Final answer:
To compare the weights of a solid and a hollow circular shaft required to transmit the same torque with equal maximum shear stresses, calculate the volumes based on their geometries and multiply by the material's density. It turns out that the hollow shaft weighs 7/16 of the solid shaft's weight.
Step-by-step explanation:
The problem of comparing the weights of a solid circular shaft and a hollow circular shaft to transmit a given torque involves concepts from mechanical engineering and physics. Since both shafts are made of the same material and have equal lengths, we need to focus on the differences in volume due to the shafts' geometries, as the weight is directly proportional to the volume for a constant density material.
For both solid and hollow circular shafts, the volume can be calculated based on their geometric shapes. Assuming that the shafts have the same outer diameter, D, the volume of the solid shaft, Vs, is just π/4 * D^2 * length. For the hollow shaft, the inside diameter is given as (3/4)D, so its volume, Vh, is π/4 * ((D^2) - ((3/4)D)^2) * length.
To compare the weights, Ws and Wh, of the solid and hollow shafts respectively, we use their volumes and the material's density, ρ: Ws = ρ * Vs and Wh = ρ * Vh. After simplification, we find that Wh = Ws * (7/16), which means the hollow shaft weighs 7/16 the weight of the solid shaft when they are designed to support the same maximum shear stress.