Question

2 small weights of M1=5-Kg. and M2=7-Kg. are mounted 4cm apart on a light rod as shown. The moment of inertia of the system
a. if rotated about the axis half between weights.
b. if rotated about the axis 0.50cm left of 7-Kg.mass

*There are 3 equations for the rod figure on our workbook. I don't know what to use.*
1. I center = mL^2/12
2. I end = mL^2/3
3. I = mR^2

Answer 1

2 small weights of M1=5-Kg. and M2=7-Kg. are mounted 4cm apart on a light rod as shown. The moment of inertia of the system

a. if rotated about the axis half between weights.

b. if rotated about the axis 0.50cm left of 7-Kg.mass

*There are 3 equations for the rod figure on our workbook. I don't know what to use.*

1. I center = mL^2/12

2. I end = mL^2/3

3. I = mR^2

Please explain and show your work

You shouldn't use equations for the rod ("on a light rod"). There are two point masses in your situation. So, the total moment of inertia will consist of two moments of inertia of the point mass:

$$I = I _ {1} + I _ {2} = M _ {1} r _ {1} ^ {2} + M _ {2} r _ {2} ^ {2}$$

a. $r_1 = r_2 = r = 0.02m$

$$I = \left(M _ {1} + M _ {2}\right) r ^ {2}$$ $$I = (5 k g + 7 k g) (0. 0 2 m) ^ {2} = 4. 8 * 1 0 ^ {- 3} k g * m ^ {2}$$

b. $r_1 = 0.035m, r_2 = 0.005m$

$$I = 5 k g * (0. 0 3 5 m) ^ {2} + 7 k g * (0. 0 0 5 m) ^ {2} = 6. 3 * 1 0 ^ {- 3} k g * m ^ {2}$$

Answer: a. $I = 4.8 * 10^{-3} kg * m^2$ , b. $I = 6.3 * 10^{-3} kg * m^2$