Statics and Dynamics Quiz: Planar Rigid Body Motion

What this quiz covers

This quiz focuses on Planar Rigid Body Motion, 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 period increases when the disk is free to spin, because the effective moment of inertia about O is larger — the disk's spin degree of freedom adds kinetic energy storage without contributing to restoring torque, making the system dynamically 'heavier.'
2. 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. (correct answer)
3. The period remains the same regardless of whether the disk spins freely or is welded, because the restoring torque $mgd\sin\phi$ and the moment of inertia about O are both unchanged — the disk's spin at C is an internal degree of freedom that does not affect the pendulum's equation of motion.
4. The period remains the same only if $d \gg R$; otherwise the period increases when the disk is free to spin, because the gyroscopic effect of the freely spinning disk introduces a coupling between spin and swing that increases the effective inertia for small $d/R$ ratios.

Question 2

1. The claim is incorrect: the work-energy equation must include the work done by the normal force from the ground, which acts over the vertical displacement of the contact point during deformation, so the equation should include a term $N \cdot \delta$ for the ground's normal work.
2. The claim is incorrect: the friction force does negative work on the disk equal to $-f \cdot s$ (where $f$ is the friction force and $s$ is the displacement of the center), so the correct equation is $(F - f) \cdot s = \frac{1}{2}mv_C^2 + \frac{1}{2}I_C\omega^2$, and the claim understates the kinetic energy.
3. The claim is correct in conclusion but flawed in reasoning: while the contact point does have zero instantaneous velocity, friction still does work in the rotational sense, so the correct justification is that the translational work of friction and its rotational work cancel exactly, leaving $F \cdot s$ as the net work.
4. 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. (correct answer)

Question 3

1. $I_{\text{corner}} = \dfrac{1}{3}ma^2$ and $H = \dfrac{1}{3}ma^2\omega_0$, obtained by treating the square plate as equivalent to a slender rod of the same mass and length $a$ rotating about one end, and applying $I_{\text{rod,end}} = \frac{1}{3}ma^2$.
2. $I_{\text{corner}} = \dfrac{7}{6}ma^2$ and $H = \dfrac{7}{6}ma^2\omega_0$, obtained by using $I_{\text{center}} = \frac{1}{6}ma^2$ for the centroidal axis perpendicular to the plate, then applying the parallel-axis theorem with the incorrect distance $d = a$ (full side length) from center to corner, giving $I_{\text{corner}} = \frac{1}{6}ma^2 + ma^2 = \frac{7}{6}ma^2$.
3. $I_{\text{corner}} = \dfrac{2}{3}ma^2$ 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 $\frac{2}{3}ma^2$. (correct answer)
4. $I_{\text{corner}} = \dfrac{1}{2}ma^2$ and $H = \dfrac{1}{2}ma^2\omega_0$, obtained by treating the square plate as a thin disk of diameter $a$ (radius $a/2$) with $I = \frac{1}{2}mR^2 = \frac{1}{8}ma^2$, then applying the parallel-axis theorem with $d = a/2$, giving $\frac{1}{8}ma^2 + \frac{1}{4}ma^2 = \frac{3}{8}ma^2$... rounded up to $\frac{1}{2}ma^2$.

Question 4

1. $d = \dfrac{2L}{3} = 0.80 \text{ m}$, 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. (correct answer)
2. $d = \dfrac{3L}{4} = 0.90 \text{ m}$, because at this distance the centripetal acceleration component of the rod point equals $g$, satisfying the condition for the block to become weightless relative to the rod surface.
3. $d = \dfrac{L}{2} = 0.60 \text{ m}$, because at the midpoint the rod's angular acceleration produces a tangential acceleration equal to the component of gravity along the rod's length, releasing the normal force.
4. $d = \dfrac{3L}{4} = 0.90 \text{ m}$, because the block's weight $Mg$ adds to the driving torque about O, shifting the critical point outward beyond $2L/3$; the heavier the block, the farther from O the critical distance moves.

Question 5

1. $\omega_{AB} = 5 \text{ rad/s}$ clockwise and $v_B = 3 \text{ m/s}$ to the right, obtained by noting that A moves purely upward at $v_A = \omega_{OA} \cdot r = 3$ m/s, and since AB is horizontal at this configuration, the vertical component of $v_A$ drives rotation of AB while B moves horizontally with $v_B = v_A - \omega_{AB} \cdot \ell$.
2. $\omega_{AB} = 5 \text{ rad/s}$ 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 $v_B = 0$. (correct answer)
3. $\omega_{AB} = 0 \text{ rad/s}$ and $v_B = 3 \text{ m/s}$ to the right, because when OA is horizontal, A moves vertically and AB is horizontal, so the velocity of A relative to B must be purely perpendicular to AB — but AB is horizontal, meaning the perpendicular direction is vertical, which matches $v_A$, requiring $\omega_{AB} = v_A/\ell$ and $v_B = 0$. This answer reflects the error of setting $\omega_{AB} = 0$ by confusing the velocity direction constraint.
4. $\omega_{AB} = 10 \text{ rad/s}$ clockwise and $v_B = 3 \text{ m/s}$ to the right, obtained by incorrectly applying $\omega_{AB} = \omega_{OA}$ (assuming the connecting rod spins at the same rate as the crank) and computing $v_B$ from $v_A + \omega_{AB} \times r_{B/A}$.

Question 6

1. $\mu_{s,\min} = \dfrac{1}{3}\tan\theta$ and $a = g\sin\theta$, derived by first computing the friction required for rolling and then substituting $a = g\sin\theta$ as the free-sliding acceleration, conflating the rolling and sliding cases.
2. $\mu_{s,\min} = \dfrac{1}{2}\tan\theta$ and $a = \dfrac{1}{2}g\sin\theta$, derived by incorrectly using the moment of inertia $I = mR^2$ (solid sphere) instead of $\frac{1}{2}mR^2$ in the rotational equation, leading to a different friction requirement and acceleration.
3. $\mu_{s,\min} = \dfrac{2}{3}\tan\theta$ and $a = \dfrac{1}{3}g\sin\theta$, derived by applying the rotational equation about the contact point and incorrectly including the friction force moment, which double-counts the friction contribution.
4. $\mu_{s,\min} = \dfrac{1}{3}\tan\theta$ 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. (correct answer)