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A solid cylinder and a hollow cylinder both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?
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- Both together
- Solid cylinder
- One with higher density
- Hollow cylinder
Correct Option: B
Torque, Iα = ƒ. R.
Using Newton's IInd law, mg sin θ – ƒ = ma
∵ pure rolling is there, a = Rα
⇒ mg sin θ - | = ma | |
R |
⇒ mg sin θ - | = ma | |
R² |
![]() | ∵ α = | ![]() | R | |
αβ |
or, acceleration, a = | ||
I/R² + m |
Using, s = ut + | at² | |
2 |
or, s = | at² ⇒ t α | ||
2 | √a |
t minimum means a should be more. This is possible when I is minimum which is the case for solid cylinder.
Therefore, solid cylinder will reach the bottom first.