Simple and Physical Pendulums - AP Physics C Mechanics

1. A simple pendulum and a physical pendulum, consisting of a uniform rod of length $L$ pivoted at one end, both have the same length $L$. How does the period of the rod, $T_{rod}$, compare to the period of the simple pendulum, $T_{simple}$?
   Answer: $T_{rod} < T_{simple}$ The period of the simple pendulum is $T_{simple} = 2\pi\sqrt{L/g}$. The period of the rod is $T_{rod} = 2\pi\sqrt{2L/3g}$. Since $\sqrt{2/3} \approx 0.816$, we have $T_{rod} \approx 0.816 \, T_{simple}$, so the rod's period is shorter.
2. A compound pendulum consists of a uniform disk of radius $R$ and mass $M$ that can pivot about a horizontal axis passing through a point on its rim. If the disk oscillates with small amplitude, what is the length of the equivalent simple pendulum that would have the same period?
   Answer: $\frac{3R}{2}$ For a physical pendulum, the equivalent simple pendulum length is $L_{eq} = \frac{I}{md}$, where $I$ is the moment of inertia about the pivot, $m$ is the mass, and $d$ is the distance from pivot to center of mass. For a disk pivoting at its rim, $d = R$. The moment of inertia about the rim is $I = I_{center} + MR^2 = \frac{MR^2}{2} + MR^2 = \frac{3MR^2}{2}$. Therefore, $L_{eq} = \frac{3MR^2/2}{M \cdot R} = \frac{3R}{2}$. Choice A uses only the center-of-mass moment of inertia. Choice B assumes the equivalent length equals the distance to center of mass. Choice D incorrectly doubles the radius without proper calculation.
3. A uniform rectangular plate of mass $M$, width $W$, and height $H$ is pivoted at its top left corner to swing as a physical pendulum. The rotational inertia about its center of mass is $I_{cm} = \frac{1}{12}M(W^2+H^2)$. What is the distance $d$ from the pivot to the center of mass?
   Answer: $\frac{1}{2}\sqrt{W^2+H^2}$ The center of mass of a uniform rectangular plate is at its geometric center. The pivot is at a corner. The coordinates of the center can be taken as $(W/2, H/2)$ relative to the corner pivot at $(0,0)$. The distance $d$ is found using the Pythagorean theorem: $d = \sqrt{(W/2)^2 + (H/2)^2} = \frac{1}{2}\sqrt{W^2+H^2}$.
4. A torsion pendulum is constructed by suspending a horizontal rod of mass $M$ and length $L$ by a wire attached to its center. The rod has a rotational inertia $I_{rod} = \frac{1}{12}ML^2$. Two small spheres, each of mass $m$, are then attached to the ends of the rod. What is the new rotational inertia of the system?
   Answer: $\frac{1}{12}ML^2 + 2m(L/2)^2$ The total rotational inertia of the system is the sum of the rotational inertias of its components. The rod's inertia is given. Each sphere of mass $m$ is at a distance $r=L/2$ from the axis of rotation. Treating the spheres as point masses, their rotational inertia is $I_{sphere} = mr^2 = m(L/2)^2$. Since there are two spheres, their total contribution is $2m(L/2)^2$. The total inertia is $I_{total} = I_{rod} + I_{spheres} = \frac{1}{12}ML^2 + 2m(L/2)^2$.
5. A uniform solid disk of radius $R$ is pivoted at its center and used as a torsion pendulum. It has a period $T_C$. The disk is then re-pivoted at its rim to oscillate as a physical pendulum in Earth's gravitational field. Its period is $T_P$. How do these periods depend on the mass of the disk?
   Answer: Both periods are independent of the mass of the disk. For the torsion pendulum, $T_C = 2\pi\sqrt{I/\kappa}$, where $\kappa$ is the torsion constant of the wire. While $I \propto M$, the torsion constant $\kappa$ for a given wire and angular displacement is also proportional to $M$ (since the restoring torque per unit angle depends on the inertial properties of the system). Thus the mass cancels and $T_C$ is independent of mass. For the physical pendulum, $T_P = 2\pi\sqrt{I'/Mgd}$. Since both $I'$ and $M$ are proportional to mass, mass cancels and $T_P$ is also independent of mass.
6. The restoring torque on a physical pendulum is given by $\tau = -mgd\sin\theta$. This relation leads to simple harmonic motion for small angles primarily because under this condition:
   Answer: the term $\sin\theta$ can be approximated as $\theta$, making the torque proportional to displacement. Simple harmonic motion requires a restoring force or torque that is directly proportional to the displacement. By approximating $\sin\theta$ as $\theta$ for small angles, the torque becomes $\tau \approx -mgd\theta$. This linear relationship between torque and angular displacement is the definition of angular simple harmonic motion.
