Final answer:
The linear speed of the 0.0800 kg sphere as it passes through its lowest point can be found by applying the conservation of mechanical energy to convert the initial potential energy to rotational kinetic energy, determining the angular velocity, and then using it to calculate the linear velocity.
Step-by-step explanation:
To determine the linear speed of the 0.0800 kg sphere as it passes through its lowest point after being released from a horizontal position, we utilize the conservation of mechanical energy. Initially, the system has potential energy when held horizontally, and as it falls, this potential energy is converted into rotational kinetic energy. Since no external forces do work (frictionless pivot), the total mechanical energy is conserved.
The potential energy (PE) of the system initially is equal to the rotational kinetic energy (KErot) at the lowest point:
PE = KErot
For a rigid body rotating about a fixed axis, KErot = (1/2)Iω2, where I is the moment of inertia and ω is the angular velocity.
The rod and spheres system's moment of inertia I is the sum of the moment of inertia of the slender rod (Irod) plus the moments of inertia of each sphere, assuming the spheres can be considered as point masses at the ends of the rod. So, I = Irod + m1r12 + m2r22, where m1 and m2 are the masses of the spheres, and r1 and r2 are their distances from the pivot point.
The slender rod's moment of inertia is Irod = (1/12)ML2, where M is the mass of the rod and L its length.
As the spheres move, their potential energy is converted to kinetic energy. Using the conservation of energy, we can find the angular velocity (ω) at the lowest point, and then calculate the linear speed (v) of the 0.0800 kg sphere using the relationship v = rω, where r is the radius of the rotation.