Three different objects—a solid sphere, a solid cylinder, and a thin-walled hollow cylinder—all have the same mass M and radius R. They are released from rest at the same height on an incline and roll without slipping. Which statement correctly describes their motion down the incline?
Correct: B
This problem requires comparing the distribution of mass, which affects the moment of inertia, and thus how mechanical energy is distributed between translational and rotational kinetic energy during rolling.
For an object rolling without slipping down an incline, mechanical energy is conserved:
Initial Potential Energy (PE_initial) = MgH
Final Kinetic Energy (KE_final) = Translational KE + Rotational KE = (1/2)Mv² + (1/2)Iω²
Since v = Rω for rolling without slipping, ω = v/R.
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)
To determine which object reaches the bottom first (i.e., has the greatest final linear speed 'v'), we need to compare the quantity (M + I/R²). The smaller this quantity, the greater the final speed.
Let's list the moments of inertia (I) and the term I/R² for each object:
1. **Solid Sphere:** I_sphere = (2/5)MR²
I_sphere/R² = (2/5)M = 0.4M
So, (M + I_sphere/R²) = M + 0.4M = 1.4M
2. **Solid Cylinder:** I_cylinder = (1/2)MR²
I_cylinder/R² = (1/2)M = 0.5M
So, (M + I_cylinder/R²) = M + 0.5M = 1.5M
3. **Thin-Walled Hollow Cylinder:** I_hollow = MR²
I_hollow/R² = M
So, (M + I_hollow/R²) = M + M = 2.0M
Comparing the denominators for v²:
- Solid Sphere: 1.4M
- Solid Cylinder: 1.5M
- Hollow Cylinder: 2.0M
The object with the smallest denominator will have the largest final speed. Therefore, the solid sphere will have the greatest speed, followed by the solid cylinder, and then the hollow cylinder.
Since all objects start from the same height, the object with the greatest final speed will complete the incline in the shortest time.
Thus, the solid sphere reaches the bottom first, then the solid cylinder, and finally the hollow cylinder.
The correct choice is B.