1406 The Physical Pendulum

A Physical Pendulum is any real pendulum that uses an extended body, as contrasted to the idealized simple pendulum with all of its mass concentrated at a point.
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Figure 14.23 shows a body of irregular shape pivoted so that it can turn without friction about an axis through point $O$ . When the body is displaced as shown, the weight mg causes a restoring torque

$\begin{aligned} \tau = -(mg) (d\sin \th) \quad &\text{(1)} \end{aligned}$

When the body is releases, it oscillates about its equilibrium position. The motion is not SHM becaues the torque $\tau$ is proportional to $\sin \th$ rather than to $\th$ . However, if $\th$ is small, we can approximate $\sin \th$ by $\th$ in radians. Then the motion is approximately SHM.

$\begin{aligned} \tau &= -mgd(\th) \quad &\text{(2)} \end{aligned}$

The torque and angular acceleration have the relation that $\sum \tau = I \alpha$ ,

$\begin{gathered} \tau = -mgd(\th) = I \alpha = I \frac{d^2 \th}{d t^2}\\ \frac{d^2 \th}{d t^2} = -\frac{mgd}{I}\th \quad &\text{(3)} \end{gathered}$

Comparing equation (3) with $a = \dfrac{d^2 x}{dt^2}= -\dfrac{k}{m}x$ for the spring-mass system, the quantity ( $\dfrac{mgd}{I}$ ) and $\dfrac{k}{m}$ play the same role. Thus the angular frequency is

$\begin{aligned} \omega &= \sqrt{\frac{mgd}{I}} \quad &\text{(4)} \end{aligned}$

where $m$ is the mass of the object, $d$ is the distance from the center of mass to the pivot point and $I$ is the moment of inertia. (can be obtained by parallel axis-theorem)
The period $T$ is

$\begin{aligned} T &= \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{I}{mgd}} \quad &\text{(5)} \end{aligned}$

Examples

(14.9) If the body in Fig 14.23 is a uniform rod with length L, pivoted at one end, what is the period of its motion as a pendulum?

Solution

According to the parallel axis theorem, the moment of inertia of a uniform rod about an axis through one end is

$\begin{aligned} I =\frac{1}{12}mL^2 + m(\frac{L}{2})^2 = \frac{1}{3} mL^2 \end{aligned}$

Let the length $L=1.00m$ , the period $T$ is

$\begin{aligned} T &= 2\pi \sqrt{\frac{I}{mgd}} = 2\pi \sqrt{\frac{\frac{1}{3} mL^2}{mgL/2}} = 2\pi \sqrt{\frac{2L}{3g}}\\ &= 2\pi \sqrt{\frac{2 \cdot 1}{3 \cdot 9.8}} = 1.64s \end{aligned}$

Exercises

53, 54, 55, 64, 66, 68

(14.53) Two pendulums have the same dimensions (length L) and total mass (m). Pendulum A is a very small ball swinging at the end of a uniform massless bar. In pendulum B, half the mass is in the ball and half is in the uniform bar.
a. Find the period of pendulum A for small oscillations.
b. Find the period of pendulum B for small oscillations.

Solution

a. The moment of inertia of the ball is $I = mL^2$ . The period $T_A$ is

$\begin{aligned} T_A = 2\pi \sqrt{\frac{I}{mgd}} = 2\pi \sqrt{\frac{mL^2}{mgL}} = 2\pi \sqrt{\frac{L}{g}} \end{aligned}$

b. For pendulum B the distance d from the axis to the center of gravity is $3L/4$ . $I_{bar}=\dfrac{1}{3} \cdot \dfrac{m}{2}L^2$ for a bar of mass $\dfrac{m}{2}$ and the axis at one end. For a small ball of mass $\dfrac{m}{2}$ at a distance L from the axis, $I_{ball}=\dfrac{m}{2} L^2$ . The distance $d = \frac{3}{4}L$

$\begin{aligned} T_B &= 2\pi \sqrt{\frac{I}{mgd}}\\ &= 2\pi \sqrt{\frac{\frac{1}{3} \cdot \frac{m}{2}L^2 + \frac{m}{2} L^2 }{md \cdot \frac{3}{4}L}}\\ &= 2\pi \sqrt{\frac{8L}{9g}} \end{aligned}$

(14.54) We want to support a thin hoop by a horizontal nail and have the hoop make one complete small-angle oscillation each 2.0 s.

