🇮🇳 UPSC ESE Mechanical Engineering · flashcards

UPSC ESE Mechanical Engineering Engineering Mechanics Flashcards

50 question-and-answer cards covering Engineering Mechanics as it is examined in UPSC ESE Mechanical Engineering. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.

50Cards in deck
24Free preview
9Syllabus topics
~172Chars per answer
FreePrice

24 sample cards from the Engineering Mechanics deck

Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.

  1. Relate linear velocity and tangential acceleration of a point to the angular motion of a rotating rigid body.

    $v = \omega r$ and tangential acceleration $a_t = \alpha r$, where $r$ is the distance from the axis of rotation; the normal acceleration is $a_n = \omega^{2} r$.

  2. State Newton's second law for a particle and for rotation of a rigid body.

    Translation: $\vec{F} = m\vec{a}$. Rotation about a fixed axis: $M = I\alpha$, where $I$ is the mass moment of inertia and $\alpha$ the angular acceleration.

  3. Define the instantaneous center of rotation (ICR) for a rigid body in plane motion.

    The point in the body (or its extension) that has zero velocity at a given instant; the body appears to rotate purely about it, so any point's velocity is $v = \omega \times (\text{distance to ICR})$, directed perpendicular to the line to the ICR.

  4. State D'Alembert's principle.

    A moving body can be treated as in equilibrium by introducing a fictitious inertia force $-m\vec{a}$ (and inertia couple $-I\alpha$): $\vec{F} - m\vec{a} = 0$, converting a dynamics problem into a statics-type problem.

  5. State the impulse-momentum principle for a particle.

    The impulse of the net force equals the change in linear momentum: $\displaystyle\int_{t_1}^{t_2}\vec{F}\,dt = m\vec{v_2} - m\vec{v_1}$.

  6. State the principle of conservation of linear momentum.

    If the net external force on a system is zero, total linear momentum is conserved: $m_1\vec{u_1} + m_2\vec{u_2} = m_1\vec{v_1} + m_2\vec{v_2}$.

  7. Define the coefficient of restitution $e$ for a direct central impact.

    $e = \dfrac{\text{relative velocity of separation}}{\text{relative velocity of approach}} = \dfrac{v_2 - v_1}{u_1 - u_2}$. $e=1$ for perfectly elastic, $e=0$ for perfectly plastic impact ($0 \le e \le 1$).

  8. Define work done by a constant force and by a variable force.

    Constant force: $W = \vec{F}\cdot\vec{d} = Fd\cos\theta$. Variable force along a path: $W = \displaystyle\int \vec{F}\cdot d\vec{s}$.

  9. Write the expressions for translational and rotational kinetic energy.

    Translational: $KE = \tfrac{1}{2}mv^{2}$. Rotational: $KE = \tfrac{1}{2}I\omega^{2}$. For general plane motion: $KE = \tfrac{1}{2}mv_G^{2} + \tfrac{1}{2}I_G\omega^{2}$.

  10. State the work-energy principle (theorem) for a particle.

    The net work done by all forces on a particle equals the change in its kinetic energy: $W_{net} = \Delta KE = \tfrac{1}{2}mv^{2} - \tfrac{1}{2}mu^{2}$.

  11. Give the expression for gravitational potential energy and elastic (spring) potential energy.

    Gravitational: $PE = mgh$. Elastic spring: $PE = \tfrac{1}{2}kx^{2}$, where $k$ is the stiffness and $x$ the deformation from the natural length.

  12. State the principle of conservation of mechanical energy and the condition for its validity.

    When only conservative forces act, total mechanical energy is constant: $KE + PE = \text{constant}$, i.e. $\tfrac{1}{2}mv_1^{2} + mgh_1 = \tfrac{1}{2}mv_2^{2} + mgh_2$. It holds only when non-conservative forces (e.g. friction) do no work.

  13. Define power and give its expression in terms of force and velocity.

    Power is the rate of doing work: $P = \dfrac{dW}{dt} = \vec{F}\cdot\vec{v} = Fv\cos\theta$. For rotation, $P = T\omega$ (torque times angular velocity).

