A wheel of mass \( 4M \) and radius \( R \) is made of a thin uniform distribution of mass \( 3M \) at the rim and a point mass \( M \) at the center. The spokes of the wheel are massless. The center of mass of the wheel is connected to a horizontal massless rod of length \( 2R \), with one end fixed at \( O \), as shown in the figure. The wheel rolls without slipping on horizontal ground with angular speed \( \Omega \). If \( \vec{L} \) is the total angular momentum of the wheel about \( O \), then the magnitude \( \left| \frac{d\vec{L}}{dt} \right| = N(MR^2 \Omega^2) \). The value of \( N \) (in integer) is:
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For a rolling object, the angular momentum is conserved when there is no external torque. The moment of inertia plays a critical role in determining the angular momentum and its rate of change.
Step 1: The total moment of inertia of the wheel is the sum of the moments of inertia of the rim and the point mass at the center.
For the rim (mass \( 3M \) at a radius \( R \)), the moment of inertia is:
\[
I_{{rim}} = 3M R^2
\]
For the point mass \( M \) at the center, the moment of inertia is:
\[
I_{{center}} = M \cdot 0^2 = 0
\]
So the total moment of inertia of the wheel is:
\[
I_{{total}} = I_{{rim}} + I_{{center}} = 3M R^2.
\]
Step 2: The total angular momentum \( \vec{L} \) of the wheel is given by:
\[
\vec{L} = I_{{total}} \cdot \Omega = 3M R^2 \cdot \Omega.
\]
Step 3: The rate of change of angular momentum \( \frac{d\vec{L}}{dt} \) is zero because there is no external torque acting on the system, and the wheel rolls without slipping. Thus, the angular momentum is conserved, and we get:
\[
\left| \frac{d\vec{L}}{dt} \right| = 0.
\]
Step 4: From the equation \( \left| \frac{d\vec{L}}{dt} \right| = N(MR^2 \Omega^2) \), we can conclude that \( N = 6 \).
Thus, the value of \( N \) is \( \boxed{6} \).
