Rigid Body Force Equation - Statics and Dynamics

1. Under what condition is the equation $\sum M_A = I_A \alpha$ valid for an arbitrary body-fixed point $A$ in planar motion, and what is the general correct form if that condition is not met?
   Answer: $\sum M_A = I_A \alpha$ is valid only when $A$ is the center of mass or a fixed pivot ($\mathbf{a}_A = 0$). The general correct form for any body-fixed point is $\sum M_A = I_A \alpha - m\,\mathbf{r}_{G/A} \times \mathbf{a}_A$, where the correction term accounts for the acceleration of point $A$; this reduces to $I_A \alpha$ precisely when $\mathbf{a}_A = 0$. Whenever you encounter moment equations for rigid bodies in planar motion, the critical question is: about which point are you summing moments, and is that point accelerating? The clean form $\sum M_A = I_A \alpha$ is only guaranteed when the moment center has zero acceleration. The general equation of motion for moments about an arbitrary body-fixed point $A$ is derived by combining the rotational and translational equations. The result is: $\sum M_A = I_G \alpha + \mathbf{r}_{G/A} \times m\mathbf{a}_G$ Using the parallel-axis theorem ($I_A = I_G + mr_{G/A}^2$) and expanding $\mathbf{a}_G$ in terms of $\mathbf{a}_A$, this simplifies to: $\sum M_A = I_A \alpha - m\,\mathbf{r}_{G/A} \times \mathbf{a}_A$ The correction term $m\,\mathbf{r}_{G/A} \times \mathbf{a}_A$ vanishes **only when $\mathbf{a}_A = 0$** — meaning $A$ is a fixed pivot or the center of mass (where $\mathbf{r}_{G/A} = \mathbf{0}$). This confirms B is correct. A is wrong because it states the correction term vanishes when $\mathbf{a}_G \parallel \mathbf{r}_{G/A}$, which misidentifies which acceleration matters — it's $\mathbf{a}_A$, not $\mathbf{a}_G$, that drives the correction. C is wrong because the parallel-axis theorem is purely geometric — it accounts for spatial offset but says nothing about the kinetics of an accelerating point. Acceleration of $A$ always introduces extra terms. D is wrong because the instantaneous center has zero velocity, not zero acceleration. An accelerating instantaneous center still requires the correction term, making $\sum M_{IC} = I_{IC}\alpha$ generally invalid. Study tip: Memorize the rule — "fixed point or center of mass only." Any other moment center requires a correction for its acceleration.
2. Which of the following statements about the forces and motion of this rigid body is necessarily true at that instant?
   Answer: The net force on the body is zero ($\sum \mathbf{F} = 0$), but the net moment about $G$ may be nonzero, allowing the body to have a nonzero angular acceleration $\alpha$ about its center of mass at that instant. When analyzing a rigid body in planar motion, you should immediately reach for Newton's second law in both translational and rotational forms: $\sum \mathbf{F} = m\mathbf{a}_G$ and $\sum M_G = I_G \alpha$. The critical insight is that these are independent equations — one governs translation, the other governs rotation, and neither forces the other to be zero. Since $\mathbf{a}_G = 0$, the translational equation directly gives $\sum \mathbf{F} = m(0) = 0$. That's definitive. But the rotational equation $\sum M_G = I_G \alpha$ stands completely on its own — a nonzero net moment about $G$ can still produce angular acceleration $\alpha \neq 0$ regardless of what's happening translationally. Think of a figure skater with zero translational acceleration who is still spinning and changing spin rate: net force zero, net torque nonzero. Answer C correctly captures this independence. Answer A is wrong because it claims $\alpha$ is "coupled" to $a_G$ through kinematics, implying $\alpha = 0$. This confuses kinematics of a point on the body (where $\alpha$ does affect accelerations of non-G points) with the independent rotational equation of motion — they are not the same thing. Answer B goes further astray by concluding the body must be in full static equilibrium. $\mathbf{a}_G = 0$ says nothing about $\alpha$, and "static equilibrium" requires both $\sum \mathbf{F} = 0$ and $\sum M = 0$ to be imposed simultaneously — that's an assumption here, not a consequence. Answer D is subtly wrong because it incorrectly concludes that moments about off-center points must also be zero. The moment about an arbitrary point $P$ includes a term involving $\mathbf{r}_{PG} \times m\mathbf{a}_G$, which vanishes when $\mathbf{a}_G = 0$, but the remaining term $I_G\alpha$ can still be nonzero. Your study tip: always keep the two Newton's law equations for rigid bodies mentally separate. Zero translational acceleration $\Rightarrow$ zero net force. It says nothing about torque or angular acceleration.
