Planar Rigid-Body Motion - Statics and Dynamics

1. Immediately after release from the horizontal position, what is the minimum distance $d$ from the pin O at which the block must be placed so that the block does not press against the rod (i.e., the normal force between the block and rod is zero)?
   Answer: $d = \dfrac{2L}{3} = 0.80 \text{ m}$, because at this distance the tangential acceleration of the rod at that point equals $g$, meaning the rod accelerates away from the block faster than gravity alone pulls the block downward. When a pinned rod is released from rest, every point on it undergoes a tangential acceleration (downward, since the rod starts horizontal) given by $a_t = \alpha \cdot d$, where $\alpha$ is the angular acceleration. The block, meanwhile, is pulled downward only by gravity $g$. If the rod's tangential acceleration at point $d$ exceeds $g$, the rod "falls away" from the block, and the normal force drops to zero. Your goal is to find the minimum $d$ where $a_t = g$. For a uniform rod pinned at one end, the rotational equation of motion (using $I = \frac{1}{3}mL^2$ and torque from gravity acting at the center) gives: $\alpha = \frac{mg \cdot \frac{L}{2}}{\frac{1}{3}mL^2} = \frac{3g}{2L}$ Setting the tangential acceleration equal to $g$: $\alpha \cdot d = g \implies \frac{3g}{2L} \cdot d = g \implies d = \frac{2L}{3}$ This confirms answer A is correct. The frictionless block doesn't affect $\alpha$ because it exerts only a normal (perpendicular) force on the rod — it contributes no torque about O. The mass $M$ and distance $d$ are irrelevant to computing $\alpha$. Answer B is wrong on two counts: the centripetal acceleration is zero at the instant of release ($\omega = 0$), so it can't govern this condition, and $3L/4$ is numerically incorrect. Answer C incorrectly identifies the midpoint as critical and confuses "along the rod" with the tangential direction. Answer D is a tempting trap — it sounds physically intuitive that the block's mass shifts the answer, but since the block exerts no torque, it cannot change $\alpha$ or the critical distance. Study tip: Whenever a block rests on a rotating body without friction, remember it only exerts a normal force — zero torque contribution. Always check whether $\omega = 0$ before invoking centripetal terms.
2. A rigid disk of radius $R$ and mass $m$ rolls without slipping on a flat horizontal surface. A horizontal force $F$ is applied at the center of the disk. A student claims: *"Because the disk rolls without slipping, the friction force at the contact point does no work, so the work-energy theorem for the system gives $F \cdot s = \frac{1}{2}mv_C^2 + \frac{1}{2}I_C\omega^2$, where $s$ is the displacement of the center."* Which statement best evaluates this claim?
   Answer: The claim is correct: static friction at the contact point does no work because the contact point has zero velocity at every instant (it is the instantaneous center of rotation), so the work-energy equation as written is valid with $F \cdot s$ as the only work input. Whenever you encounter rolling-without-slipping problems, the key question is: does friction do work? The answer hinges on where friction acts and what velocity that point has. For a disk rolling without slipping, the contact point is the instantaneous center of rotation — its velocity is exactly zero at every instant. Since work is defined as $dW = \vec{F} \cdot \vec{v} \, dt$, a force applied at a point with zero velocity does zero work, regardless of the force's magnitude. Static friction at the contact point therefore contributes nothing to the work-energy balance. This makes D correct: the only external force doing work on the disk is $F$ acting at the center (which moves with velocity $v_C$), giving the clean result $F \cdot s = \frac{1}{2}mv_C^2 + \frac{1}{2}I_C\omega^2$. A is wrong because the normal force acts vertically while the contact point has no vertical displacement on a rigid horizontal surface — it does zero work. Introducing a term $N \cdot \delta$ conflates rigid-body analysis with contact-deformation models that don't apply here. B is wrong because it treats friction as though it acts at a point moving with the center's velocity $s$. That's only true for a sliding (slipping) contact. For rolling without slipping, the contact point is stationary, so friction does no work — not negative work. C is wrong because it invents a fictional cancellation between "translational" and "rotational" work of friction. This is unnecessary and misleading; the correct reason friction does no work is simply that its point of application has zero velocity. Study tip: Always ask yourself "What is the velocity of the point where this force is applied?" before computing work — this single check resolves most rolling-friction work questions instantly.
3. Compared to the same system but with the disk welded (fixed) to the rod so it cannot spin at C, how does the period of small oscillations change when the disk is free to spin at the pin?
