
#1
Jan209, 11:33 PM

P: 74

1. The problem statement, all variables and given/known data
A rod of negligible mass is pivoted about one end. Masses can be attached to the rod at various positions along the rod. Currently, there is a mass (m) attached a distance (L) from the pivot. To increase the moment of inertia about the pivot by a factor of 5, you must attach... 2. Relevant equations I=mr^2 3. The attempt at a solution Io= mL^2 If= 5(Io) = mL^2(original mass)+4(mL^20) All I did was add four more masses of equal size at a distance L from the pivot. Is there another solution? 



#2
Jan309, 01:05 AM

PF Gold
P: 619

You are correct. The definition of moment of Inertia is [tex]\sum_{i} m_{i} r^{2}_{i}[/tex]




#3
Jan309, 06:36 AM

Mentor
P: 40,905

For example: What if you could only add a single mass of equal size. How could you solve the problem then? 


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