# Two objects joined by a rectilinear cable rotating

• Guillem_dlc
Guillem_dlc
Homework Statement
We have two objects of mass ##3\, \textrm{kg}## joined by a rectilinear cable of ##3\, \textrm{m}## and negligible mass. The axis of rotation is normal to the cable and passes through it ##1\, \textrm{m}## from one of the objects, the moment of inertia of this object being ##7\, \textrm{kg m}^2##. If we want ##\omega =6\, \textrm{rad}/\textrm{s}## and ##L=10\, \textrm{kg m}^2/\textrm{s}##, what must be the moment of inertia of the second object with respect to the axis passing through its center of mass and parallel to the axis of rotation? Hints: It is advisable to draw a schematic of the system described. In addition, Steiner's theorem must be used in the calculations. Sol: ##15,67\, \textrm{kg m}^2##.
Relevant Equations
##L=I\omega##, Steiner's theorem
I've tried the following, but I don't get the correct result:

The moment of inertia of the system with respect to the axis of rotation is:

$$L=I\omega \Rightarrow I=\dfrac{L}{\omega}=\dfrac53 \, \textrm{kg m}^2$$

Then,

$$I=I_1+I_2\Rightarrow I_2=I-I_1=\dfrac53 -7=-\dfrac{16}3\, \textrm{kg m}^2$$

Finally, applying the Steiner's theorem:

$$I_2=I_{c2}+m_2d_2^2 \Rightarrow -\dfrac{16}{3}=I_{c2}+3\cdot 2^2 \Rightarrow I_{c2}=-17,33\, \textrm{kg m}^2$$

What have I done wrong in my reasoning?

Guillem_dlc said:
Hints: It is advisable to draw a schematic of the system described.

berkeman said:
The problem description makes no sense to me. I cannot come up with a diagram that matches the numbers, and it looks like @Guillem_dlc can't either.
I read the 7kg m2 as being the MoI about the object's centre, which is not how it is interpreted in the OP. But that only makes the numbers crazier.

haruspex said:
The problem description makes no sense to me.
Same here for at least 2 read-throughs. After a couple more, I'm thinking that the 2 masses are joined by a massless cable that goes over a pulley at a 90 degree angle and the pulley has an MOI that is involved in the problem. But that could be wrong assumptions on my part.

@Guillem_dlc -- Is this really how the whole problem was presented to you? There was no diagram, and you are supposed to come up with a sketch of your interpretation of the problem statement?

berkeman said:
I'm thinking that the 2 masses are joined by a massless cable that goes over a pulley at a 90 degree angle
It says the cable is "rectilinear", i.e. straight.

berkeman said:
Same here for at least 2 read-throughs. After a couple more, I'm thinking that the 2 masses are joined by a massless cable that goes over a pulley at a 90 degree angle and the pulley has an MOI that is involved in the problem. But that could be wrong assumptions on my part.

@Guillem_dlc -- Is this really how the whole problem was presented to you? There was no diagram, and you are supposed to come up with a sketch of your interpretation of the problem statement?
There was no diagram, no.

Guillem_dlc said:
If we want ##\omega =6\, \textrm{rad}/\textrm{s}## and ##L=10\, \textrm{kg m}^2/\textrm{s}## ...
The system's total moment of inertia would be ##I = \frac L{\omega} = \frac {10}6 \approx 1.67 kg~m^2##.

This is smaller than the given MoI for the 1st object, which is impossible.

Guillem_dlc
Guillem_dlc said:
There was no diagram, no.
Okay, then showing us your diagram is doubly-important.

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