Question

A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be. 

(A) (2/R) √(gh) 

(B) (2/R) √(gh / 2) 

(C) (2/R) √(gh / 3) 

(D) (1 / 2R) √(gh)

Answer 1

Correct option: (C) (2/R) √(gh / 3)

Explanation:

MI of solid cylinder = [(MR²) / 2]

∵ PE = KE

Mgh = (1/2)MV² + (1/2)Iω² 

= (1/2)MV² + (1/2)[(RM²) / 2] (V² / R²)------v = rω

Mgh = (1/2)MV² + [(MV²) / 4]

gh = (V² / 2) + (V² / 4)

gh = [(3V²) / 4]

hence V² = [(4gh) / 3] i.e. V = 2√[(gh) / 3]

ω = (V / r) = (2/R)√(gh / 3)