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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

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

Classical Dynamics of Particles and Systems
Classical Dynamics of Particles and Systems
5th Edition
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Stephen T. Thornton, Jerry B. Marion
Chapter11: Dynamics Of Rigid Bodies
Section: Chapter Questions
Problem 11.15P
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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
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: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.
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).
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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