MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 4a Equilibrium Torque Rotational Stability

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This deck focuses on 4a Equilibrium Torque Rotational Stability, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Find $\alpha$ if $\sum\tau=12\,\text{N}\cdot\text{m}$ acts on an object with $I=3.0\,\text{kg}\cdot\text{m}^2$.

Answer: $\alpha=4.0\,\text{rad}\cdot\text{s}^{-2}$. Angular acceleration equals net torque divided by moment of inertia, per the rotational second law.

Flashcard 2: State the relationship between torque and angular acceleration for rotation about a fixed axis.

Answer: $\sum \tau=I\alpha$. This rotational analog of Newton's second law relates net torque to angular acceleration via moment of inertia.

Flashcard 3: Identify the torque about the hinge from a $20\,\text{N}$ force applied $0.30\,\text{m}$ away at $0^\circ$ to the door.

Answer: $\tau=0\,\text{N}\cdot\text{m}$. At $\theta=0^\circ$, $\sin(\theta)=0$, so the lever arm is zero, producing no torque.

Flashcard 4: Find the net torque if $+6\,\text{N}\cdot\text{m}$ (CCW) and $9\,\text{N}\cdot\text{m}$ (CW) act on the same axis.

Answer: $\sum\tau=-3\,\text{N}\cdot\text{m}$ (clockwise). Net torque is the algebraic sum, with clockwise negative, resulting in $6 - 9 = -3\,\text{N}\cdot\text{m}$.

Flashcard 5: State the moment of inertia of a uniform rod of mass $M$ and length $L$ about its center (axis perpendicular to rod).

Answer: $I=\frac{1}{12}ML^2$. Derived from integrating mass elements along the rod's length about its midpoint.

Flashcard 6: Find the torque magnitude if $r=0.40\,\text{m}$, $F=10\,\text{N}$, and $\theta=90^\circ$.

Answer: $\tau=4.0\,\text{N}\cdot\text{m}$. With $\theta=90^\circ$, $\sin(\theta)=1$, so torque equals the product of $r$ and $F$.

Flashcard 7: State the static friction inequality that prevents slipping for a body in equilibrium on a surface.

Answer: $f_s\le\mu_s N$. Static friction provides the necessary force up to its maximum to maintain no relative motion in equilibrium.

Flashcard 8: State the moment of inertia of a point mass $m$ at distance $r$ from the rotation axis.

Answer: $I=mr^2$. For a point mass, inertia depends on mass and squared distance from the rotation axis.

Flashcard 9: State the moment of inertia of a thin hoop (ring) of mass $M$ and radius $R$ about its center.

Answer: $I=MR^2$. All mass elements of the hoop are at distance $R$ from the central axis, yielding this inertia value.

Flashcard 10: State the formula for torque magnitude using lever arm and force magnitude.

Answer: $\tau=rF\sin(\theta)$. Torque magnitude arises from the cross product, incorporating the perpendicular component via $\sin(\theta)$.

Flashcard 11: What is the lever arm (moment arm) in the torque expression $\tau=rF\sin(\theta)$?

Answer: $r_\perp=r\sin(\theta)$. The lever arm represents the perpendicular distance from the axis to the line of action of the force.

Flashcard 12: What is the condition for rotational equilibrium in terms of net torque about an axis?

Answer: $\sum \tau=0$. Rotational equilibrium is achieved when the sum of all torques about any axis is zero, preventing angular acceleration.

Flashcard 13: Which force component produces torque about a pivot: parallel or perpendicular to the radius vector?

Answer: Perpendicular component: $F_\perp=F\sin(\theta)$. Torque is produced only by the force component perpendicular to the position vector from the pivot.

Flashcard 14: What is the definition of moment of inertia for point masses about an axis?

Answer: $I=\sum m_ir_i^2$. Moment of inertia quantifies rotational inertia as the sum of each mass times its squared distance from the axis.

Flashcard 15: Find the lever arm $r_\perp$ if $r=0.50\,\text{m}$ and the force makes $30^\circ$ with $\vec{r}$.

Answer: $r_\perp=0.25\,\text{m}$. Lever arm is $r \sin(\theta)$, where $\theta$ is between $\vec{r}$ and $\vec{F}$, yielding $0.50 \times \sin(30^\circ)=0.25\,\text{m}$.

Flashcard 16: What is the rotational kinetic energy of a rigid body with moment of inertia $I$ and angular speed $\omega$?

Answer: $K=\frac{1}{2}I\omega^2$. Rotational kinetic energy is analogous to translational, substituting $I$ for $m$ and $\omega$ for $v$.

Flashcard 17: Identify the direction of torque given by the right-hand rule for $\vec{\tau}=\vec{r}\times\vec{F}$.

Answer: Along $\vec{r}\times\vec{F}$ (thumb direction of right-hand rule). The right-hand rule determines the direction of the cross product $\vec{r} \times \vec{F}$, with thumb pointing along torque.

Flashcard 18: State the moment of inertia of a solid disk (or solid cylinder) of mass $M$ and radius $R$ about its center.

Answer: $I=\frac{1}{2}MR^2$. Integration over the disk's uniform mass distribution results in half the inertia of an equivalent hoop.

Flashcard 19: What is the parallel-axis theorem for moment of inertia?

Answer: $I=I_{\text{cm}}+Md^2$. The theorem allows calculation of inertia about a parallel axis displaced by distance $d$ from the center of mass.

Flashcard 20: What sign convention is commonly used for planar torques in statics problems?

Answer: Counterclockwise $+$, clockwise $-$. This convention assigns positive to counterclockwise and negative to clockwise for consistent torque summation in 2D problems.

Flashcard 21: What is the SI unit of torque?

Answer: $\text{N}\cdot\text{m}$. Torque units derive from force times perpendicular distance, yielding newton-meters in SI.

Flashcard 22: What is angular momentum for a rigid body rotating about a fixed axis?

Answer: $\vec{L}=I\vec{\omega}$. For rigid body rotation about a fixed axis, angular momentum is the product of inertia and angular velocity.

Flashcard 23: State the moment of inertia of a uniform rod of mass $M$ and length $L$ about one end (axis perpendicular to rod).

Answer: $I=\frac{1}{3}ML^2$. Obtained using the parallel-axis theorem from the center-of-mass inertia, adding $M(L/2)^2$.

Flashcard 24: What is the relationship between net external torque and angular momentum?

Answer: $\sum \vec{\tau}_{\text{ext}}=\frac{d\vec{L}}{dt}$. This equation is the rotational equivalent of Newton's second law, linking torque to change in angular momentum.

Flashcard 25: What is the condition for translational equilibrium in terms of net force?

Answer: $\sum \vec{F}=\vec{0}$. Translational equilibrium requires the vector sum of all forces acting on an object to be zero, ensuring no net linear acceleration.