Simple and Physical Pendulums
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AP Physics C: Mechanics › Simple and Physical Pendulums
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?
The period remains the same because the effects of mass in inertia and torque cancel.
The period is halved because it is inversely proportional to mass.
The period increases by a factor of $$\sqrt{2}$$ because it is proportional to the square root of mass.
The period is doubled because it is directly proportional to mass.
Explanation
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.
A torsion pendulum consists of a solid disk with rotational inertia $$I$$ attached to a wire with torsion constant $$\kappa$$. It oscillates with a period $$T_0$$. If the disk is replaced with a hoop of the same mass and radius, what is the new period? The rotational inertia of a hoop is twice that of a solid disk of the same mass and radius.
$$2T_0$$
$$T_0/2$$
$$T_0/\sqrt{2}$$
$$\sqrt{2}T_0$$
Explanation
The period of a torsion pendulum is $$T = 2\pi\sqrt{I/\kappa}$$. Since the period is proportional to the square root of the rotational inertia ($$T \propto \sqrt{I}$$), and the new rotational inertia is $$I' = 2I$$, the new period will be $$T' = \sqrt{2}T_0$$.
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, $$T_{simple}$$?
$$T_{rod} = T_{simple}$$
$$T_{rod} < T_{simple}$$
The comparison depends on the mass of the rod and the pendulum bob.
$$T_{rod} > T_{simple}$$
Explanation
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.
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?
$$\frac{1}{12}ML^2 + m(L/2)^2$$
$$\frac{1}{12}(M+2m)L^2$$
$$\frac{1}{12}ML^2 + 2mL^2$$
$$\frac{1}{12}ML^2 + 2m(L/2)^2$$
Explanation
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 $$I_{total} = I_{rod} + I_{spheres} = \frac{1}{12}ML^2 + 2m(L/2)^2$$.
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?
The period $$T_C$$ depends on mass, while $$T_P$$ is independent of mass.
Both periods are independent of the mass of the disk.
Both periods depend on the mass of the disk.
The period $$T_C$$ is independent of mass, while $$T_P$$ depends on mass.
Explanation
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.
A uniform rod of mass $$M$$ and length $$L$$ is pivoted at one end and oscillates as a physical pendulum. The rotational inertia of the rod about its end is $$\frac{1}{3}ML^2$$. What is the period of the rod for small-amplitude oscillations?
$$2\pi\sqrt{\frac{L}{2g}}$$
$$2\pi\sqrt{\frac{2L}{3g}}$$
$$2\pi\sqrt{\frac{L}{3g}}$$
$$2\pi\sqrt{\frac{L}{g}}$$
Explanation
The period of a physical pendulum is given by $$T = 2\pi\sqrt{\frac{I}{mgd}}$$. For a uniform rod pivoted at one end, the rotational inertia is $$I = \frac{1}{3}ML^2$$ and the distance from the pivot to the center of mass is $$d = L/2$$. Substituting these values gives $$T = 2\pi\sqrt{\frac{\frac{1}{3}ML^2}{Mg(L/2)}} = 2\pi\sqrt{\frac{2L}{3g}}$$.
A physical pendulum consists of a rigid body oscillating about a fixed pivot point. For the pendulum's motion to be accurately modeled as simple harmonic motion, which of the following conditions is required?
The period of oscillation must be independent of the mass of the object.
The mass of the object must be concentrated at its center of mass for the model to apply.
The pivot point must be located at one end of the object for the motion to be harmonic.
The amplitude of the oscillation must be small enough that $$\sin\theta \approx \theta$$.
Explanation
The restoring torque for a physical pendulum is $$\tau = -mgd\sin\theta$$. Simple harmonic motion requires the restoring torque to be proportional to the angular displacement, $$\tau = -k\theta$$. This condition is met when the oscillation amplitude is small, allowing for the approximation $$\sin\theta \approx \theta$$.
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?
$$2T$$
$$\sqrt{2}T$$
$$T/2$$
$$T/\sqrt{2}$$
Explanation
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 $$T' = 2\pi\sqrt{L/g'} = 2\pi\sqrt{L/(2g)} = (1/\sqrt{2})T$$.
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?
$$\frac{2}{5}MR^2$$
$$\frac{7}{5}MR^2$$
$$MR^2$$
$$\frac{3}{5}MR^2$$
Explanation
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, $$I = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$$.
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?
$$\frac{mgd}{I}$$
$$\sqrt{\frac{mgd}{I}}$$
$$\frac{I}{mgd}$$
$$\sqrt{\frac{I}{mgd}}$$
Explanation
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 $$\omega = \sqrt{\frac{mgd}{I}}$$.