7. A simple pendulum of length $L$ oscillates with small amplitude in a region where the gravitational acceleration is $g$. If the pendulum bob is replaced with one having twice the mass and the length is increased by a factor of 4, what is the ratio of the new period to the original period?
   Answer: $2$ The period of a simple pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$, which is independent of the mass of the bob. When the length increases by a factor of 4, the new period becomes $T_{new} = 2\pi\sqrt{\frac{4L}{g}} = 2 \cdot 2\pi\sqrt{\frac{L}{g}} = 2T_{original}$. The mass change has no effect on the period. Choice A results from incorrectly thinking the period decreases with length. Choice B assumes both mass and length changes cancel out. Choice D incorrectly applies the length factor directly without taking the square root.
8. A simple pendulum of length $L$ has a period $T$ on Earth. The pendulum is moved to a planet with a radius twice that of Earth and a mass eight times that of Earth. What is the new period of the pendulum on this planet?
   Answer: $T/\sqrt{2}$ The acceleration due to gravity on a planet's surface is $g = GM/R^2$. The new gravity is $g' = G(8M)/(2R)^2 = G(8M)/(4R^2) = 2(GM/R^2) = 2g$. The period of a simple pendulum is $T = 2\pi\sqrt{L/g}$. The new period is $T' = 2\pi\sqrt{L/g'} = 2\pi\sqrt{L/(2g)} = (1/\sqrt{2})T$.
9. A uniform rod of length $L=1.0$ m is pivoted at a distance $d=0.25$ m from its center and oscillates as a physical pendulum. The rotational inertia of a rod about its center is $\frac{1}{12}ML^2$. Which of the following expressions correctly gives the period of small oscillations?
   Answer: $2\pi\sqrt{\frac{(1/12)L^2 + (0.25)^2}{g(0.25)}}$ The period is $T = 2\pi\sqrt{I/Mgd}$. The distance from the pivot to the center of mass is given as $d=0.25$ m. By the parallel-axis theorem, the rotational inertia about the pivot is $I = I_{cm} + Md^2 = \frac{1}{12}ML^2 + M(0.25)^2$. Substituting this into the period formula and cancelling M gives $T = 2\pi\sqrt{\frac{(1/12)L^2 + (0.25)^2}{g(0.25)}}$.
10. Which of the following modifications to a simple pendulum of length $L$ and mass $m$ would result in a doubling of its period of oscillation?
    Answer: Quadrupling the length $L$ while keeping the mass $m$ constant. The period of a simple pendulum is given by $T = 2\pi\sqrt{L/g}$. To double the period $T$, the argument of the square root, $L/g$, must be quadrupled. This can be achieved by quadrupling the length $L$.
11. A uniform meter stick is pivoted at one end. A small object of the same mass as the meter stick is attached to the other end. The system oscillates as a physical pendulum. Let L be the length of the stick. The rotational inertia of the stick about the pivot is $\frac{1}{3}ML^2$. What is the rotational inertia of the combined system about the pivot?
    Answer: $\frac{4}{3}ML^2$ The total rotational inertia is the sum of the inertias of the stick and the small object. The stick's inertia is $I_{stick} = \frac{1}{3}ML^2$. The object can be treated as a point mass $M$ at a distance $L$ from the pivot, so its inertia is $I_{object} = ML^2$. The total inertia is $I_{total} = I_{stick} + I_{object} = \frac{1}{3}ML^2 + ML^2 = \frac{4}{3}ML^2$.
12. A pendulum clock is found to be running too fast. To correct its timing, a small adjustment must be made. Which of the following adjustments would slow the clock down?
    Answer: Slightly increasing the length of the pendulum arm by moving the bob down. If the clock is running too fast, its period is too short. To slow the clock down, the period must be increased. The period of a pendulum is given by $T = 2\pi\sqrt{L/g}$ (or a similar form for a physical pendulum). Increasing the length $L$ will increase the period. This is typically done by moving the pendulum bob down along a threaded rod.
13. A hoop of radius $R$ and mass $M$ is pivoted at its rim and oscillates as a physical pendulum. The rotational inertia of a hoop about its center is $MR^2$. What is the period of the hoop for small-amplitude oscillations?
    Answer: $2\pi\sqrt{\frac{2R}{g}}$ Using the parallel-axis theorem, the rotational inertia about the rim is $I = I_{cm} + Md^2 = MR^2 + MR^2 = 2MR^2$, where $d=R$. The period is $T = 2\pi\sqrt{\frac{I}{Mgd}} = 2\pi\sqrt{\frac{2MR^2}{MgR}} = 2\pi\sqrt{\frac{2R}{g}}$.
14. A simple pendulum oscillates with a period $T$ when its maximum angular displacement is $5^\circ$. According to the idealized model of simple harmonic motion, what would be the period if the maximum angular displacement were increased to $10^\circ$?
