Anonymous
Anonymous asked in Science & MathematicsPhysics · 2 months ago

What is the rotational kinetic energy of the cylinder when it reaches the bottom of the ramp?

A 0.20-kg solid cylinder is released from rest at the top of a ramp 1.8 m long. The cylinder has a radius of 0.15 m, and the ramp is at an angle of 15 degrees with the horizontal.

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  • 2 months ago
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    the height of the ramp is 1.8 sin 15 = 0.466 m

    mgh = ½mV² + ½Iω² = 0.2•9.8•0.466 = 0.913 J

    mV² + Iω² = 1.826 J

    I = ½0.2•0.15² = 0.00225 kg•m²

    ω = V/r = V/0.15

    mV² + Iω² = 1.826 J

    0.2V² + (0.00225)(V/0.15)² = 1.826 J

    0.2V² + 0.1V² = 1.826 J

    0.3V² = 1.826 

    V = 2.47 m/s

    ω = 2.47/0.15 = 16.4 rad/s

    rot KE = ½Iω² = ½(0.00225)(16.4)² = 0.303 J

    I is moment of inertia in kg•m²

    I = cMR²

        M is mass (kg), R is radius (meters)

        c = 1 for a ring or hollow cylinder

        c = 2/5 solid sphere around a diameter

        c = 7/5 solid sphere around a tangent

        c = ⅔ hollow sphere around a diameter

        c = ½ solid cylinder or disk around its center

    cylinder or ball rolling down a slope

    PE at top results in both linear KE and rotational KE

       at the bottom.

    mgh = ½mV² + ½Iω²

    V = rω r is radius

    V² = (2gh) / (1 + I/mr²)

    for a solid cylinder that reduces to V² = (4/3)gh (1.33)

    for a solid sphere, that is V² = (10/7)gh (1.43)

    for a ring or hollow cylinder, that is V² = 2gh

    for a hollow sphere, that is V² = (6/5)gh (1.2)

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  • 2 months ago

    For a solid cylinder it requires half as much energy as the kinetic energy of a moving mass at the same speed.

    ie mgh = 2*Er + Er -> Er = 1/3 mgh = 1/3 * 0.20 * 9.8 * 1.8*sin(15)  = 0.30 J  ( and of course Ek = 2*Er = 0.61J   total = 0.91 J )

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