- #1

**Mariam**

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- 1

## Homework Statement

Two particles of mass m1 = 14 kg and m2 = 24 kg are connected by a massless rod of length 1.4 m. Find the moment of inertia of this system for rotations about the following pivot points. Assume the rotation axis is perpendicular to the rod.

A) center of rod

B) end of m1

C) end of m2

D) at center of mass

## Homework Equations

I=1/12 m l^2

I=1/3 m l^2

I'=Icm+md

## The Attempt at a Solution

first I calculate the center of mass :

Xcm= ((x1m1)*(x2m2))/(m1+m2)

= (1.4*24)/(14+24)=

Xcm= 0.88 m from m1

for A) and D)

(i am confused which to use for the formula of I=1/12 m l^2 and which to use parallel axis theorem.)

for B)

I=1/3(24)(1.4)^2

I=15.68 kgm^2

(here i am a bit confused as to whether or not to add the 14 kg mass. So is it correct or not?)

for C)

I=1/3(14)(1.4)^2

I=9.15 kgm^2

(same problem as B)

for D)

first I calculate the center of mass :

xcm= ((x1m1)*(x2m2))/(m1+m2)

= (1.4*24)/(14+24)=0.88 m from m1

Extra question: What if the rod had a mass? How will moment of inertia be then?Sorry, if my questions are basic but I have a physics test tomorrow and my teacher didn't explain moment of inertia a lot.

Thanks in advance.