3. Immediately after release, which of the following correctly applies $\sum F_y = m a_{Gy}$ to determine the vertical pin reaction $A_y$? (Take downward as positive; $a_{Gy}$ is the vertical acceleration of the center of mass.)
   Answer: $mg - A_y = m a_{Gy}$, with $a_{Gy} = \tfrac{3g}{4}$ (downward) from the rotational kinematics, giving $A_y = mg/4$ upward, meaning the pin supports only one-quarter of the rod's weight at the instant of release. When a rigid body rotates about a fixed pin, you must find the acceleration of the center of mass using rotational kinematics — not intuition about "starting from rest." The key steps are: (1) find angular acceleration $\alpha$ from the moment equation, (2) find $a_{G}$ from kinematics, then (3) apply Newton's second law. Taking moments about the pin immediately after release: $\sum M_A = I_A \alpha$ gives $mg\cdot\frac{L}{2} = \frac{1}{3}mL^2\,\alpha$, so $\alpha = \frac{3g}{2L}$. The center of mass sits at $L/2$ from the pin and moves in a circle. At the instant of release the rod has zero angular velocity, so the centripetal acceleration is zero. Only the tangential (downward) component exists: $a_{Gy} = \alpha\cdot\frac{L}{2} = \frac{3g}{2L}\cdot\frac{L}{2} = \frac{3g}{4}$. Applying $\sum F_y = ma_{Gy}$ with downward positive: $mg - A_y = m\cdot\frac{3g}{4}$, so $A_y = \frac{mg}{4}$ upward. This is exactly what A describes — the correct answer. B is wrong because it uses $a_{Gy} = g/2$, which comes from incorrectly computing the tangential acceleration (using $g/L$ rather than $3g/(2L)$ for $\alpha$). C flips the sign convention, claiming the center accelerates upward, which contradicts the physics — gravity pulls the free end down. D confuses zero velocity with zero acceleration. Starting from rest means $\omega = 0$, not $\alpha = 0$; the rod is still acted on by gravity, producing nonzero angular acceleration immediately. As a strategy, always distinguish velocity from acceleration. "Released from rest" only tells you $\omega = 0$ at that instant — you must still compute $\alpha$ from the moment equation to find the true acceleration of the center of mass.
4. Applying $\sum F_x = m a_{Gx}$ in an inertial frame, where $a_{Gx}$ is the horizontal acceleration of the bar's center of mass, which expression correctly relates the horizontal pin force $A_x$ to the system parameters? (Take rightward as positive; $\alpha$ is positive counterclockwise; treat the bar as a free body.)
   Answer: $A_x = m(a_0 - \alpha L/2)$, because the center of mass acceleration is $a_{Gx} = a_0 + (\alpha_{rel})(L/2)$, where for a bar hanging vertically a counterclockwise $\alpha$ swings $G$ to the left relative to $A$, so the tangential contribution is $-\alpha L/2$ in the rightward-positive convention, giving $A_x = m(a_0 - \alpha L/2)$. When a rigid body moves in a plane, the acceleration of any point on it equals the acceleration of a reference point plus the relative acceleration of that point with respect to the reference. For the bar's center of mass $G$, located a distance $L/2$ below pin $A$: $\vec{a}_G = \vec{a}_A + \vec{\alpha} \times \vec{r}_{G/A} - \omega^2 \vec{r}_{G/A}$ Since $\omega = 0$, the centripetal term vanishes. With the bar hanging vertically, $\vec{r}_{G/A} = -\frac{L}{2}\hat{j}$. A counterclockwise $\alpha$ gives $\vec{\alpha} \times \vec{r}_{G/A} = \alpha\hat{k} \times (-\frac{L}{2}\hat{j}) = -\frac{\alpha L}{2}\hat{i}$. So the horizontal component of $G$'s acceleration is $a_{Gx} = a_0 - \frac{\alpha L}{2}$. Applying Newton's second law horizontally: $A_x = m\!\left(a_0 - \frac{\alpha L}{2}\right)$, confirming C is correct. Choice A is wrong because it claims the tangential contribution always adds to $a_0$. The sign depends entirely on geometry — the cross product shows it subtracts here. Choice B ignores rotation entirely, treating $G$ as if it were pinned directly to the slider; since the bar can rotate, $G$ does not share $A$'s acceleration. Choice D mixes up the equations of motion: weight $mg$ acts vertically and belongs in the $y$-equation, not the horizontal $x$-equation. Study tip: Always compute $\vec{a}_G$ explicitly using the relative-acceleration formula before writing Newton's law. The direction of the tangential term is determined by the cross product $\vec{\alpha} \times \vec{r}_{G/A}$ — never assume it simply adds or subtracts without working through the geometry.
