A solid cylinder of mass 2 kg and radius 50 cm rolls up an inclined plane of angle of inclination 30°. The centre of mass of cylinder has speed of 4 m/s. The distance travelled by the cylinder on the incline surface will be (Take g=10 m/s²)

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(4) 2.4 m. Given: solid cylinder, v=4 m s⁻¹, θ=30°, g=10 m s⁻². Idea: rolling without slipping means static friction does no work, so all the kinetic energy becomes gravitational potential energy. K=1/2mv²+1/2(1/2mR²)(v²)/(R²)=3/4mv² 3/4mv²=mgh⇒ h=(3v²)/(4g)=(3×16)/(40)=1.2 m s=h/(sin θ)=(1.2)/(0.5) s=2.4 m. The mass and the radius both cancel.

A solid cylinder of mass 2 kg and radius 50 cm rolls up an inclined plane of angle of inclination 30°. The centre of mass of cylinder has speed of 4 m/s. The distance travelled by the cylinder on the incline surface will be (Take g=10 m/s²)

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