Answer:
The length of the rod must be 1.226 meter
Explanation:
Step 1: Data given
mass of rod = 4.17 kg
mass of the bodies = each 0.519 kg
length of the rod = L
I = 0.913 kg*m²
Step 2: Calculate moment of inertia of the rod
Consider L = length of the rod.
⇒ Moment of inertia of the rod = (1/12)*4.17*L²
Step 3: Moment of inertia of the 2 bodies
Moment of inertia of each of the two bodies = 0.519*(L/2)²
Step 4: Total moment of inertia
Total moment of inertia of the three-body system =
(1/12)*4.17*L² + 2*0.519*(L/2)² = 0.913 kg*m²
0.3475* L² +0.2595 *L²= 0.913
0.607 L² = 0.913
L² = 1.504
L = 1.226 m
The length of the rod must be 1.226 meter
The length L of the rod must be about 1.23 m
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Centripetal Acceleration of circular motion could be calculated using following formula:
[tex]\boxed {a_s = v^2 / R}[/tex]
where:
a = centripetal acceleration ( m/s² )
v = velocity ( m/s )
R = radius of circle ( m )
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Let's recall Moment of Inertia formula as follows:
[tex]\boxed {I = m R^2}[/tex]
where:
I = moment of inertia ( kg m² )
m = mass of object ( kg )
R = radius ( m )
Let us now tackle the problem!
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Given:
mass of uniform thin rod = M = 4.17 kg
mass of small body = m = 0.519 kg
moment of inertia = I = 0.913 kg m²
Asked:
length of the rod = L = ?
Solution:
Total Moment of Inertia = ( Moment of Inertia of thin rod) + 2 ( Moment of Inertia of small body )
[tex]I = \frac{1}{12}ML^2 + 2 ( m R^2 )[/tex]
[tex]I = \frac{1}{12}ML^2 + 2 ( m (\frac{L}{2})^2 )[/tex]
[tex]I = \frac{1}{12}ML^2 + \frac{1}{2}mL^2[/tex]
[tex]I = \frac{1}{12}L^2(M + 6m)[/tex]
[tex]L^2 = 12I \div ( M + 6m )[/tex]
[tex]L = \sqrt{ 12I \div ( M + 6m ) }[/tex]
[tex]L = \sqrt{ 12( 0.913 ) \div ( 4.17 + 6(0.519) ) }[/tex]
[tex]\boxed{L \approx 1.23 \texttt{ m}}[/tex]
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Grade: High School
Subject: Physics
Chapter: Rotational Motion
