This document describes the simple pendulum and physical pendulum. It defines a simple pendulum as a point mass attached to a string of negligible mass that swings back and forth under gravity. Forces acting on the pendulum are its weight and the tension in the string. For small oscillations, the pendulum's motion is simple harmonic. The period of a simple pendulum depends only on its length and gravity. A physical pendulum is an extended object that oscillates about a pivot point, and its period depends on its moment of inertia as well as gravity and length. The document provides examples to calculate pendulum periods.

1. The Simple Pendulum
Prachyawanich (Daniel) Khotawanich
PHYS 101 Lab section: L2M Blue
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LO

2. Simple Pendulum
• Consists of a point mass attached to a string of negligible mass
that doesn’t stretch.
• The other end of the string is tied to a frictionless and rigid
support
• The pendulum swings back and forth under the influence of
gravity where the motion of the pendulum is periodic.
• Not simple harmonic motion

3. Diagram
Forces acting on the mass are:
• Mg
• The weight
• Tension (T)
Radial Axis
• Along the length of the string
Tangent Axis
• Tangent to the circular motion of the
mass

4. The Radial Axis
• Weight (mg) makes an angle θ with radial axis
• mgcos(θ) = point away from suspension point
• Tention (T) is oriented towards suspension point
• No motion along the radial axis  radial
components of the weight and the tension are
equal
• T - mgcos(θ) = 0  T=mgcos(θ)

5. The Tangential Axis
• Tangential component of weight  mgsin(θ)
• Is the restoring force that pulls the mass toward the
equilibrium position
• Newton’s second law  Fnet= -mgsin(θ) = ma
• “a” represents the tangential component of the
acceleration
• a = -gsin(θ) = -gsin(s/L)
• “s” is displacement, “L” is length

6. Small Angle
Approximation/Displacement
• If angle is small, use small-angle approximation
for sine function  sin(x) = x  sin (s/L) = s/L
• a = -(g/L)s
• For small angular displacements, the acceleration
of the mass is proportional to the displacement
and opposite in sign  motion of pendulum is
simple harmonic motion
• ω= √(g/L) ω is omega

7. Period and Frequency of
Oscillation of Pendulum
• T = 2π/ω  T= 2π√(L/g)
• f = ω/2π  f = 1/2π√(g/L)
• Period of simple pendulum depends on length of
pendulum and acceleration due to gravity
• Period of pendulum will vary slightly with value of
“g” at a given location

8. Physical Pendulum
• Extended object that, when displaced from its
equilibrium position, oscillates about an axis
• Figure (a) shows the pendulum in equilibrium
position
• Figure (b) shows that the rod is displaced by an
angle θ from equilibrium position

9. Physical Pendulum
• When the sphere is in the displaced position, the
gravitational force exerts a torque (τ) on the sphere about
the pivot point.
• τ= -(Mg)(Lcm)sinθ
• Where Lcm = L + R is the distance between pivot point and
the center of mass.
• α= -(MgLcm/Ip)sinθ  angular acceleration, Ip = inertia
• ω= √(MgL/I)
• T= 2π√(I/MgL)
• L= I/ML

10. Questions
1. A pendulum takes 8 seconds to swing in each
direction. Find the length of the pendulum.
2. A square cardboard of 20.0 g and length of 28.0
cm is free to rotate about a pivot point, P, that is
located above the center of the cardboard and
1.00 cm from the top edge. Calculate the period
for the small angular displacement about the
pivot point, P.

11. Solutions
1. T=2π√(L/g)
8=2π√(L/9.8)
L= 15.9 m
2. T= 2π√(Ip/MgLcm)
Ip= Icm +ML^2  moment of inertia about point P
Icm = (1/12)(ML^2)  moment of inertia of square sheet about its center
Lcm = 14 cm – 1 cm = 13.0 cm  distance from center of mass to pivot point
Ip = (1/12)*(0.02 kg)(0.280 m^2) + (0.02 kg)(0.13 m^2) = 4.69*10^-4 kgm^2
T= 2π√((4.69*10^-4 kgm^2)/ (0.02 kg)(9.80 m/s^2)(0.13 m))
T= 0.852 s