Solution

The period $T$ is

$\begin{aligned} T &= 2\pi \sqrt{\frac{I}{mgd}} &\text{(1)} \end{aligned}$

The moment of inertia of the thin hoop about the axis through one end is

$\begin{aligned} I &= mR^2 + mR^2 = 2mR^2 &\text{(2)} \end{aligned}$

The distance $d$ from the center of mass to the pivot point is $d = R$ .
Combining (1) and (2), we have

$\begin{aligned} T &= 2\pi \sqrt{\frac{2mR^2}{mgR}} = 2\pi \sqrt{\frac{2R}{g}}\\ \To R &= (\frac{T}{2\pi})^2 \cdot \frac{g}{2} = 0.496m \end{aligned}$

(14.55) A 1.80-kg connecting rod from a car engine is pivoted about a horizontal knife edge as shown in Fig. E14.55. The center of gravity of the rod was located by balancing and is 0.200 m from the pivot. When the rod is set into small-amplitude oscillation, it makes 100 complete swings in 120 s. Calculate the moment of inertia of the rod about the rotation axis through the pivot.
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Solution

1. Period $T = \dfrac{120s}{100} = 1.2s$
2. The moment of inertia can be obtained by

$\begin{aligned} T &= 2\pi \sqrt {\dfrac{I}{mgd}}\\ \To I &= mgd \cdot (\frac{T}{2\pi})^2\\ &= 1.8 \cdot 9.8 \cdot 0.2 \cdot (\frac{1.2}{2\pi})^2\\ &= 0.129 kg \cdot m^2 \end{aligned}$

(14.64) An object is undergoing SHM with period 0.300 s and amplitude 6.00 cm. At t = 0 the object is instantaneously at rest at x = 6.00 cm. Calculate the time it takes the object to go from x = 6.00 cm to x = -1.50 cm.

Solution

The position vs time equation is $x(t) = A\cos(\omega t + \phi)$ . When $t = 0, x = A$ , we have $\phi = 0$ . Therefore

$\begin{aligned} \omega &= \frac{2\pi}{T} = 20.94 rad/s\\ A &= 0.06m\\ x(t) &= 0.06\cos(20.94 t)\\ x(t_1) &= 0.06 \To t_1 = 0s\\ x(t_2) &= -0.015 \To t_2 = \frac{\arccos(-\frac{1}{4})}{20.94}= 0.087s\\ \Delta t &= t_2 - t_1 = 0.087s \end{aligned}$

(14.66) Four passengers with combined mass 250 kg compress the springs of a car with worn-out shock absorbers by 4.00 cm when they get in. Model the car and passengers as a single body on a single ideal spring. If the loaded car has a period of vibration of 1.92 s, what is the period of vibration of the empty car?

Solution

1. The force constant of the spring is

$\begin{aligned} k &= \frac{F}{x} = \frac{250 \cdot 9.8}{0.04} = 61250N/m \end{aligned}$

2. The period of the loaded car is $T_{loaded} = 1.92s$ . Calculate the mass of the loaded car $m_{loaded}$

$\begin{aligned} T_{loaded} &= 2\pi\sqrt{\frac{m_{loaded}}{k}}\\ \To m_{loaded} &= k \cdot (\frac{T_{loaded}}{2\pi})^2\\ &= 61250 \cdot (\frac{1.92}{2\pi})^2\\ &= 5719.4kg \end{aligned}$

3. The period of vibration of the empty car $T_{empty}$ is

$\begin{aligned} m_{empty} &= m_{loaded} - 250 = 5469.4kg\\ T &= 2\pi \sqrt{\frac{m_{empty}}{k}}\\ &= 2\pi \sqrt{\frac{5469.4}{61250}}\\ &= 1.88s \end{aligned}$

(14.68) A block with mass $M$ rests on a frictionless surface and is connected to a horizontal spring of force constant $k$ . The other end of the spring is attached to a wall (Fig. P14.68). A second block with mass $m$ rests on top of the first block. The coefficient of static friction between the blocks is $\mu_s$ . Find the maximum amplitude of oscillation such that the top block will not slip on the bottom block.
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Solution

Todo