   Answer: The period decreases when the disk is free to spin, because the disk's rotational inertia about C does not contribute to the moment of inertia about O in the free-spin case ($I_{\text{disk about O}} = md^2$ only, versus $md^2 + \frac{1}{2}mR^2$ when welded), reducing the effective inertia and increasing the natural frequency. When analyzing compound pendulums, your key tool is the rotational equation of motion: $\tau = I_O \ddot{\phi}$. The period depends on both the restoring torque and the effective moment of inertia about the pivot O, so you need to carefully identify what contributes to each. For the free-spin case, the disk can rotate independently at pin C. This means the disk's orientation in space is not coupled to the rod's angular displacement — the disk doesn't rotate with the rod. Consequently, the disk contributes only its translational (point-mass) inertia to the swing: $I_O = md^2$. The restoring torque remains $mgd\sin\phi \approx mgd\phi$, giving a natural frequency $\omega = \sqrt{mgd/(md^2)} = \sqrt{g/d}$. For the welded case, the disk is forced to rotate with the rod, so its spin inertia adds via the parallel axis theorem: $I_O = md^2 + \frac{1}{2}mR^2$. This larger inertia produces a lower frequency and longer period. Since the free-spin case has smaller effective inertia, its frequency is higher and its period is shorter — confirming B is correct. A is wrong because it claims the free-spin disk increases effective inertia. In reality, the disk's spin is unconstrained, so it contributes zero rotational inertia about O — not extra inertia. C is wrong because the moment of inertia about O does change between the two cases: welding adds $\frac{1}{2}mR^2$, which is not negligible. D invents a gyroscopic coupling that doesn't exist in this planar, small-oscillation scenario. Study tip: When a component can spin freely at an internal pin, ask yourself — is its rotation forced by the system's coordinate? If not, it drops out of the effective inertia entirely.
4. A thin uniform square plate of side $a$ and mass $m$ rotates freely (no external torques) about a fixed pivot at one corner. The pivot bearing is frictionless. At time $t = 0$, the plate is in the horizontal plane with angular velocity $\omega_0$ about the vertical axis through the pivot corner. Which expression gives the moment of inertia of the square plate about the vertical axis through the pivot corner, and what is the angular momentum about that axis?
   Answer: $I_{\text{corner}} = \dfrac{2}{3}ma^2$ and $H = \dfrac{2}{3}ma^2\omega_0$, obtained by summing the moments of inertia of two rectangular halves of the plate about the pivot corner: each half is a rectangle of mass $m/2$, side $a$, and depth $a/2$, and direct integration about the corner gives $\frac{1}{3}ma^2$ per half, totaling $\frac{2}{3}ma^2$. When finding the moment of inertia of a 2D plate about an out-of-plane axis through a corner, your most reliable approach is the parallel-axis theorem: start from a known centroidal moment of inertia, then shift to the new axis using $I = I_{cm} + md^2$. For a thin square plate, the moment of inertia about the centroidal axis perpendicular to the plate is $I_{cm} = \frac{1}{6}ma^2$. The distance from the center of the square to any corner is the half-diagonal: $d = \frac{a\sqrt{2}}{2}$, so $d^2 = \frac{a^2}{2}$. Applying the parallel-axis theorem: $I_{\text{corner}} = \frac{1}{6}ma^2 + m\cdot\frac{a^2}{2} = \frac{1}{6}ma^2 + \frac{3}{6}ma^2 = \frac{2}{3}ma^2$. Since there are no external torques, angular momentum is conserved and equals $H = I_{\text{corner}}\,\omega_0 = \frac{2}{3}ma^2\omega_0$, confirming C. A is wrong because a square plate is not equivalent to a slender rod — a rod's mass is distributed along one dimension only, whereas the plate extends in two dimensions, giving a larger moment of inertia. B uses the correct centroidal formula but substitutes $d = a$ (the full side length) instead of the true center-to-corner distance $d = \frac{a\sqrt{2}}{2}$. This is a classic trap: the center-to-corner distance is the half-diagonal, not the side length. D incorrectly models a square plate as a disk — these shapes have fundamentally different mass distributions and non-interchangeable formulas. Study tip: Always identify the correct centroid-to-axis distance before applying the parallel-axis theorem — for a square, the center-to-corner distance is $\frac{a\sqrt{2}}{2}$, not $\frac{a}{2}$ or $a$.
