Statics and Dynamics Quiz: Rigid Body Moment Equation

What this quiz covers

This quiz focuses on Rigid Body Moment Equation, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The claim is entirely correct: the pin at G exerts no force because a couple produces pure rotation with no net force, the center of mass does not accelerate, and $\alpha = 12\mathcal{M}/(mL^2)$ follows directly from $\Sigma M_G = I_G\alpha$.
2. The claim is incorrect because $\Sigma M_G = I_G\alpha$ cannot be applied when the rod is pinned; instead, the student must use $\Sigma M_{\text{pin}} = I_{\text{pin}}\alpha$, where $I_{\text{pin}} = mL^2/3$, giving $\alpha = 3\mathcal{M}/(mL^2)$.
3. The claim is incorrect because applying a couple to a rod pinned at its center violates the assumption of rigid-body motion; couples can only be applied to free bodies, and the pin constraint changes $\mathcal{M}$ into an equivalent force-couple system that alters $I_G$.
4. 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. (correct answer)

Question 2

1. Only gravity contributes a moment about G because the cable tension, acting at the upper-left corner, passes through a point that is equidistant from G in both directions, causing its moment to vanish by symmetry of the rectangular plate geometry.
2. 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. (correct answer)
3. Both the remaining cable tension and gravity contribute moments about G: the cable tension acts at the upper-left corner (nonzero distance from G), and gravity acts at G with zero moment arm, so only tension produces a moment — but this moment must equal zero at the instant of release because the system was previously in equilibrium.
4. Only gravity contributes a moment about G because the tension in the remaining cable is an internal force of the plate-cable system and internal forces produce no net moment about the center of mass in rigid-body dynamics.

Question 3

1. $Pr - fR = I_G\alpha$, 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. (correct answer)
2. $PR - fr = I_G\alpha$, where $P$ acts at the outer radius $R$ because the cord exits from the outer flange, and friction at the contact point acts at the inner hub radius $r$, reversing the conventional radius assignments for this geometry.
3. $Pr + fR = I_G\alpha$, where both $P$ and $f$ contribute clockwise moments: $P$ at the hub pulls right creating clockwise rotation, and friction acts to the right at the contact point because the spool's base tends to slip backward, also producing a clockwise moment about G.
4. $fR - Pr = I_G\alpha$, where friction at the outer contact radius $R$ drives clockwise rotation and $P$ at the inner hub radius $r$ resists it, reflecting an assumption that the spool rolls in the direction of friction rather than in the direction of the applied cord force.

Question 4

1. $a_G = \frac{F}{2m}$, found by applying $\Sigma M_G = I_G\alpha$ for a disk (using $I_G = \frac{1}{2}mR^2$ instead of $mR^2$): force $F$ at the top gives moment $FR$, friction $f$ at the bottom gives moment $fR$ in the opposite direction, and solving with $\Sigma F_x = ma_G$ yields $a_G = F/(2m)$.
2. $a_G = \frac{2F}{3m}$, calculated by using the moment of inertia of a disk ($I_G = \frac{1}{2}mR^2$) and then applying the no-slip constraint incorrectly as $a_G = 2R\alpha/3$, which underestimates the rotational inertia and gives an intermediate value between the disk and hoop results.
3. $a_G = \frac{F}{m}$, 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 $a_G = F/m$. (correct answer)
4. $a_G = \frac{F}{3m}$, obtained by taking moments about the contact point using $I_{\text{contact}} = 2mR^2$ and moment arm $2R$ for $F$, giving $F(2R) = 2mR^2\alpha$, but then incorrectly substituting $\alpha = a_G/(2R)$ instead of the correct no-slip relation $\alpha = a_G/R$.

Question 5

1. $\alpha = -\frac{6g}{L}$, found by taking the moment of gravity (acting downward at G, located $L/2$ from A) about G using moment arm $L/2$, giving $-mg(L/2) = I_G\alpha$, so $\alpha = -mg(L/2)/(mL^2/12) = -6g/L$.
2. $\alpha = -\frac{3g}{2L}$, obtained by applying $\Sigma M_A = I_A\alpha$ about the pin at A instead of about G: gravity acts at G, which is $L/2$ from A horizontally when the rod is horizontal, giving $-mg(L/2) = \frac{1}{3}mL^2\alpha$, so $\alpha = -3g/(2L)$.
3. $\alpha = -\frac{3g}{2L}$, 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 $\alpha = -3g/(2L)$. (correct answer)
4. $\alpha = -\frac{g}{L}$, found by applying $\Sigma M_G = I_G\alpha$ and including both the vertical pin reaction $A_y$ and the horizontal pin reaction $A_x$ as moment contributors about G, with $A_x$ assigned a moment arm of $L/2$ equal to the vertical distance from G to the line of action of $A_x$, which overcounts the pin's contribution.

Question 6

1. 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 $\Sigma F = ma_G$. (correct answer)
2. Taking moments about B is simpler because $I_B < I_G$ for any rigid body by the parallel-axis theorem, so the rotational inertia is minimized at the fixed pin, reducing the magnitude of the right-hand side and making the algebra easier to check numerically.
3. Taking moments about B is only valid when B coincides with the center of mass. Since G is at $(L/4, L/4)$ and not at B, $\Sigma M_B = I_B\alpha$ is an approximation, whereas $\Sigma M_G = I_G\alpha$ is the exact form of Newton's second law for rotation and should always be preferred.
4. Taking moments about B avoids computing the location of G, since no formula for the position of G is needed to write $\Sigma M_B = I_B\alpha$. The method about G also has this advantage, so both approaches are equally efficient; the real benefit of using B is simply that $I_B$ is tabulated directly for common shapes.