A) \[\frac{4\omega }{5}\]
B) \[\frac{2\omega }{5}\]
C) \[\frac{3\omega }{2}\]
D) \[\frac{2\omega }{3}\]
Correct Answer: D
Solution :
First, we find out the moment of inertia of the disc about the axis passing the centre and normal to through the plane is \[I=\frac{M{{R}^{2}}}{2}\] So, for first disc \[{{I}_{1}}=\frac{M{{R}^{2}}}{2}\] Similarly, for second disc \[{{I}_{2}}=\frac{2M}{2}{{\left( \frac{R}{2} \right)}^{2}}=\frac{M{{R}^{2}}}{4}\] Total moment of inertia of the whole system \[I={{I}_{1}}+{{I}_{2}}=\frac{M{{R}^{2}}}{2}+\frac{M{{R}^{2}}}{4}=\frac{3}{4}M{{R}^{2}}\] According to the conservation of angular momentum \[{{I}_{1}}\omega =I\omega \] \[\frac{M{{R}^{2}}}{2}\times \omega =\frac{3}{4}M{{R}^{2}}\times \omega \] So, \[\omega =\frac{2}{3}\omega \]
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