Two weights on a bar: different axis, different I. Two small "weights," of mass 5.2 kg and 8.3 kg, are mounted 6.0 m apart on a light rod (whose mass can be ignored), as shown in Fig. 8–19.  Calculate the moment of inertia of the system when rotated about an axis halfway between the weights, Fig. 8–19a. Units = kg*m^2

Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter13: Universal Gravitation
Section: Chapter Questions
Problem 13.79CP: The oldest artificial satellite still in orbit is Vanguard I, launched March 3, 1958. It mass is...
icon
Related questions
icon
Concept explainers
Question

Two weights on a bar: different axis, different I. Two small "weights," of mass 5.2 kg and 8.3 kg, are mounted 6.0 m apart on a light rod (whose mass can be ignored), as shown in Fig. 8–19. 

Calculate the moment of inertia of the system when rotated about an axis halfway between the weights, Fig. 8–19a.

Units = kg*m^2

 

Two weights on a bar: different axis, different I. Two small "weights," of mass 5.6 kg and 6.9 kg, are mounted 4.0 m apart on a light rod (whose mass can be ignored), as shown in Fig. 8–19. 

Calculate the moment of inertia of the system when rotated about an axis halfway between the weights, Fig. 8–19a.

 

Two weights on a bar: different axis, different I. Two small "weights," of mass 7.2 kg and 5.6 kg, are mounted 5.0 m apart on a light rod (whose mass can be ignored), as shown in Fig. 8–19. 

Calculate the moment of inertia of the system when rotated about an axis 1.83 m to the left of the 7.2-kg mass (Fig. 8–19b).

 

Two weights on a bar: different axis, different I. Two small "weights," of mass 8.5 kg and 8.5 kg, are mounted 8.0 m apart on a light rod (whose mass can be ignored), as shown in Fig. 8–19. 

Calculate the moment of inertia of the system when rotated about an axis 0.95 m to the left of the 8.5-kg mass (Fig. 8–19b).

EXAMPLE 8-9 Two weights on a bar: different axis, different I. Two
small "weights," of mass 5.0 kg and 7.0 kg, are mounted 4.0 m apart on a light
rod (whose mass can be ignored), as shown in Fig. 8-19. Calculate the moment
of inertia of the system (a) when rotated about an axis halfway between the
weights, Fig. 8-19a, and (b) when rotated about an axis 0.50 m to the left of the
5.0-kg mass (Fig. 8-19b).
APPROACH In each case, the moment of inertia of the system is found by
summing over the two parts using Eq. 8-13.
SOLUTION (a) Both weights are the same distance, 2.0 m, from the axis of rota-
tion. Thus
I =
Emr² = (5.0 kg)(2.0 m)² + (7.0 kg)(2.0 m)²
= 20 kg m² + 28 kg-m² = 48 kg-m².
(b) The 5.0-kg mass is now 0.50 m from the axis, and the 7.0-kg mass is 4.50 m
from the axis. Then
I = Emr² = (5.0 kg)(0.50 m)2 + (7.0 kg)(4.5 m)²
= 1.3 kg-m² + 142 kg m² = 143 kg-m².
NOTE This Example illustrates two important points. First, the moment of
inertia of a given system is different for different axes of rotation. Second, we
see in part (b) that mass close to the axis of rotation contributes little to the total
moment of inertia; here, the 5.0-kg object contributed less than 1% to the total.
FIGURE 8-19 Example 8-9:
calculating the moment of inertia.
4.0 m-
5.0 kg
7.0 kg
Axis
(a)
-4.0 m
(b)
0.50 m
it
| 5.0 kg
7.0 kg
1
Axis
CAUTION
I depends on axis of rotation
and on distribution of mass
Transcribed Image Text:EXAMPLE 8-9 Two weights on a bar: different axis, different I. Two small "weights," of mass 5.0 kg and 7.0 kg, are mounted 4.0 m apart on a light rod (whose mass can be ignored), as shown in Fig. 8-19. Calculate the moment of inertia of the system (a) when rotated about an axis halfway between the weights, Fig. 8-19a, and (b) when rotated about an axis 0.50 m to the left of the 5.0-kg mass (Fig. 8-19b). APPROACH In each case, the moment of inertia of the system is found by summing over the two parts using Eq. 8-13. SOLUTION (a) Both weights are the same distance, 2.0 m, from the axis of rota- tion. Thus I = Emr² = (5.0 kg)(2.0 m)² + (7.0 kg)(2.0 m)² = 20 kg m² + 28 kg-m² = 48 kg-m². (b) The 5.0-kg mass is now 0.50 m from the axis, and the 7.0-kg mass is 4.50 m from the axis. Then I = Emr² = (5.0 kg)(0.50 m)2 + (7.0 kg)(4.5 m)² = 1.3 kg-m² + 142 kg m² = 143 kg-m². NOTE This Example illustrates two important points. First, the moment of inertia of a given system is different for different axes of rotation. Second, we see in part (b) that mass close to the axis of rotation contributes little to the total moment of inertia; here, the 5.0-kg object contributed less than 1% to the total. FIGURE 8-19 Example 8-9: calculating the moment of inertia. 4.0 m- 5.0 kg 7.0 kg Axis (a) -4.0 m (b) 0.50 m it | 5.0 kg 7.0 kg 1 Axis CAUTION I depends on axis of rotation and on distribution of mass
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps

Blurred answer
Knowledge Booster
Moment of inertia
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Physics for Scientists and Engineers, Technology …
Physics for Scientists and Engineers, Technology …
Physics
ISBN:
9781305116399
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Physics for Scientists and Engineers with Modern …
Physics for Scientists and Engineers with Modern …
Physics
ISBN:
9781337553292
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
College Physics
College Physics
Physics
ISBN:
9781938168000
Author:
Paul Peter Urone, Roger Hinrichs
Publisher:
OpenStax College