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27 November, 13:38

A solid cylinder of mass 3.0 kg and radius 0.2 m starts from rest at the top of a ramp, inclined 15°, and rolls to the bottom without slipping. (For a cylinder I = MR2.) The upper end of the ramp is 1.5 m higher than the lower end. Find the linear speed of the cylinder when it reaches the bottom of the ramp. (g = 9.8 m/s2)

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  1. 27 November, 13:53
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    v = 4.4271 m/s

    Explanation:

    Given

    m = 3 Kg

    R = 0.2 m

    ∅ = 15°

    h = 1.5 m

    g = 9.8 m/s²

    v = ?

    Ignoring frictional losses, at the bottom of the plane

    Total kinetic energy is = Potential Energy at the top of plane

    Using Law of conservation of energy we have

    U = Kt + Kr

    m*g*h = 0.5*m*v² + 0.5*I*ω²

    knowing that

    Icylinder = 0.5*m*R²

    ω = v/R

    we have

    m*g*h = 0.5*m*v² + 0.5 * (0.5*mR²) * (v/R) ² = 0.75*m*v²

    ⇒ v = √ (g*h/0.75) = √ (9.8 m/s²*1.5 m/0.75)

    ⇒ v = 4.4271 m/s
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