Assume that a physical pendulum can be modeled as a long, thin cylindrical rod, pivoting around one end. The mass of the pendulum is 630. g. Its period is 2.444 s. If its moment of inertia can be calculated by I = 1/3mL2, what is the length, L, of the rod? (The answer it’s supposed to be 1.06m) in the image there are two possible formulas
Assume that a physical pendulum can be modeled as a long, thin cylindrical rod, pivoting around one end. The mass of the pendulum is 630. g. Its period is 2.444 s. If its moment of inertia can be calculated by I = 1/3mL2, what is the length, L, of the rod? (The answer it’s supposed to be 1.06m) in the image there are two possible formulas
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Assume that a physical pendulum can be modeled as a long, thin cylindrical rod, pivoting around one end. The mass of the pendulum is 630. g. Its period is 2.444 s. If its moment of inertia can be calculated by I = 1/3mL2, what is the length, L, of the rod?
(The answer it’s supposed to be 1.06m)
in the image there are two possible formulas

Transcribed Image Text:Period of a Pendulum-Your Leg
What is the natural period of a human leg? This is an important consider-
ation for "smart" prosthetic legs that adjust to the motion of the user.
Solution:
First, model an average human leg as a thin solid rod of length L
and mass m 16.0 kg. From Appendix F, the moment of inertia for a thin
solid rod is I mL/3. The pendulum arm is the distance from the pivot
point to the center of mass. Assume for this example that the mass is uni-
formly distributed throughout the leg; then L/2.
= 1.5 m
Use Equation 12.10 to determine the period of a human leg.
Substitute the known information into the formula and solve for T..
T= 2m
mgle
V lg|/2)
= 2T
T-2V3191
2L
3|g|
2(1.5 m)
31-9.81 m/s
T 2.00 s= 2.0 s
The natural period for an entire human leg is about 2 seconds.

Transcribed Image Text:EXAMPLE 12-1
SHM in a Spring-Mass System
A 0.190 kg mass is hung from an ideal spring with a spring constant
k 76.0 N/m. It is set in motion by displacing it 3.00 cm downward, then
releasing it. (a) What is the period of its motion? (b) What is the
frequency?
Solution:
a. Use Equation 12.6 to solve for period:
T= 2TTK
0.190 kg
76.0 N/m
=2T
kg
(kg-m.)/(m-s)
(unit cancellation)
T 0.3141.
A,
T 0.314 s
b. Take the reciprocal of the period to find the frequency:
f==0.3141 s
f= 3.183 s = 3.18 Hz
Notice that the initial displacement, or amplitude, of the mass has no
effect on period (T) or frequency (f).
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