correct inversion is specific to which link is the frame opposite the slider. Common Pitfalls Confusing the Whitworth inversion with the crank-and-rod quick return (drag link or slotted disc types). Thinking link length alone determines inversion; it is the fixed link choice that matters. Final Answer Frame opposite the slider is fixed

Question 14

For a rotating shaft with a mounted disc and small eccentricity, according to classical whirling theory, which parameter(s) influence the critical (whirling) speed of the system?

Accepted Answer

Introduction / Context Critical or whirling speed is the rotational speed at which a rotating shaft–disc system experiences resonance due to coincidence of the running speed with a natural frequency, leading to large lateral vibrations. Designers must understand the variables that set this speed. Given Data / Assumptions Uniform shaft with a mounted disc. Small eccentricity between mass center and geometric center. Linear, classical whirling theory applies (small deflections). Concept / Approach The lateral natural frequency depends on shaft stiffness (a function of length, diameter, and boundary conditions) and the attached mass distribution. Eccentricity introduces a rotating unbalance force that excites the mode strongly near the natural frequency. Step-by-Step Solution 1) Shaft span/length enters stiffness k ∝ EI/L³ for typical beam-like behavior; longer spans reduce stiffness and lower critical speed.2) Disc mass (often related to its diameter and thickness) changes the effective mass m at the critical section; higher mass lowers natural frequency and critical speed.3) Eccentricity e creates an unbalance force m·e·ω² that peaks near resonance; while e does not change the natural frequency itself, it determines the response amplitude and practical safe speed, effectively influencing usable critical speed considerations. Verification / Alternative check Rayleigh or Dunkerley estimates show explicit dependence on stiffness (length) and mass; design charts account for unbalance via eccentricity. Why Other Options Are Wrong Single-parameter answers ignore coupled dependence on stiffness (length), mass (disc), and excitation via eccentricity. None of these contradicts theory and practice. Common Pitfalls Confusing effect on natural frequency (stiffness, mass) with excitation amplitude (eccentricity); both matter to practical critical speed operation. Final Answer All of these

Question 15

In the coupling rod assembly of a steam locomotive, each of the four joints idealized with pins corresponds to which kinematic pair type?

Accepted Answer

Introduction / Context Kinematic pairs classify how two machine elements constrain each other's motion. Locomotive coupling rods connect driving wheels and transmit torque and phase, relying on joints that allow rotation but restrict relative translation. Given Data / Assumptions Ideal pin joints at the four coupling rod connections. Planar mechanism with no significant clearance or compliance assumed. Concept / Approach A turning (revolute) pair permits relative rotation about a fixed axis while preventing relative translation along or perpendicular to that axis. A pin in a hole is the canonical realiztion of a revolute pair, which models the coupling rod connections. Step-by-Step Solution 1) Identify motion permitted at each joint: rotation about the pin.2) Confirm constraints: no axial sliding intended; the rod length remains fixed.3) Conclude the pair type is a turning (revolute) pair. Verification / Alternative check Mechanism texts universally model such wheel–rod pin joints as revolute pairs in planar kinematics. Why Other Options Are Wrong Sliding pair: allows translational motion, not appropriate here. Rolling pair: requires pure rolling contact, not pin–hole rotation. Screw pair: couples rotation and translation helically, absent here. Cylindrical pair: allows rotation and translation along one axis; over-permissive for this joint. Common Pitfalls Confusing real-world clearances with intended kinematic constraints; the idealized model is a revolute pair. Final Answer Turning (revolute) pair

Question 16

In an ideal slider–crank mechanism for a reciprocating steam engine, what is the piston (slider) velocity exactly at the inner dead centre (IDC), which is an extreme stroke position?

