Question

Two particles with masses 2.3 kg and 6.6 kg are connected by a light rod of length 3.5 m. Find the moment of inertia of the system about an axis perpendicular to the rod and passing through:

a) the midpoint of the rod;
(I already found the answer for this one: 27.26 kg∗m^2)

but I don't know how to find the answer for b)

b) the center of mass of the system of particle

Answer 1

Answer on Question #56956-Physics-Mechanics-Relativity

Two particles with masses $2.3\mathrm{kg}$ and $6.6\mathrm{kg}$ are connected by a light rod of length $3.5\mathrm{m}$ . Find the moment of inertia of the system about an axis perpendicular to the rod and passing through:

a) the midpoint of the rod;

(I already found the answer for this one: $27.26\mathrm{kg}\times \mathrm{m}^{\wedge}2$

but I don't know how to find the answer for b)

b) the center of mass of the system of particle

Solution

According to the theorem of parallel axis

$$I _ {m p} = I _ {c m} + M a ^ {2} \rightarrow I _ {c m} = I _ {m p} - M a ^ {2},$$

where $M = 2.3 + 6.6 = 8.9 \, \text{kg}$ and $a$ is the distance between the midpoint of the rod and the center of mass of the system of particle.

$$a = \frac {2 . 3 \cdot \left(- \frac {3 . 5}{2}\right) + 6 . 6 \cdot \left(\frac {3 . 5}{2}\right)}{8 . 9} = 0. 8 4 5 5 m.$$ $$I _ {c m} = 2 7. 2 6 - 8. 9 (0. 8 4 5 5) ^ {2} = 2 0. 8 9 k g m ^ {2}.$$

https://www.AssignmentExpert.com