- #1
Guillem_dlc
- 188
- 17
- 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?
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?
Can you upload your sketch please? Thanks.