From newton's second law 40+f=m(Rα) ...(i) Taking torque about 0 we get 40×R−f×R=Iα 40×R−f×R=mR2α 40−f=mRα ...(ii) Solving equation (i) and (ii) α=mR40=16rad/s2
a string is wound around a hollow cylinder of mass 5 ~kg and radius 0.5 ~m. If the string is now pulled with a horizontal force of 40 ~N, and the…
JEE Main 2019 — Physics Mechanics
2019mcqmedium
a string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string) 
Official previous-year question
Held on 11 Jan 2019 · Verified 6 Jul 2026.
Options
- A
20rad/s2
- B
16rad/s2
- C
12rad/s2
- D
10rad/s2
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