  14. Define simple harmonic motion (SHM) and write its governing differential equation.

    Motion where acceleration is proportional to displacement and directed toward the mean position: $\ddot{x} + \omega_n^{2}x = 0$, with solution $x = A\sin(\omega_n t + \phi)$.

  15. For a spring-mass system, give the natural frequency of free undamped vibration.

    $\omega_n = \sqrt{\dfrac{k}{m}}$ (rad/s); natural frequency $f_n = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$ (Hz); time period $T = 2\pi\sqrt{\dfrac{m}{k}}$.

  16. Distinguish between free and forced vibrations.

    Free vibration occurs at the natural frequency after an initial disturbance with no continuing external force. Forced vibration is sustained by a continuous external periodic excitation, and the body vibrates at the forcing frequency.

  17. Write the equation of motion for a damped free single-degree-of-freedom system.

    $m\ddot{x} + c\dot{x} + kx = 0$, where $m$ is mass, $c$ the viscous damping coefficient, and $k$ the stiffness.

  18. Define the damping factor (damping ratio) $\zeta$ and the critical damping coefficient $c_c$.

    Critical damping $c_c = 2\sqrt{km} = 2m\omega_n$. Damping ratio $\zeta = \dfrac{c}{c_c} = \dfrac{c}{2\sqrt{km}}$.

  19. Classify the damping of a free vibration system by the value of $\zeta$.

    $\zeta = 0$: undamped; $0 < \zeta < 1$: underdamped (oscillatory decay); $\zeta = 1$: critically damped (fastest non-oscillatory return); $\zeta > 1$: overdamped (sluggish, non-oscillatory).

  20. Give the damped natural frequency for an underdamped system.

    $\omega_d = \omega_n\sqrt{1 - \zeta^{2}}$, which is always less than the undamped natural frequency $\omega_n$.

  21. Define the logarithmic decrement $\delta$ and relate it to the damping ratio.

    $\delta = \ln\dfrac{x_n}{x_{n+1}}$, the natural log of the ratio of successive amplitudes. It relates to damping by $\delta = \dfrac{2\pi\zeta}{\sqrt{1-\zeta^{2}}}$.

  22. Define resonance and state the condition at which it occurs.

    Resonance is the condition where the forcing frequency equals the system's natural frequency ($\omega = \omega_n$), producing maximum amplitude (theoretically infinite for an undamped system). Damping limits the peak amplitude.

  23. Define the magnification factor (dynamic amplification factor) for forced vibration.

    $M.F. = \dfrac{1}{\sqrt{(1 - r^{2})^{2} + (2\zeta r)^{2}}}$, the ratio of dynamic to static deflection, where $r = \dfrac{\omega}{\omega_n}$ is the frequency ratio.

  24. Define transmissibility in forced vibration and give the frequency ratio beyond which isolation occurs.

    Transmissibility is the ratio of force transmitted to the foundation to the applied force: $T_r = \dfrac{\sqrt{1+(2\zeta r)^{2}}}{\sqrt{(1-r^{2})^{2}+(2\zeta r)^{2}}}$. Isolation ($T_r < 1$) occurs only when $r > \sqrt{2}$.

What this deck covers

The Engineering Mechanics deck follows the UPSC ESE Mechanical Engineering Engineering Mechanics syllabus — 3 chapters and 9 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 16.7 cards per chapter.

Answers are written to be recallable, not just readable — averaging about 172 characters, which is long enough to carry the reasoning and short enough to say out loud.

A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.

Engineering Mechanics flashcards FAQ

How many Engineering Mechanics flashcards are in this UPSC ESE Mechanical Engineering deck?

50 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.

Are these UPSC ESE Mechanical Engineering flashcards free?

Yes. The preview here is free to read with no signup, and the full 50-card deck is free inside the Examius app.

What do the Engineering Mechanics cards cover?

They follow the UPSC ESE Mechanical Engineering Engineering Mechanics syllabus — 3 chapters and 9 topics — so the questions track what is actually examinable.

How should I use these flashcards?

Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.