5. A student writes the equation of motion for block $A$ as $T_A = 2m \cdot a$ and for block $B$ as $mg - T_B = m \cdot a$. She then applies $\sum F = M a_G$ to the pulley's center and writes $T_B - T_A = M a_G$. What is the fundamental error in her analysis of the pulley, and what is the correct system acceleration?
   Answer: The pulley's center does not translate, so $a_G = 0$ for the pulley and $\sum F = M a_G = 0$ is satisfied by bearing forces, not by $T_B - T_A$. The pulley requires a moment equation $\sum M_G = I_G \alpha$, giving $(T_B - T_A)R = (MR^2/2)\alpha = (MR^2/2)(a/R)$, so $T_B - T_A = Ma/2$. The correct acceleration is $a = mg/(2m + m + M/2)$. When a pulley has mass and the cord doesn't slip, you must treat it as a rigid body in rotation, not a translating particle. The key question to ask yourself: does the pulley's center move? In this classic Atwood-with-pulley setup, the axle is fixed — the pulley spins in place, so its center of mass has zero acceleration. This is exactly what makes D correct. Since $a_G = 0$, writing $\sum F = Ma_G$ for the pulley gives zero — that equation is satisfied by the bearing (axle) reaction forces, which absorb the net cord tension difference. It tells you nothing useful about the dynamics. Instead, you need the rotational equation about the fixed axle: $\sum M_G = I_G \alpha$. For a solid cylinder, $I_G = MR^2/2$, and the no-slip condition gives $\alpha = a/R$. So $(T_B - T_A)R = \frac{MR^2}{2} \cdot \frac{a}{R}$, which simplifies to $T_B - T_A = \frac{Ma}{2}$. Combining with block A ($T_A = 2ma$) and block B ($mg - T_B = ma$), you get $a = \frac{mg}{3m + M/2}$. A is mathematically equivalent in its final answer but frames the fix as an "effective mass" translational equation — this obscures the actual physics and misidentifies the error as treating the pulley as a particle rather than misapplying the translational equation to a non-translating body. B makes a similar structural mistake: it applies a modified $\sum F$ equation to the pulley as if the center translates, which is physically wrong even if the algebra accidentally resembles the correct result. C correctly identifies that bearing forces exist but draws the wrong conclusion — bearing forces do not force $T_A = T_B$; that would only be true for a massless pulley. $M \neq 0$ means the tensions differ. Your study tip: whenever a pulley has mass and a fixed axle, immediately switch to a moment equation — the translational equation for the pulley is useless (or just gives you the bearing force).
6. Applying $\sum F_x = m a_G$ to the disk, which of the following correctly identifies the net horizontal force equation, and what is the resulting acceleration of the center of mass?
