MIT 8.01 - Classical Mechanics & Newtonian Dynamics Flashcards

Question 1

Newton's First Law of Motion

Accepted Answer

A body remains at rest or continues to move in a straight line at a constant velocity unless acted upon by a net external force ($
\sum \vec{F} = 0 \implies \vec{v} = 	ext{constant}$). This establishes the concept of inertial reference frames.

Question 2

Newton's Second Law of Motion

Accepted Answer

The net external force on an object equals the rate of change of its linear momentum: $\vec{F}_{	ext{net}} = \frac{d\vec{p}}{dt}$. For constant mass $m$, this reduces to $\vec{F}_{	ext{net}} = m\vec{a}$.

Question 3

Newton's Third Law of Motion

Accepted Answer

When two objects interact, they exert equal and opposite forces on each other: $\vec{F}_{A 	o B} = -\vec{F}_{B 	o A}$. Action-reaction pairs act on different bodies simultaneously.

Question 4

1D Kinematics Equations (Constant Acceleration)

Accepted Answer

For motion with constant acceleration $a$:
1. $v(t) = v_0 + at$
2. $x(t) = x_0 + v_0 t + \frac{1}{2}at^2$
3. $v^2 = v_0^2 + 2a(x - x_0)$
4. $x - x_0 = \frac{v_0 + v}{2}t$

Question 5

Centripetal Acceleration

Accepted Answer

Acceleration directed radially inward toward the center of curvature for an object undergoing circular motion: $a_c = \frac{v^2}{r} = \omega^2 r$, where $v$ is linear speed, $r$ is radius, and $\omega$ is angular velocity.

Question 6

Work-Kinetic Energy Theorem

Accepted Answer

The net work $W_{	ext{net}}$ done by all forces on a particle equals the change in its kinetic energy $K$:
$W_{	ext{net}} = \Delta K = \int_{\vec{r}_1}^{\vec{r}_2} \vec{F}_{	ext{net}} \cdot d\vec{r} = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2$

Question 7

Conservative Force & Potential Energy

Accepted Answer

A force $\vec{F}$ is conservative if work done around any closed path is zero ($\oint \vec{F} \cdot d\vec{r} = 0$). Potential energy $U$ is defined relative to conservative forces by $\vec{F} = -
abla U$, or $\Delta U = -W_{	ext{internal}} = -\int_{x_i}^{x_f} F_x dx$.

Question 8

Conservation of Mechanical Energy

Accepted Answer

In a system where only conservative forces do work, total mechanical energy is conserved:
$E_{	ext{mech}} = K + U = 	ext{constant} \implies K_i + U_i = K_f + U_f$
If non-conservative forces act: $W_{	ext{nc}} = \Delta E_{	ext{mech}}$.

Question 9

Impulse-Momentum Theorem

Accepted Answer

The total impulse $\vec{J}$ delivered by a net force $\vec{F}$ over time interval $[t_1, t_2]$ equals the change in linear momentum $\Delta \vec{p}$:
$\vec{J} = \int_{t_1}^{t_2} \vec{F}_{	ext{net}} dt = \Delta \vec{p} = \vec{p}_f - \vec{p}_i$

Question 10

Conservation of Linear Momentum

Accepted Answer

If the net external force on a system is zero ($\sum \vec{F}_{	ext{ext}} = 0$), the total linear momentum of the system remains constant:
$\sum \vec{p}_i = \sum \vec{p}_f$

Question 11

Elastic vs. Inelastic Collisions

Accepted Answer

- **Elastic**: Linear momentum AND kinetic energy are conserved ($K_i = K_f$).
- **Inelastic**: Linear momentum is conserved, but kinetic energy is lost ($K_f < K_i$). In a **completely inelastic** collision, objects stick together ($v_f$ is identical for both).