5. What is the minimum value of $\mu_s$ required for the cylinder to roll without slipping down the incline, and what is the linear acceleration of the center if rolling without slipping occurs?
   Answer: $\mu_{s,\min} = \dfrac{1}{3}\tan\theta$ and $a = \dfrac{2}{3}g\sin\theta$, derived by applying Newton's second law for translation and rotation simultaneously and solving for the friction force required, then dividing by the normal force. When a cylinder rolls without slipping, you must apply both Newton's second law for translation and the rotational equation simultaneously — the two are linked by the rolling constraint $a = \alpha R$. For translation down the incline: $mg\sin\theta - f = ma$. For rotation about the center: $fR = I_C\alpha = \frac{1}{2}mR^2\cdot\frac{a}{R}$, giving $f = \frac{1}{2}ma$. Substituting back: $mg\sin\theta - \frac{1}{2}ma = ma$, so $a = \frac{2}{3}g\sin\theta$. The required friction force is then $f = \frac{1}{3}mg\sin\theta$. Since the normal force is $N = mg\cos\theta$, the minimum static friction coefficient is $\mu_{s,\min} = \frac{f}{N} = \frac{1}{3}\tan\theta$. This is exactly D. Choice A makes a conceptual error by plugging in the free-sliding acceleration $a = g\sin\theta$ (no friction case) into the friction equation — it mixes two physically distinct scenarios. Rolling and frictionless sliding are not the same situation, so you cannot borrow that acceleration. Choice B uses $I = mR^2$, which is the moment of inertia for a thin cylindrical shell or hollow cylinder, not a solid cylinder ($I_C = \frac{1}{2}mR^2$). Using the wrong $I$ propagates through both the acceleration and friction results. Choice C takes torques about the contact point but incorrectly re-includes the friction moment, effectively double-counting friction's contribution. Taking torques about the contact point should eliminate friction from the equation, not add it twice. Study tip: Always apply the rolling constraint $a = \alpha R$ to couple your two equations, and keep your moment of inertia formulas straight — confusing solid vs. hollow cylinders is one of the most common traps on dynamics problems.
6. Using the velocity analysis of the slider-crank mechanism at the instant when OA is horizontal (A is directly to the right of O), which of the following correctly gives $\omega_{AB}$ and the velocity of slider B?
   Answer: $\omega_{AB} = 5 \text{ rad/s}$ clockwise and $v_B = 0$ m/s, obtained by recognizing that when OA is horizontal, A moves vertically upward, and since AB must remain rigid with B constrained to horizontal motion, the only consistent solution is $\omega_{AB} = v_A/\ell = 5$ rad/s clockwise and $v_B = 0$. Whenever you see a velocity analysis problem for a slider-crank mechanism, your go-to tool is the rigid-body velocity equation: $\vec{v}_B = \vec{v}_A + \vec{\omega}_{AB} \times \vec{r}_{B/A}$. The key is applying kinematic constraints carefully — B can only move horizontally, and AB is rigid. At the instant OA is horizontal, point A sits directly right of O. Since OA rotates counterclockwise at $\omega_{OA} = 10$ rad/s, the velocity of A is purely vertical (upward): $v_A = \omega_{OA} \cdot r = 10 \times 0.3 = 3$ m/s upward. At this same instant, AB is horizontal, so $\vec{r}_{B/A}$ points in the $+x$ direction. Writing out the velocity equation: $v_B \hat{x} = 3\hat{j} + \omega_{AB}(-\hat{k}) \times (0.6\hat{i}) = 3\hat{j} - 0.6\omega_{AB}\hat{j}$. Matching components: the $\hat{j}$ equation gives $0 = 3 - 0.6\omega_{AB}$, so $\omega_{AB} = 5$ rad/s clockwise. The $\hat{i}$ equation gives $v_B = 0$. This confirms B is correct. Choice A gets $\omega_{AB}$ right but incorrectly computes $v_B = 3$ m/s by misapplying the relative velocity formula — it adds rather than properly enforcing the horizontal constraint on B. Choice C sets $\omega_{AB} = 0$, confusing "the velocity of A is perpendicular to AB" with "AB has no rotation," which is backwards — that perpendicularity is precisely what drives the rotation. Choice D assumes $\omega_{AB} = \omega_{OA}$, a classic error of treating the connecting rod like it's pinned at a fixed center. Study tip: Always enforce all constraint directions simultaneously. Write out both the $\hat{i}$ and $\hat{j}$ components — each gives you one equation, and together they uniquely solve for both unknowns.

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