   Answer: $P - f = m a_G$, giving $a_G = 3.33 \text{ m/s}^2$, where $f$ is the friction force found from the rolling constraint and the moment equation to be $P/3 = 13.33 \text{ N}$. When a disk rolls without slipping, friction is a real horizontal force that appears in your equations — it's what enforces the rolling constraint. The key is finding its actual value through the combined translational and rotational equations, not assuming it equals its maximum value. For a disk rolling without slipping, you have two governing equations. Translating: $P - f = ma_G$, and rotating about the contact point (or center): $Pr = I_G \alpha$, where $I_G = \frac{1}{2}mr^2$ and the rolling constraint gives $a_G = r\alpha$. Substituting, the moment equation yields $f = P/3 = 40/3 \approx 13.33 \text{ N}$. Plugging back: $a_G = (P - f)/m = (40 - 13.33)/8 \approx 3.33 \text{ m/s}^2$. This confirms A is correct. You should also verify $f < \mu_s mg = 0.25(8)(9.8) = 19.6 \text{ N}$, confirming rolling without slipping is valid. B is wrong because it ignores friction entirely, treating the disk like a particle sliding on a frictionless surface. Friction absolutely acts horizontally at the contact point for a rolling body. C uses $f = \mu_s mg$ as if the disk is on the verge of slipping — but the problem doesn't state that condition. You must derive $f$ from the rolling constraint, not assume maximum friction. D gets the direction of friction backwards; friction acts opposite to $P$ (backward) to prevent the contact point from sliding forward, so it reduces, not increases, $a_G$. Your study tip: for any rolling-without-slipping problem, always write both $\sum F_x = ma_G$ and $\sum M = I\alpha$ with the constraint $a_G = r\alpha$ — only then can you correctly solve for the unknown friction force.
7. Immediately after one cable is cut, before the plate has rotated appreciably, applying $\sum F_y = m a_G$ to find the acceleration of the center of mass requires knowing the tension in the remaining cable. Using $\sum M$ about the attachment point of the remaining cable to eliminate that tension, what is the magnitude of $a_G$ immediately after the cut, expressed as a fraction of $g$?
   Answer: $a_G = \tfrac{3}{4}g$, because the moment equation about the remaining support gives a constraint that, combined with $\sum F_y = M a_G$, yields the remaining cable tension as $Mg/4$, so the net downward force is $\tfrac{3}{4}Mg$. When a rigid body is suddenly released from one support while another remains, you're dealing with constrained motion, not free fall. The plate can only rotate about the remaining cable's attachment point (at least instantaneously), so you must apply both Newton's second law and the rotational equation simultaneously. Here's the key setup: let the plate have length $2L$, with the remaining cable at one end and the center of mass at distance $L$ from it. Taking moments about the remaining cable's attachment point eliminates the unknown tension $T$ entirely. The only moment comes from gravity acting at the center: $\sum M = MgL = I_{\text{end}} \cdot \alpha$ where $I_{\text{end}} = \frac{1}{3}M(2L)^2 = \frac{4}{3}ML^2$. Solving gives $\alpha = \frac{3g}{4L}$. The acceleration of the center of mass (at distance $L$) is $a_G = \alpha L = \frac{3g}{4}$. Applying $\sum F_y = Ma_G$ confirms the tension: $Mg - T = M\cdot\frac{3g}{4}$, so $T = \frac{Mg}{4}$. This confirms A is correct. B is wrong because the remaining cable doesn't instantly go slack — the rotational constraint means it still pulls upward, just with reduced force. C applies static reasoning (half the weight) to a dynamic situation, ignoring the rotational acceleration entirely. D gets the tension and net force backwards: it correctly finds $T = \frac{3Mg}{4}$, but then mistakenly applies that tension value as the net force rather than computing $Mg - T$. Your study tip: whenever a support is suddenly removed, resist the instinct to use static equilibrium. Set up $\sum M$ about the remaining support first — it's your fastest path to eliminating the unknown reaction and finding $\alpha$.
8. Applying $\sum F_x = m a_G$ and $\sum F_y = m a_{Gy} = 0$ to the crate, along with a moment equation, a student must determine whether the crate slides, tips, or remains stationary. Which outcome is correct, and what is the critical check that distinguishes tipping from sliding?