Accepted Answer

Introduction / Context The slider–crank mechanism converts rotary motion to reciprocation. Dead centres (inner and outer) are extreme positions of the slider where the connecting rod, crank, and line of stroke align in specific ways. Velocity at these positions is a fundamental kinematic property. Given Data / Assumptions Ideal joints without clearance or compliance. Constant angular velocity of the crank. IDC is an end-of-stroke position. Concept / Approach Slider velocity is the time derivative of displacement along the line of stroke. At an extreme position of a periodic displacement, instantaneous velocity is zero because displacement has a local maximum or minimum there. This follows both from geometry and harmonic approximations. Step-by-Step Solution 1) Write slider displacement s(θ); for small obliquity, s ≈ r cosθ + (r²/4l)cos2θ + …2) Differentiate: v = ds/dt = (ds/dθ)·ω.3) At IDC, θ equals the dead-center angle, where ds/dθ = 0; hence v = 0. Verification / Alternative check Vector geometry shows slider motion reverses direction at dead centres; reversal implies zero instantaneous velocity. Why Other Options Are Wrong Minimum (but not zero): at the turning point of displacement, the exact velocity is zero. Maximum: occurs near mid-stroke for uniform crank speed. None of these / Indeterminate: contradicts deterministic kinematics. Common Pitfalls Confusing acceleration (which is maximum in magnitude at dead centres) with velocity (which is zero). Final Answer Zero

Question 17

In vibration isolation for a lightly damped single-degree-of-freedom system, what is the force transmissibility (transmitted force divided by exciting force) when the frequency ratio r = ω/ωₙ equals 2?

Accepted Answer

Introduction / Context Force transmissibility T in vibration isolation quantifies how much dynamic force passes through a mount from a vibrating machine to its foundation. Designers aim for T < 1 (isolation) by operating above resonance (r > √2) with appropriate damping. Given Data / Assumptions Single-degree-of-freedom mass–spring–damper. Light damping (typical isolation mounts). Frequency ratio r = ω/ωₙ = 2. Concept / Approach The standard formula for force transmissibility is T = √(1 + (2ζr)²) / √((1 − r²)² + (2ζr)²), where ζ is damping ratio. For light damping (say ζ ≈ 0.05–0.1), r = 2 places the system well into the isolation region, making T significantly below unity and near one third. Step-by-Step Solution 1) Substitute r = 2 into the formula: numerator √(1 + (4ζ)²); denominator √((1 − 4)² + (4ζ)²) = √(9 + (4ζ)²).2) For ζ = 0.05: numerator ≈ √(1 + 0.04) ≈ 1.02; denominator ≈ √(9 + 0.04) ≈ 3.00; hence T ≈ 1.02 / 3.00 ≈ 0.34.3) For ζ = 0.1: numerator ≈ √(1 + 0.16) ≈ 1.08; denominator ≈ √(9 + 0.16) ≈ 3.03; hence T ≈ 0.36.4) Light-damping values cluster near one third, commonly rounded to 0.33. Verification / Alternative check As ζ → 0, T → 1/|1 − r²| = 1/3 ≈ 0.333 for r = 2, matching the isolation-limit estimate. Why Other Options Are Wrong 1, 1.5, 2: imply no isolation or amplification; inconsistent with r = 2 in light damping. 0.5: higher than the theoretical undamped value 1/3; too conservative. Common Pitfalls Using displacement transmissibility instead of force transmissibility; forms differ near resonance. Assuming heavy damping always improves isolation; excessive ζ increases numerator and can degrade high-frequency isolation. Final Answer 0.33

Question 18

Clock mechanisms and gear trains — which gear arrangement is most commonly employed in mechanical clocks and timepieces to keep the hour and minute shafts coaxial while producing a large speed reduction?

Accepted Answer

Introduction / Context: Mechanical clocks and traditional timepieces require the minute hand shaft and the hour hand shaft to be coaxial, while the minute shaft must rotate 12 times faster than the hour shaft. The gear train that achieves this large, precise speed reduction and maintains coaxial shafts is central to clock design. Given Data / Assumptions: Two output shafts (minute and hour) must be coaxial. Required speed ratio ≈ 12:1 between minute and hour hands. Compact layout, high accuracy, and ease of assembly are desired. Concept / Approach: A reverted gear train is a compound train in which the input and output shafts lie on the same line (coaxial). The intermediate gear centers are arranged so the first and last gears share a common axis, enabling a large overall gear ratio in a compact envelope—ideal for clock mechanisms. Step-by-Step Solution: 1) Identify the need for coaxial input–output: clock hands share an axis.2) Recognize that simple trains rarely keep the first and last gears coaxial without large center distances.3) A reverted compound train places idlers so that the net center distance from first to last equals zero, giving coaxial shafts.4) The compound nature allows stacking ratios to reach 12:1 (or more) with modest gear sizes, improving accuracy and packaging. Verification / Alternative Check: Historical and modern clockworks commonly use reverted arrangements in the motion-work to deliver 12:1 while preserving a shared axis. Why Other Options Are Wrong: Simple gear train — Does not inherently ensure coaxial input and output for large reductions. Sun and planet gear — More typical of Watt’s engine; not standard for coaxial clock motion-work. Differential gear — Used to combine motions; unnecessary and overly complex here. Common Pitfalls: Confusing any compound train with a reverted train; only reverted layouts guarantee coaxial input–output. Final Answer: Reverted gear train.

