15-4 Pendulums, Circular Motion

15-4 Pendulums, Circular Motion#

Prompts

  • For a simple pendulum, what provides the restoring torque? Why do we need the small-angle approximation?

  • Derive the period \(T = 2\pi\sqrt{L/g}\) for a simple pendulum. Why does \(T\) not depend on the bob’s mass?

  • What is a physical pendulum? How does its period differ from a simple pendulum? What is \(h\)?

  • Three physical pendulums have the same shape and pivot but masses \(m\), \(2m\), and \(3m\). How do their periods compare? Why?

  • Explain: “Simple harmonic motion is the projection of uniform circular motion onto a diameter.” How does this connect \(x(t)\), \(v(t)\), and \(a(t)\) to circular motion?

Lecture Notes#

Overview#

  • Pendulums are SHM oscillators where gravity provides the restoring torque (not a spring or twisted wire).

  • Simple pendulum: point mass on a string. Physical pendulum: extended body; period depends on \(I\) and distance \(h\) from pivot to center of mass.

  • SHM = projection of uniform circular motion onto a diameter—a deep connection that explains why \(x\), \(v\), and \(a\) are sinusoidal.

Type

Inertia \(\mu\)

Stiffness \(\kappa\)

Equation of motion

Period \(T\)

Spring–block

\(m\)

\(k\)

\(m\frac{d^2x}{dt^2} = -kx\)

\(2\pi\sqrt{m/k}\)

Torsion

\(I\)

\(\kappa\)

\(I\frac{d^2\theta}{dt^2} = -\kappa\theta\)

\(2\pi\sqrt{I/\kappa}\)

Simple pendulum

\(mL^2\)

\(mgL\)

\(mL^2\frac{d^2\theta}{dt^2} = -mgL\theta\)

\(2\pi\sqrt{L/g}\)

Physical pendulum

\(I\)

\(mgh\)

\(I\frac{d^2\theta}{dt^2} = -mgh\theta\)

\(2\pi\sqrt{I/(mgh)}\)

The simple pendulum#

A simple pendulum is a particle (bob) of mass \(m\) on a massless string of length \(L\), pivoted at the top.

  • Restoring torque: The tangential component of gravity, \(mg\sin\theta\), produces a torque about the pivot. For small \(\theta\), \(\sin\theta \approx \theta\) (in radians):

(68)#\[ \tau = -L(mg\sin\theta) \approx -mgL\theta \]

  • Energy (for small \(\theta\), \(1-\cos\theta \approx \theta^2/2\)):

(69)#\[\begin{split} \begin{split} E &= \frac{1}{2}mL^2\left(\frac{d\theta}{dt}\right)^2 + mgL(1-\cos\theta) \\ &\approx \frac{1}{2}mL^2\left(\frac{d\theta}{dt}\right)^2 + \frac{1}{2}mgL\,\theta^2 \end{split} \end{split}\]

  • Inertia: \(\mu = mL^2\)

  • Stiffness: \(\kappa = mgL\)

  • By section 15-2, Eq. (59):

(70)#\[ \omega = \sqrt{\frac{g}{L}}, \qquad T = 2\pi\sqrt{\frac{L}{g}} \]

  • Independent of mass: \(T\) depends only on \(L\) and \(g\). Heavier and lighter bobs swing at the same rate.

Caution

Small-angle only. The approximation \(\sin\theta \approx \theta\) holds for \(\theta \lesssim 10°\). For larger amplitudes, the motion is periodic but not sinusoidal; \(T\) increases slightly.

Why \(g \approx \pi^2\) numerically?

The meter was originally proposed as the length of a seconds pendulum (half-period 1 s, so \(T = 2\) s). From \(T = 2\pi\sqrt{L/g}\) with \(L = 1\) m and \(T = 2\) s: \(g = \pi^2\,\text{m/s}^2\). That definition was later abandoned (gravity varies with location; the meter was instead based on Earth’s geometry), but the meter was sized similarly, so \(g \approx 9.8\,\text{m/s}^2\) and \(\pi^2 \approx 9.87\) remain close.

The physical pendulum#

A physical pendulum is any rigid body that swings about a pivot (not at its center of mass).

  • Restoring torque: Gravity acts at the center of mass, distance \(h\) from the pivot. The moment arm of \(mg\sin\theta\) is \(h\):

(71)#\[ \tau = -h(mg\sin\theta) \approx -mgh\theta \]

  • Energy (for small \(\theta\), \(1-\cos\theta \approx \theta^2/2\)):

(72)#\[\begin{split} \begin{split} E &= \frac{1}{2}I\left(\frac{d\theta}{dt}\right)^2 + mgh(1-\cos\theta) \\ &\approx \frac{1}{2}I\left(\frac{d\theta}{dt}\right)^2 + \frac{1}{2}mgh\,\theta^2 \end{split} \end{split}\]

  • Inertia: \(\mu = I\) rotational inertia about the pivot

  • Stiffness: \(\kappa = mgh\) (\(h\) = distance from pivot to center of mass)

  • By section 15-2, Eq. (59):

(73)#\[ \omega = \sqrt{\frac{mgh}{I}}, \qquad T = 2\pi\sqrt{\frac{I}{mgh}} \]

  • If the pivot is at the center of mass, \(h = 0\) and \(T \to \infty\)—no oscillation.

Simple vs physical

A simple pendulum is a special case: all mass at distance \(L\), so \(I = mL^2\) and \(h = L\). Then Eq. (73) reduces to Eq. (70).

Center of oscillation#

For any physical pendulum with period \(T\), there is an equivalent simple pendulum of length \(L_0\) with the same period:

(74)#\[ T = 2\pi\sqrt{\frac{L_0}{g}} = 2\pi\sqrt{\frac{I}{mgh}} \quad \Rightarrow \quad L_0 = \frac{I}{mh} \]

The point at distance \(L_0\) from the pivot is the center of oscillation. For a uniform rod pivoted at one end: \(I = \frac{1}{3}mL^2\), \(h = L/2\), so \(L_0 = \frac{2}{3}L\).

Summary#

  • Simple pendulum: \(T = 2\pi\sqrt{L/g}\); independent of mass; valid for small angles.

  • Physical pendulum: \(T = 2\pi\sqrt{I/(mgh)}\); \(h\) = pivot-to-CM distance.

  • Center of oscillation: \(L_0 = I/(mh)\)—equivalent simple-pendulum length.