Question 12

Center of Mass

Accepted Answer

The mass-weighted average position of a discrete or continuous system:
$\vec{R}_{	ext{cm}} = \frac{1}{M} \sum_{i} m_i \vec{r}_i = \frac{1}{M} \int \vec{r} dm$
Motion of the center of mass follows $M \vec{a}_{	ext{cm}} = \vec{F}_{	ext{ext, net}}$.

Question 13

Rotational Inertia (Moment of Inertia)

Accepted Answer

The measure of an object's resistance to rotational acceleration about a specific axis:
$I = \sum_{i} m_i r_i^2 = \int r^2 dm$
where $r$ is the perpendicular distance from the axis of rotation to the mass element $dm$.

Question 14

Parallel Axis Theorem

Accepted Answer

Relates the moment of inertia $I$ about any arbitrary axis to the moment of inertia $I_{	ext{cm}}$ about a parallel axis passing through the center of mass:
$I = I_{	ext{cm}} + M d^2$
where $d$ is the distance between the two parallel axes and $M$ is total mass.

Question 15

Perpendicular Axis Theorem

Accepted Answer

For any flat planar (two-dimensional) object lying in the $xy$-plane, the moment of inertia about the perpendicular $z$-axis equals the sum of moments of inertia about the $x$ and $y$ axes:
$I_z = I_x + I_y$

Question 16

Torque

Accepted Answer

The rotational analog of force measuring the tendency of a force to cause rotation:
$\vec{	au} = \vec{r} 	imes \vec{F} \implies |	au| = r F \sin	heta$
Newton's 2nd Law for Rotation: $\vec{	au}_{	ext{net}} = \frac{d\vec{L}}{dt} = I\vec{\alpha}$ (for fixed moment of inertia $I$).

Question 17

Rotational Kinetic Energy & Rolling Motion

Accepted Answer

Rotational kinetic energy is $K_{	ext{rot}} = \frac{1}{2} I \omega^2$.
For an object rolling without slipping on a fixed surface:
$v_{	ext{cm}} = R\omega, \quad a_{	ext{cm}} = R\alpha$
Total kinetic energy is $K_{	ext{total}} = \frac{1}{2} M v_{	ext{cm}}^2 + \frac{1}{2} I_{	ext{cm}} \omega^2$.

Question 18

Angular Momentum and Its Conservation

Accepted Answer

For a single particle: $\vec{L} = \vec{r} 	imes \vec{p}$.
For a rigid body rotating about a fixed axis: $L = I\omega$.
If external torque is zero ($\sum \vec{	au}_{	ext{ext}} = 0$), total angular momentum is conserved: $\vec{L}_i = \vec{L}_f$.

Question 19

Simple Harmonic Motion (SHM) Differential Equation

Accepted Answer

The fundamental second-order differential equation for SHM:
$\frac{d^2 x}{dt^2} + \omega^2 x = 0$
General solution: $x(t) = A \cos(\omega t + \phi)$, where $A$ is amplitude, $\omega$ is angular frequency, and $\phi$ is the phase constant.

Question 20

Period and Frequency of Simple Harmonic Motion

Accepted Answer

For a mass $m$ on a spring with force constant $k$ ($F = -kx$):
Angular frequency: $\omega = \sqrt{\frac{k}{m}}$
Period: $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k}}$
Frequency: $f = \frac{1}{T} = \frac{\omega}{2\pi}$

Question 21

Simple Pendulum Period

Accepted Answer

For a point mass $m$ suspended by a massless string of length $L$ in uniform gravity $g$, under small-angle approximation ($\sin	heta \approx 	heta$):
$\frac{d^2	heta}{dt^2} + \frac{g}{L}	heta = 0 \implies T = 2\pi \sqrt{\frac{L}{g}}$

Question 22

Physical Pendulum Period

Accepted Answer

For an extended rigid body of mass $m$ pivoting about an axis at distance $d$ from its center of mass:
$T = 2\pi \sqrt{\frac{I}{m g d}}$
where $I$ is the moment of inertia of the body about the pivot axis.

MIT 8.01 - Classical Mechanics & Newtonian Dynamics

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