A solid cylinder and a thin-walled hollow cylinder, both of identical mass M and radius R, are released from rest at the top of an incline of height H. They both roll without slipping down the incline. Which of the following statements is true about their linear speeds (v) at the bottom of the incline?
Correct: A
This problem can be solved using the principle of conservation of mechanical energy. Since both objects roll without slipping, the static friction does no work (the point of contact is instantaneously at rest), and thus mechanical energy is conserved.
Initial Potential Energy (PE_initial) at height H = MgH (for both).
Final Kinetic Energy (KE_final) at the bottom consists of translational kinetic energy (1/2 Mv²) and rotational kinetic energy (1/2 Iω²).
Since the objects roll without slipping, v = Rω, so ω = v/R.
Conservation of Energy: PE_initial = KE_final
MgH = (1/2)Mv² + (1/2)I(v/R)²
MgH = (1/2)Mv² + (1/2)(I/R²)v²
MgH = (1/2)v² (M + I/R²)
v² = (2gH) / (M + I/R²)
Now, let's compare the moments of inertia (I) for the two objects:
1. **Solid Cylinder:** I_solid = (1/2)MR²
So, I_solid / R² = (1/2)M
v_solid² = (2gH) / (M + (1/2)M) = (2gH) / ((3/2)M) = (4gH) / (3)
v_solid = sqrt( (4/3)gH )
2. **Thin-Walled Hollow Cylinder:** I_hollow = MR²
I_hollow / R² = M
v_hollow² = (2gH) / (M + M) = (2gH) / (2M) = gH
v_hollow = sqrt( gH )
Comparing the speeds:
v_solid = sqrt( (4/3)gH ) and v_hollow = sqrt( gH )
Since 4/3 > 1, it means v_solid > v_hollow. The solid cylinder has a smaller fraction of its kinetic energy tied up in rotation (smaller moment of inertia for the same mass and radius), meaning more energy is available for translational motion.
Therefore, the solid cylinder reaches the bottom with a greater linear speed.
The correct choice is A.