Rigid Body Moment Equation - Statics and Dynamics

1. Applying $\Sigma M_G = I_G \alpha$ about the center of the spool, and assuming the spool rolls without slipping, which equation is correct (taking clockwise as positive $\alpha$)?
   Answer: $Pr - fR = I_G\alpha$, where the applied force $P$ at the hub (radius $r$) creates a clockwise moment and friction $f$ at the outer flange contact (radius $R$) creates a counterclockwise moment, with friction assumed to act to the left opposing the tendency of the contact point to slip rightward. When analyzing the rotation of a spool about its center of mass, your first job is to identify each force, where it acts, and what rotational direction its moment creates. Treat clockwise as positive and carefully assign the correct moment arm (perpendicular distance from G to the line of action). The horizontal cord is wrapped around the inner hub, so force $P$ acts at radius $r$. Pulling right on the top side of the hub produces a clockwise moment: $+Pr$. The friction force acts at the outer contact point, radius $R$. Because the cord pulls the spool rightward, the base of the spool tends to slip to the right, so static friction acts to the left — which produces a counterclockwise moment about G: $-fR$. Applying Newton's second law for rotation gives $Pr - fR = I_G\alpha$, confirming A is correct. Choice B incorrectly swaps the radii, assigning $P$ to $R$ and $f$ to $r$ — a straightforward geometry error that ignores where the cord is actually wrapped. Choice C claims both $P$ and $f$ produce clockwise moments. This is wrong because leftward friction at the bottom of the spool creates a counterclockwise moment about G, not clockwise. Getting the sign of the friction moment is the most common mistake on spool problems. Choice D reverses the roles entirely, treating friction as the driver and $P$ as the resistor — a misread of the physical setup that has no basis in the given geometry. Study tip: For any spool or disk problem, always sketch the force, mark its point of application, and use the right-hand rule (or your sign convention) to determine moment direction before writing the equation. Sign errors on friction are the #1 source of mistakes here.
2. After the impulse, the rod swings. At the instant when the rod has rotated $90°$ from vertical to horizontal, a student applies $\Sigma M_G = I_G\alpha$ to find the angular acceleration $\alpha$ at that instant. Which expression correctly gives $\alpha$ at this horizontal position (taking counterclockwise as positive, with the rod extending to the right from pin A)?
   Answer: $\alpha = -\frac{3g}{2L}$, found by applying $\Sigma M_G = I_G\alpha$ where only the vertical pin reaction $A_y$ at A (located $L/2$ from G) contributes a moment about G; gravity acts at G and contributes no moment. After solving $\Sigma F_y = ma_{Gy}$ for $A_y$ and substituting, one obtains $\alpha = -3g/(2L)$. When a rigid body rotates about a fixed pin, you can apply Newton's second law for rotation about any point — but your choice of reference point changes which forces appear as moment contributors. This question tests whether you understand that taking moments about the center of mass G eliminates gravity (which acts at G) but keeps pin reactions in play. When the rod is horizontal, gravity acts downward at G and produces zero moment about G (zero moment arm). The pin at A, located $L/2$ from G, exerts reactions $A_x$ (horizontal) and $A_y$ (vertical). With the rod horizontal and extending rightward, $A_y$ acts vertically at a horizontal distance $L/2$ from G, so it does create a moment. $A_x$ acts horizontally at A, which is at the same height as G — zero vertical moment arm, zero moment contribution. Applying $\Sigma F_y = ma_{Gy}$, where $a_{Gy} = -\alpha(L/2)$ (centripetal terms vanish at $\omega=0$ if we consider the instant after impulse dies, but in general you solve for $A_y$), substitution into $\Sigma M_G = I_G\alpha$ yields $\alpha = -\frac{3g}{2L}$. That's C, the correct answer. A is wrong because it applies $\Sigma M_G = I_G\alpha$ but incorrectly treats gravity as having a moment arm of $L/2$ about G — gravity acts at G, so its moment arm is zero. B uses $\Sigma M_A = I_A\alpha$, which is valid and gives the same $\alpha = -\frac{3g}{2L}$, but the question asks specifically about applying the $\Sigma M_G = I_G\alpha$ formulation — so B arrives at a correct numerical answer through a method that doesn't answer what's being asked. D incorrectly assigns $A_x$ a nonzero moment arm about G. Since A and G are at the same height when the rod is horizontal, $A_x$ is parallel to the rod with no perpendicular distance to G — its moment is zero. Your study tip: always identify the moment arm geometrically before plugging in. Forces acting through your reference point or parallel to the position vector contribute zero moment — don't let their magnitudes tempt you into including them.
