Two masses are joined by a massless rod of length 0.8m. One mass is 1kg and the

Question

Two masses are joined by a massless rod of length 0.8m. One mass is 1kg and the other is 3kg. They are rotating with a period of 0.25s. What is their moment of inertia and what is their angular momentum?

Where the system is rotating aound in this question? How can you deltermin the r to calculate moment of intertia (I = sum(mr)?

Somebody asnwered as below:

he found center of mass first,then said

So the system is rotating in circular part with the radius 0.6m (aroud where? How did he know it?)                                

Explanation / Answer

The length of rod, L = 0.8 m

masses m1 = 1 kg ; m2 = 3 kg ;

The time period, T = 0.25 s

Let the centre of mass of the system is at a distance ' x ' from m1 and 0.8-x

from m2

Then, m1x = m2(0.8 - x)

m1x + m2x = m2 * 0.8

From the above, x = 0.6 m

So the system is rotating in circular part with the radius 0.6m

So the moment of inertia, I = mr^2 = (m1 + m2)r^2 = 1.44 kg m^2

The angular velocity, ? = 2p/T = 25.13 rad/s

The angular momentum, L = I? = 36.18 Js

b. Conservation of angular momentum

[ 1 * ( 0.8 /2 ) ^2 + 3 * ( 0.8 / 2 ) ^2 ] ( 2 p / 0.25 ) = [ 1 * ( 0.8 /2 ) ^2 + 3 * ( 0.8 / 2 ) ^2 ] ( 2 p / 0.25 )

new

I = [ 3 * 0.2^2 + 1 * ( 0.8 - 0.2 )^2 ] = 0.48 ;

L = I ? = 0.48 * ( 2p / 0.25 ) = 12.0637

-----------------------------------------------------------------------------------------

The linear density = 400gm/m

So the mass of given rod, m3 = 400gm * 0.8 = 320 gm = 0.32 kg

Then the moment of inertia, I = (m1 + m2 + m3)r^2 = 1.56 kg m^2

The angular momentum, L = I? = 39.2 Js

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