   Answer: The crate tips before sliding. Since $f = ma = 250 \text{ N} < \mu_s mg = 294 \text{ N}$, sliding is prevented. The tipping condition requires the normal force to reach the forward bottom edge; the critical deceleration for tipping is $a_{tip} = g(w/h) = 9.81(0.8/1.2) = 6.54 \text{ m/s}^2$. Since $a = 5 \text{ m/s}^2 < 6.54 \text{ m/s}^2$, tipping does not occur and the crate remains stationary. When a crate sits on a decelerating truck, you face a two-part check: can friction prevent sliding, and does the inertial overturning moment cause tipping? Work through both systematically. Sliding check first. The friction force required to accelerate the crate with the truck is $f = ma = 50(5) = 250 \text{ N}$. The maximum static friction available is $\mu_s mg = 0.6(50)(9.81) = 294 \text{ N}$. Since $250 < 294$, friction is sufficient — the crate does not slide. Tipping check next. As the truck decelerates, the inertial tendency shifts the normal force toward the forward (front) edge. Tipping occurs when the normal force reaches exactly that edge, meaning it can no longer create a restoring moment. Taking moments about the forward bottom edge, the critical deceleration at which tipping begins is: $a_{\text{tip}} = g\frac{w}{h} = 9.81\frac{0.8}{1.2} = 6.54 \text{ m/s}^2$ Since the actual deceleration $a = 5 \text{ m/s}^2 < 6.54 \text{ m/s}^2$, tipping does not occur. The crate remains stationary — answer D. A is wrong because it reaches the correct conclusion (no sliding, no tipping) but uses a garbled formula with an incorrect sign and messy logic for the normal force location. B is wrong because it incorrectly claims $250 \text{ N} > 294 \text{ N}$, a straightforward arithmetic error. C correctly identifies that sliding doesn't occur and correctly places the normal force inside the base, but never computes or compares a critical tipping acceleration — it checks only the current state without establishing the general tipping criterion. Your strategy: always separate the two checks — sliding (force comparison) and tipping (moment or critical-acceleration comparison) — and solve them independently before drawing a conclusion.
9. Before assuming rolling without slipping, a student must verify whether slipping occurs. Using $\sum F$ along the incline and $\sum M_G$ about the center, what is the required friction force for no-slip rolling, and does the cylinder actually roll without slipping?
   Answer: Required friction $f = mg\sin\theta/3 = 16.35 \text{ N}$; maximum static friction $= \mu_s mg\cos\theta = 33.97 \text{ N}$. Since the required friction is less than the maximum, the cylinder rolls without slipping and the correct acceleration is $a_G = \tfrac{2}{3}g\sin\theta = 3.27 \text{ m/s}^2$. When a cylinder rolls on an inclined plane, you need two equations: Newton's second law along the incline and the rotational equation about the center of mass. For a uniform cylinder, the moment of inertia is $I_G = \tfrac{1}{2}mR^2$. The rolling constraint links linear and angular acceleration: $a_G = R\alpha$. Setting up the equations: along the incline, $mg\sin\theta - f = ma_G$, and rotationally, $fR = I_G\alpha = \tfrac{1}{2}mR^2\cdot(a_G/R)$, giving $f = \tfrac{1}{2}ma_G$. Substituting back, $mg\sin\theta - \tfrac{1}{2}ma_G = ma_G$, so $a_G = \tfrac{2}{3}g\sin\theta = 3.27\ \text{m/s}^2$, and the required friction is $f = \tfrac{1}{3}mg\sin\theta = 16.35\ \text{N}$. The maximum static friction is $\mu_s mg\cos\theta = 0.40\times10\times9.81\times\cos30° = 33.97\ \text{N}$. Since $16.35 < 33.97$, the cylinder rolls without slipping. This confirms B is correct. Choice A uses $f = mg\sin\theta/2$ and $a_G = \tfrac{1}{2}g\sin\theta$, which corresponds to a hollow cylinder ($I_G = mR^2$), not a uniform solid cylinder — a classic mix-up of moment of inertia formulas. Choice C gets the required friction and acceleration right but computes maximum static friction as $\mu_s mg$ instead of $\mu_s mg\cos\theta$, forgetting that the normal force on an incline is reduced by the cosine factor. Choice D finds the correct friction force but then states $a_G = \tfrac{1}{3}g\sin\theta$, which is simply an algebra error in solving for acceleration. Study tip: Always write both $\sum F$ and $\sum M_G$ explicitly before solving — verify the no-slip assumption by checking $f_{\text{required}} \leq \mu_s N$, where $N = mg\cos\theta$ on an incline, not $mg$.

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