    Answer: $T$ In the model for simple harmonic motion of a pendulum (which relies on the small-angle approximation), the period is independent of the amplitude of oscillation. Therefore, changing the maximum displacement from $5^\circ$ to $10^\circ$ (both of which are considered small angles) does not change the period.
15. The differential equation of motion for a physical pendulum is $I\frac{d^2\theta}{dt^2} = -mgd\sin\theta$. For small oscillations, this equation is approximated to represent simple harmonic motion. What is the angular frequency $\omega$ of this simple harmonic motion?
    Answer: $\sqrt{\frac{mgd}{I}}$ For small angles, $\sin\theta \approx \theta$, so the equation becomes $I\frac{d^2\theta}{dt^2} = -mgd\theta$. This can be rewritten as $\frac{d^2\theta}{dt^2} = -(\frac{mgd}{I})\theta$. The general form for SHM is $\frac{d^2x}{dt^2} = -\omega^2 x$. By comparison, $\omega^2 = \frac{mgd}{I}$, so $\omega = \sqrt{\frac{mgd}{I}}$.
16. A physical pendulum has a period $T$. If the entire apparatus is placed in an elevator that is accelerating upward with a constant acceleration of magnitude $a$, what is the new period of oscillation, $T'$?
    Answer: $T' = T \sqrt{\frac{g}{g+a}}$ In the accelerating frame of the elevator, the effective gravitational acceleration is $g_{eff} = g+a$. The period of a physical pendulum is $T = 2\pi\sqrt{I/(mgd)}$, so it is inversely proportional to the square root of $g$. The new period will be $T' = 2\pi\sqrt{I/(mg_{eff}d)} = 2\pi\sqrt{I/(m(g+a)d)} = T\sqrt{g/(g+a)}$.
17. A physical pendulum has a certain period of oscillation. If the mass of the pendulum is doubled, but its shape, size, and pivot point remain exactly the same, what is the effect on its period?
    Answer: The period remains the same because the effects of mass in inertia and torque cancel. The period of a physical pendulum is $T = 2\pi\sqrt{I/(mgd)}$. The rotational inertia $I$ is directly proportional to the mass $m$ (e.g., $I = \beta m R^2$ for some constant $\beta$). Therefore, the mass $m$ in the numerator (within $I$) cancels with the mass $m$ in the denominator, making the period independent of mass.
18. A solid sphere of mass $M$ and radius $R$ is pivoted to oscillate as a physical pendulum about an axis tangent to its surface. The rotational inertia of a solid sphere about its center of mass is $\frac{2}{5}MR^2$. What is the rotational inertia of the sphere about the pivot axis?
    Answer: $\frac{7}{5}MR^2$ The parallel-axis theorem states $I = I_{cm} + Md^2$. For a sphere pivoted at its surface, the distance $d$ from the center of mass to the pivot is the radius $R$. Therefore, $I = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$.
19. A simple pendulum oscillates with period $T_0$ when the amplitude is very small. When the maximum angular displacement is increased to $30°$, the period becomes approximately $T$. Which of the following best represents the relationship between $T$ and $T_0$?
    Answer: $T = 1.04 T_0$ For large amplitude oscillations, the period of a simple pendulum increases beyond the small angle approximation value. The exact formula involves elliptic integrals, but the first-order correction gives $T \approx T_0(1 + \frac{1}{16}\theta_{max}^2)$ where $\theta_{max}$ is in radians. For $30° = \frac{\pi}{6}$ radians ≈ 0.524 radians, we get $T \approx T_0(1 + \frac{1}{16} \times 0.524^2) ≈ T_0(1 + 0.017) ≈ 1.04 T_0$. Choice A assumes the small angle approximation still holds. Choice C incorrectly suggests the period decreases. Choice D overestimates the correction factor.
20. A uniform rod of mass $M$ and length $L$ is pivoted at a point located at distance $\frac{L}{4}$ from one end. The rod oscillates as a physical pendulum with small amplitude. What is the period of oscillation?
    Answer: $2\pi\sqrt{\frac{7L}{12g}}$ For a physical pendulum, $T = 2\pi\sqrt{\frac{I}{mgd}}$, where $I$ is the moment of inertia about the pivot, $m$ is the mass, and $d$ is the distance from pivot to center of mass. The center of mass is at $\frac{L}{2}$ from either end, so $d = |\frac{L}{2} - \frac{L}{4}| = \frac{L}{4}$. Using parallel axis theorem: $I = I_{cm} + Md^2 = \frac{ML^2}{12} + M(\frac{L}{4})^2 = \frac{ML^2}{12} + \frac{ML^2}{16} = \frac{7ML^2}{48}$. Therefore, $T = 2\pi\sqrt{\frac{7ML^2/48}{Mg \cdot L/4}} = 2\pi\sqrt{\frac{7L}{12g}}$. Choice B uses incorrect moment of inertia calculation. Choice C forgets the parallel axis theorem. Choice D has an algebraic error in the final simplification.

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