The figure above shows two barbells, which can be modeled as two point masses connected by a massless rod. In the figure, L = 0.77 m and m = 20.1 kg. The barbells can rotate about an axis that is perpendicular to the rod and passes through the midpoint of each rod. For the barbell with the larger moment of inertia, give the moment of inertia (in units of kg m2). The diagram provides a visual representation of two systems involving masses and rods. **Top Diagram:** - Two spheres, each labeled with mass \( m \), are connected by a horizontal rod.- The total distance between the centers of the two masses is marked as \( L \).**Bottom Diagram:**- Two spheres with different masses are shown. The left sphere is labeled with mass \( 2m \), and the right sphere is labeled with mass \( m \).- These are also connected by a horizontal rod.- The distance between the centers of these two masses is marked as \( L/2 \).Both diagrams illustrate how different allocations of mass and their separations are represented in simple mechanical systems, which could be used to study concepts such as center of mass or rotational inertia.

Given : L = 0.77 m Mass m = 20.1 kg Moment of inertia of barebells is given by I=m1m2m1+m2L2 Plugging value of L and m we get I=20.1×20.120.1+20.10.772 By solving above equation we get I=404.0140.2×0.5929 I=5.96 kg.m2 ≈6 kg.m2For second figure m1=2mm2=mL'=L/2 Moment of inertia of barebells is given by I=m1m2m1+m2L2 Plugging value of L and m we get I=2m×m2m+m×L22 ⇒I=2m23m×L24 =mL26 By solving above equation we get I=20.1×0.7726 =11.926=1.99 ≈2 kg.m2 Therefore larger moment of inertia is for figure 1 which is 6 kg.m2 .