Peoples Physics Concepts

http://www.ck12.org Chapter 9. Rotational Motion

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Example 2

Two 2 kg masses is placed at either end of a rod that has a mass of .5 kg and a length of 3 m. What is the moment of
inertia if the system it is rotated about (a) one end of the rod and (b) the center of the rod?

Solution

For both parts of this problem we will be summing the moment of inertia’s for each component of the system based
on the table above. We will treat the masses as satellites when calculating their moment of inertia’s.
(a): For this part, we’ll only add the value for the rod being rotated about one end and one of the masses because
one of the masses is at the axis of rotation.

I=ΣIi=

1

3

mL^2 +ML^2

I=

1

3

.5kg∗(3 m)^2 +2 kg∗(3 m)^2

I= 19 .5 kg*m^2

(b):For the second part, we’ll sum the moment of inertia’s of both masses and the rod being rotated about it’s center.

I=

1

12

m(

L

2

)^2 + 2 M(

L

2

)^2

I=

1

12

.5 kg∗(

3m

2

)^2 + 2 ∗2 kg∗(

3m

2

)^2

I= 9 .1 kg*m^2

Watch this Explanation

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