Equilibrium, Torque, and Rotational Stability (4A) - MCAT Chemical and Physical Foundations of Biological Systems
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What is the parallel-axis theorem for moment of inertia?
What is the parallel-axis theorem for moment of inertia?
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$I=I_{\text{cm}}+Md^2$. The theorem allows calculation of inertia about a parallel axis displaced by distance $d$ from the center of mass.
$I=I_{\text{cm}}+Md^2$. The theorem allows calculation of inertia about a parallel axis displaced by distance $d$ from the center of mass.
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State the moment of inertia of a point mass $m$ at distance $r$ from the rotation axis.
State the moment of inertia of a point mass $m$ at distance $r$ from the rotation axis.
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$I=mr^2$. For a point mass, inertia depends on mass and squared distance from the rotation axis.
$I=mr^2$. For a point mass, inertia depends on mass and squared distance from the rotation axis.
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State the static friction inequality that prevents slipping for a body in equilibrium on a surface.
State the static friction inequality that prevents slipping for a body in equilibrium on a surface.
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$f_s\le\mu_s N$. Static friction provides the necessary force up to its maximum to maintain no relative motion in equilibrium.
$f_s\le\mu_s N$. Static friction provides the necessary force up to its maximum to maintain no relative motion in equilibrium.
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Find the torque magnitude if $r=0.40,\text{m}$, $F=10,\text{N}$, and $\theta=90^\circ$.
Find the torque magnitude if $r=0.40,\text{m}$, $F=10,\text{N}$, and $\theta=90^\circ$.
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$\tau=4.0,\text{N}\cdot\text{m}$. With $\theta=90^\circ$, $\sin(\theta)=1$, so torque equals the product of $r$ and $F$.
$\tau=4.0,\text{N}\cdot\text{m}$. With $\theta=90^\circ$, $\sin(\theta)=1$, so torque equals the product of $r$ and $F$.
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Find the lever arm $r_\perp$ if $r=0.50,\text{m}$ and the force makes $30^\circ$ with $\vec{r}$.
Find the lever arm $r_\perp$ if $r=0.50,\text{m}$ and the force makes $30^\circ$ with $\vec{r}$.
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$r_\perp=0.25,\text{m}$. Lever arm is $r \sin(\theta)$, where $\theta$ is between $\vec{r}$ and $\vec{F}$, yielding $0.50 \times \sin(30^\circ)=0.25,\text{m}$.
$r_\perp=0.25,\text{m}$. Lever arm is $r \sin(\theta)$, where $\theta$ is between $\vec{r}$ and $\vec{F}$, yielding $0.50 \times \sin(30^\circ)=0.25,\text{m}$.
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Identify the torque about the hinge from a $20,\text{N}$ force applied $0.30,\text{m}$ away at $0^\circ$ to the door.
Identify the torque about the hinge from a $20,\text{N}$ force applied $0.30,\text{m}$ away at $0^\circ$ to the door.
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$\tau=0,\text{N}\cdot\text{m}$. At $\theta=0^\circ$, $\sin(\theta)=0$, so the lever arm is zero, producing no torque.
$\tau=0,\text{N}\cdot\text{m}$. At $\theta=0^\circ$, $\sin(\theta)=0$, so the lever arm is zero, producing no torque.
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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.
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.
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$\sum\tau=-3,\text{N}\cdot\text{m}$ (clockwise). Net torque is the algebraic sum, with clockwise negative, resulting in $6 - 9 = -3,\text{N}\cdot\text{m}$.
$\sum\tau=-3,\text{N}\cdot\text{m}$ (clockwise). Net torque is the algebraic sum, with clockwise negative, resulting in $6 - 9 = -3,\text{N}\cdot\text{m}$.
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Find $\alpha$ if $\sum\tau=12,\text{N}\cdot\text{m}$ acts on an object with $I=3.0,\text{kg}\cdot\text{m}^2$.
Find $\alpha$ if $\sum\tau=12,\text{N}\cdot\text{m}$ acts on an object with $I=3.0,\text{kg}\cdot\text{m}^2$.
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$\alpha=4.0,\text{rad}\cdot\text{s}^{-2}$. Angular acceleration equals net torque divided by moment of inertia, per the rotational second law.
$\alpha=4.0,\text{rad}\cdot\text{s}^{-2}$. Angular acceleration equals net torque divided by moment of inertia, per the rotational second law.
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What is the condition for translational equilibrium in terms of net force?
What is the condition for translational equilibrium in terms of net force?
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$\sum \vec{F}=\vec{0}$. Translational equilibrium requires the vector sum of all forces acting on an object to be zero, ensuring no net linear acceleration.
$\sum \vec{F}=\vec{0}$. Translational equilibrium requires the vector sum of all forces acting on an object to be zero, ensuring no net linear acceleration.
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What is the condition for rotational equilibrium in terms of net torque about an axis?
What is the condition for rotational equilibrium in terms of net torque about an axis?
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$\sum \tau=0$. Rotational equilibrium is achieved when the sum of all torques about any axis is zero, preventing angular acceleration.
$\sum \tau=0$. Rotational equilibrium is achieved when the sum of all torques about any axis is zero, preventing angular acceleration.
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State the formula for torque magnitude using lever arm and force magnitude.
State the formula for torque magnitude using lever arm and force magnitude.
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$\tau=rF\sin(\theta)$. Torque magnitude arises from the cross product, incorporating the perpendicular component via $\sin(\theta)$.
$\tau=rF\sin(\theta)$. Torque magnitude arises from the cross product, incorporating the perpendicular component via $\sin(\theta)$.
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What is the lever arm (moment arm) in the torque expression $\tau=rF\sin(\theta)$?