3. Immediately after the right cable is cut, the plate begins to rotate. A student applies $\Sigma M_G = I_G \alpha$ about the center of mass G. Which of the following statements correctly identifies the forces that contribute moments about G in this equation at the instant of cable cutting?
   Answer: Both the remaining cable tension and gravity contribute moments about G, since gravity acts at G (zero moment arm) but the cable tension acts at the upper-left corner at a perpendicular distance from G, making tension the only force producing a nonzero moment about G. When analyzing rigid-body motion immediately after a constraint is removed, you need to carefully inventory every external force acting on the body and determine each force's moment arm about your chosen reference point — in this case, the center of mass G. At the instant the right cable is cut, exactly two external forces act on the plate: the remaining cable tension $T$ pulling upward at the upper-left corner, and gravity $mg$ pulling downward at G. Since gravity acts at G, its moment arm about G is zero — it contributes no moment. The cable tension, however, acts at the upper-left corner, which is displaced from G by both horizontal ($b/2$) and vertical ($h/2$) distances, giving a nonzero perpendicular distance and therefore a nonzero moment. So $\Sigma M_G = T \cdot d \neq 0$, which drives the angular acceleration $\alpha$. Answer B captures this correctly. Answer A is wrong because the cable does not pass through G — the upper-left corner is offset from G in both directions, so claiming the moment vanishes by "symmetry" is geometrically false. Answer C contains correct reasoning up to a point but then makes a critical error: equilibrium before the cut tells you the net force and moment were zero under two cables. After one cable is cut, equilibrium no longer holds, and the remaining tension's moment about G is definitely nonzero — the system is now accelerating. Answer D confuses internal and external forces. The cable is attached to the plate but is external to the plate itself; it absolutely contributes to $\Sigma M_G$. Study tip: Always re-evaluate your free-body diagram the instant a constraint changes — prior equilibrium conditions do not carry over to the new loading scenario.
4. A student wants to find $\alpha$ using $\Sigma M_G = I_G\alpha$. The center of mass G of the L-shaped body is located at $(L/4, L/4)$ from B. Which of the following correctly states why it is more efficient (and less error-prone) to apply $\Sigma M_B = I_B\alpha$ about the fixed pin B instead of $\Sigma M_G = I_G\alpha$?
   Answer: Taking moments about B eliminates the unknown pin reaction forces from the moment equation, since the pin forces act at B and have zero moment arm about B. By contrast, $\Sigma M_G = I_G\alpha$ requires the pin reaction forces explicitly — they act at B, which is offset from G, so they contribute moments about G that must first be found from the translational equations $\Sigma F = ma_G$. Whenever you see a rigid body rotating about a fixed pin, your first instinct should be to ask: "Can I take moments directly about that pin?" The answer is almost always yes — and the payoff is enormous. When you apply $\Sigma M_B = I_B\alpha$ about the fixed pin B, the unknown reaction forces at the pin (which can have both x- and y-components) act at B. Because their moment arm about B is zero, they vanish from the equation entirely. You're left with only the applied force $F$ contributing a moment, and you can solve for $\alpha$ in one clean step. This is exactly what answer A captures: the pin reactions are eliminated automatically, making the equation self-contained. Contrast this with $\Sigma M_G = I_G\alpha$. Here, G is offset from B at $(L/4, L/4)$, so the pin reaction forces do have a nonzero moment arm about G — they produce moments you must account for. That forces you to first solve the translational equations $\Sigma \vec{F} = m\vec{a}_G$ to find those reaction components, then substitute them into the moment equation. It's solvable, but it's a two-step process with more algebra and more opportunities for sign errors. Answer B is wrong because the parallel-axis theorem actually guarantees $I_B > I_G$ (not less), since B is not the center of mass. Answer C is a dangerous misconception — $\Sigma M_B = I_B\alpha$ is perfectly exact for any fixed pivot, not an approximation. Answer D is wrong because using G does require locating G, and the two methods are decidedly not equally efficient. The study tip here: whenever a problem has a fixed pivot, go straight to $\Sigma M_{\text{pivot}} = I_{\text{pivot}}\alpha$. Reaction forces at a fixed point always have zero moment arm about that point — exploit that to bypass the translational equations entirely.
