Two objects joined by a rectilinear cable rotating

[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?

 

[berkeman]

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

That sounds like a good hint. Can you upload your sketch please? Thanks.

 

[haruspex]

That sounds like a good hint. Can you upload your sketch please? Thanks.

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.

 

[berkeman]

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?

 

[haruspex]

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]

It says the cable is "rectilinear", i.e. straight.

Dagnabit, you are so literal. Okay, delete the pulley from my interpretation. Ack!

https://en.wikipedia.org/wiki/Rectilinear

 

[Guillem_dlc]

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.

 

[Steve4Physics]

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.

 

[berkeman]

There was no diagram, no.

Okay, then showing us your diagram is doubly-important.