Question 19

Governor characteristics — for a spring-controlled governor with controlling force Fc = a·r + b (where r is radius of rotation), what is the nature of stability?

Accepted Answer

Introduction / Context: In a spring-controlled centrifugal governor, the controlling force–radius relationship determines whether speed increases cause balls to move inward or outward. Stability requires speed to rise with radius (or equivalently, controlling force to increase with radius appropriately). Given Data / Assumptions: Fc = a·r + b, with a > 0 typically for practical springs. r is ball radius of rotation; Fc balances the required centripetal force m·ω²·r. At equilibrium, Fc = m·ω²·r. Concept / Approach: From equilibrium, m·ω²·r = a·r + b ⇒ ω² = (a + b/r)/m. If slope a is positive, Fc increases with r, producing a rising equilibrium speed with radius—the hallmark of a stable governor. Isochronism requires the Fc–r line to pass through the origin (b = 0) with the correct slope so that equilibrium speed is constant for all r. Step-by-Step Solution: 1) Write equilibrium: m·ω²·r = a·r + b.2) Rearrange: ω² = (a + b/r)/m → as r increases, the term a dominates; ω increases.3) Therefore, the equilibrium speed rises with radius, indicating stability. Verification / Alternative Check: Graphically, a straight Fc–r line with positive slope intersecting the Fc axis at b gives a rising characteristic and thus a stable governor. Why Other Options Are Wrong: Unstable — Would require a decreasing Fc with r (negative slope) or inappropriate characteristic. Isochronous — Needs b = 0 and a specific slope so speed is same for all radii, which is not guaranteed with b ≠ 0. None of these — Not applicable. Common Pitfalls: Assuming any affine (linear) law is isochronous; isochronism demands b = 0 and precise tuning. Final Answer: Stable.

Question 20

Static friction — select the most complete and correct set of properties of the limiting static friction between two contacting surfaces.

Accepted Answer

Introduction / Context: Limiting static friction is the maximum frictional force that resists the impending motion of one surface over another just before sliding begins. Understanding its properties is fundamental to solving problems in engineering mechanics and machine design. Given Data / Assumptions: Dry surfaces with no significant adhesion or lubrication effects beyond the classical Coulomb model. Normal reaction R acts perpendicular to the interface. Motion is impending but not yet occurring. Concept / Approach: Coulomb’s friction law states that the limiting static friction Fmax = μs R, where μs is the coefficient of static friction. The direction of static friction is such that it opposes the tendency of motion. Additionally, within the idealized model, Fmax is largely independent of the apparent (macroscopic) area of contact. Step-by-Step Solution: 1) Relate magnitude: Fmax ∝ R ⇒ constant ratio μs.2) Area effect: Microscopic asperities govern real area; apparent area changes do not significantly alter Fmax in the model.3) Direction: Friction acts opposite the impending motion to maintain equilibrium. Verification / Alternative Check: Empirical tests across many materials confirm approximately linear proportionality to R and weak dependence on apparent area over practical ranges. Why Other Options Are Wrong: Each of (a), (b), (c) is individually true, but incomplete when taken alone. None of these — Incorrect because (a), (b), and (c) are valid under the Coulomb model. Common Pitfalls: Confusing apparent with real contact area; misdirecting friction opposite actual motion instead of impending motion. Final Answer: All of the above.

Theory of machines Questions