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29 November, 06:27

A thin cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, without slipping, from the top of an inclined plane that is 1.8 m above the ground. Find the final linear velocity of the thin cylindrical shell. The acceleration of gravity is 9.8 m/s 2. Answer in units of m/s.

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  1. 29 November, 07:32
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    Let the velocity be v

    Total energy at the bottom

    = rotational + linear kinetic energy

    = 1/2 Iω² + 1/2 mv² (I moment of inertia of shell = mr²)

    = 1/2 mr²ω² + 1/2 mv² (v = ω r)

    = 1/2 mv² + 1/2 mv²

    = mv²

    mv² = mgh (conservation of energy)

    v² = gh

    v = √gh

    = √9.8 x 1.8

    = 4.2 m / s
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