Part 1:
Consider the given figure.
Write the equation of mass moment of inertia of the rod about vertical axis passing
through its Centre of gravity G.
 m  Mass of rod 


 L  Length of rod 
mL2
IG 
12
Calculate the moment of inertia for the rod of mass 5 kg about the axis shown in
figure using Parallel axis theorem.
2
2
 L2  L
mL2
L

 
I P  I G  md 
 m   10 m   m     10 m  
12
2

 
12  2
2
2
 10 m + 50 m  2  10 m + 50 m 
 
  5 kg  

 10 m  
12
2


 
  60 m  2

2
  5 kg  
  30 m  10 m  
 12

 3500 kg-m 2
Part 2:
Consider the given figure.
Write the equation of mass moment of inertia of the rod about vertical axis passing
through its Centre of gravity G.
IG 
 m  Mass of rod 


 L  Length of rod 
mL2
12
Calculate the moment of inertia for the rod of mass 6.25 kg about the axis shown in
figure using Parallel axis theorem.
2
2
 L2  L
mL2
L

 
I Q  I G  md 
 m   21 m   m     21 m  
12
2

 
 12  2
2
2
  21 m + 42 m  2   21 m + 42 m 
 
  6.25 kg  

 21 m  
12
2


 
  63 m  2

2
  6.25 kg  
  31.5 m  21 m  
 12

 2756.25 kg-m 2
Part 3:
Write the equation for Newton second law of rotation.
T  I
 T  Applied torque



 I  Mass moment of inertia of rod 
   Angular acceleration of the rod 


Torque applied on both of the rod is same. Hence,
TP  TQ
 I P P  I Q Q

 P IQ

Q IP
 P 2756.25 kg-m 2

 0.7875
Q
3500 kg-m 2
P
1
Q
  P  Q
Hence, larger magnitude of angular acceleration result for the rod in Part (2).
Part 4:
Determine the moment of inertia for this system.
I axis  md 2  md 2  2md 2  2  5 kg  2 m   40 kg-m 2
2
 m  Mass of barbells



 d  Horizontal distance of each mass from axis of rotation 
Part 5:
Determine the moment of inertia for this system.
I axis  md L 2  md R 2   5 kg  0 m    5 kg  4 m   0  80 kg-m 2  80 kg-m 2
2
2
 m  Mass of barbells



 d L  Horizontal distance of left mass from axis of rotation 
 d  Horizontal distance of right mass from axis of rotation 
 R

Part 6:
Write the equation for Newton second law of rotation.
T  I
 T  Applied torque



 I  Mass moment of inertia about axis of rotation 
   Angular acceleration of the mass system



TI
 For constant angular acceleration ,   constant 
For same angular acceleration for each parts, required applied torque is directly
proportional to the mass moment of inertia of the system.
TI
 I axis  I axis
I5  I 4
 T5  T4
 I   80 kg-m , I  40 kg-m 
2
axis
2
axis
As required torque for part 5 is larger as compared to required torque for part 4.
Hence, for part 5, it would be more difficult to rotate the barbells.
References
Alrasheed S. , (2019, January). Fundamental university physics: Principles of
Mechanics.