5. The rod is initially at rest. After the couple $\mathcal{M}$ is applied, a student claims: *'Because the pin is at the center of mass, the pin exerts no force on the rod, so the translational equations $\Sigma F = ma_G$ give $a_G = 0$ and the moment equation $\Sigma M_G = I_G\alpha$ gives $\alpha = \frac{12\mathcal{M}}{mL^2}$.'* Which of the following best evaluates this claim?
   Answer: The claim is partially correct: $\alpha = 12\mathcal{M}/(mL^2)$ is correct, but the conclusion that the pin exerts no force is wrong — the pin must exert a force equal to $mg$ upward to prevent gravitational acceleration, though this force does not affect $\alpha$ since it acts at G. When a rigid body is pinned at its center of mass, you need to carefully separate two independent equations: the translational equation $\Sigma F = ma_G$ and the rotational equation $\Sigma M_G = I_G\alpha$. The key insight is that these equations govern different aspects of motion and must each be satisfied simultaneously. The student's rotational analysis is perfectly sound. A pure couple $\mathcal{M}$ produces a net moment but zero net force, so $\Sigma M_G = \mathcal{M} = I_G\alpha$ gives $\alpha = \frac{\mathcal{M}}{mL^2/12} = \frac{12\mathcal{M}}{mL^2}$. That result is correct. However, the claim that the pin exerts no force is wrong. Gravity still acts downward on the rod's center of mass with magnitude $mg$. For the center of mass to remain stationary ($a_G = 0$), the pin must supply an upward reaction force equal to $mg$. This is a static equilibrium condition in the vertical direction, not something the couple eliminates. Crucially, because this pin force acts at point G, it contributes zero moment to $\Sigma M_G$, so it does not affect $\alpha$ at all. This makes D correct. A is wrong because it incorrectly concludes the pin exerts no force — gravity demands a vertical reaction regardless of the applied couple. B is wrong because $\Sigma M_G = I_G\alpha$ is always valid for moments about the center of mass, even for pinned bodies; using $I_\text{pin}$ would only apply if you summed moments about the pin instead. C is wrong because couples can absolutely be applied to constrained bodies — the pin constraint does not alter $I_G$ or invalidate the couple. Study tip: Always check both translational and rotational equations separately. A pin at G eliminates translational acceleration but still supplies whatever reaction force equilibrium requires — never assume a pin at G means zero pin force.
6. Using $\Sigma M_G = I_G\alpha$ and the no-slip constraint $a_G = R\alpha$, what is the acceleration $a_G$ of the hoop's center?
   Answer: $a_G = \frac{F}{m}$, found by correctly applying $\Sigma M_G = I_G\alpha$ with $I_G = mR^2$: force $F$ at the top (moment $FR$) and friction $f$ at the bottom (moment $fR$, opposite direction) combine with $\Sigma F_x = F + f = ma_G$ to give $f = 0$ and $a_G = F/m$. When a hoop rolls without slipping, you need two equations working together: the translational equation $\Sigma F_x = ma_G$ and the rotational equation $\Sigma M_G = I_G\alpha$, linked by the no-slip constraint $a_G = R\alpha$. For the hoop, force $F$ acts at the top (moment arm $R$, producing moment $FR$ clockwise), and friction $f$ acts at the contact point (moment arm $R$, producing moment $fR$ counterclockwise). The rotational equation gives $FR - fR = mR^2\alpha$. The translational equation gives $F + f = ma_G$. Substituting $\alpha = a_G/R$ into the first equation: $FR - fR = mR^2(a_G/R) = mRa_G$, which simplifies to $F - f = ma_G$. Adding this to $F + f = ma_G$ gives $2F = 2ma_G$, so $a_G = F/m$ and $f = 0$. This is answer C, and the striking result is that friction vanishes entirely for a hoop under this loading. Choice A uses $I_G = \frac{1}{2}mR^2$, which is the moment of inertia for a disk, not a hoop — a critical substitution error. Choice B compounds two mistakes: wrong $I_G$ (disk value) and a fabricated no-slip constraint, neither of which applies here. Choice D attempts a moment-about-contact-point approach correctly at first ($I_\text{contact} = 2mR^2$, moment arm $2R$), but then corrupts the no-slip relation by writing $\alpha = a_G/(2R)$ instead of $\alpha = a_G/R$. Your go-to strategy: always confirm which body you're analyzing (hoop vs. disk vs. sphere) and write down the correct $I_G$ before setting up any moment equation — that single substitution determines your entire solution.

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