What is the lever arm (moment arm) in the torque expression $\tau=rF\sin(\theta)$?
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$r_\perp=r\sin(\theta)$. The lever arm represents the perpendicular distance from the axis to the line of action of the force.
$r_\perp=r\sin(\theta)$. The lever arm represents the perpendicular distance from the axis to the line of action of the force.
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Which force component produces torque about a pivot: parallel or perpendicular to the radius vector?
Which force component produces torque about a pivot: parallel or perpendicular to the radius vector?
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Perpendicular component: $F_\perp=F\sin(\theta)$. Torque is produced only by the force component perpendicular to the position vector from the pivot.
Perpendicular component: $F_\perp=F\sin(\theta)$. Torque is produced only by the force component perpendicular to the position vector from the pivot.
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What is the SI unit of torque?
What is the SI unit of torque?
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$\text{N}\cdot\text{m}$. Torque units derive from force times perpendicular distance, yielding newton-meters in SI.
$\text{N}\cdot\text{m}$. Torque units derive from force times perpendicular distance, yielding newton-meters in SI.
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Identify the direction of torque given by the right-hand rule for $\vec{\tau}=\vec{r}\times\vec{F}$.
Identify the direction of torque given by the right-hand rule for $\vec{\tau}=\vec{r}\times\vec{F}$.
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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.
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.
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What sign convention is commonly used for planar torques in statics problems?
What sign convention is commonly used for planar torques in statics problems?
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Counterclockwise $+$, clockwise $-$. This convention assigns positive to counterclockwise and negative to clockwise for consistent torque summation in 2D problems.
Counterclockwise $+$, clockwise $-$. This convention assigns positive to counterclockwise and negative to clockwise for consistent torque summation in 2D problems.
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State the relationship between torque and angular acceleration for rotation about a fixed axis.
State the relationship between torque and angular acceleration for rotation about a fixed axis.
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$\sum \tau=I\alpha$. This rotational analog of Newton's second law relates net torque to angular acceleration via moment of inertia.
$\sum \tau=I\alpha$. This rotational analog of Newton's second law relates net torque to angular acceleration via moment of inertia.
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What is the definition of moment of inertia for point masses about an axis?
What is the definition of moment of inertia for point masses about an axis?
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$I=\sum m_ir_i^2$. Moment of inertia quantifies rotational inertia as the sum of each mass times its squared distance from the axis.
$I=\sum m_ir_i^2$. Moment of inertia quantifies rotational inertia as the sum of each mass times its squared distance from the axis.
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State the moment of inertia of a thin hoop (ring) of mass $M$ and radius $R$ about its center.
State the moment of inertia of a thin hoop (ring) of mass $M$ and radius $R$ about its center.
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$I=MR^2$. All mass elements of the hoop are at distance $R$ from the central axis, yielding this inertia value.
$I=MR^2$. All mass elements of the hoop are at distance $R$ from the central axis, yielding this inertia value.
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State the moment of inertia of a solid disk (or solid cylinder) of mass $M$ and radius $R$ about its center.
State the moment of inertia of a solid disk (or solid cylinder) of mass $M$ and radius $R$ about its center.
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$I=\frac{1}{2}MR^2$. Integration over the disk's uniform mass distribution results in half the inertia of an equivalent hoop.
$I=\frac{1}{2}MR^2$. Integration over the disk's uniform mass distribution results in half the inertia of an equivalent hoop.
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State the moment of inertia of a uniform rod of mass $M$ and length $L$ about its center (axis perpendicular to rod).
State the moment of inertia of a uniform rod of mass $M$ and length $L$ about its center (axis perpendicular to rod).
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$I=\frac{1}{12}ML^2$. Derived from integrating mass elements along the rod's length about its midpoint.
$I=\frac{1}{12}ML^2$. Derived from integrating mass elements along the rod's length about its midpoint.
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State the moment of inertia of a uniform rod of mass $M$ and length $L$ about one end (axis perpendicular to rod).
State the moment of inertia of a uniform rod of mass $M$ and length $L$ about one end (axis perpendicular to rod).
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$I=\frac{1}{3}ML^2$. Obtained using the parallel-axis theorem from the center-of-mass inertia, adding $M(L/2)^2$.
$I=\frac{1}{3}ML^2$. Obtained using the parallel-axis theorem from the center-of-mass inertia, adding $M(L/2)^2$.
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What is the rotational kinetic energy of a rigid body with moment of inertia $I$ and angular speed $\omega$?
What is the rotational kinetic energy of a rigid body with moment of inertia $I$ and angular speed $\omega$?
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$K=\frac{1}{2}I\omega^2$. Rotational kinetic energy is analogous to translational, substituting $I$ for $m$ and $\omega$ for $v$.
$K=\frac{1}{2}I\omega^2$. Rotational kinetic energy is analogous to translational, substituting $I$ for $m$ and $\omega$ for $v$.
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What is angular momentum for a rigid body rotating about a fixed axis?
What is angular momentum for a rigid body rotating about a fixed axis?
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$\vec{L}=I\vec{\omega}$. For rigid body rotation about a fixed axis, angular momentum is the product of inertia and angular velocity.
$\vec{L}=I\vec{\omega}$. For rigid body rotation about a fixed axis, angular momentum is the product of inertia and angular velocity.
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What is the relationship between net external torque and angular momentum?
What is the relationship between net external torque and angular momentum?
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$\sum \vec{\tau}_{\text{ext}}=\frac{d\vec{L}}{dt}$. This equation is the rotational equivalent of Newton's second law, linking torque to change in angular momentum.
$\sum \vec{\tau}_{\text{ext}}=\frac{d\vec{L}}{dt}$. This equation is the rotational equivalent of Newton's second law, linking torque to change in angular momentum.
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