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16 February, 21:20

Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1. solid sphere 2. spherical shell 3. hoop 4. cylinder m/s

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  1. 16 February, 21:28
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    1. 14.4 m/s 2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

    Explanation:

    Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

    ω = v/r

    If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

    ∆K = - ∆U

    ½ * m*v² + ½ * I * ω² = m*g*h

    Simplifying and replacing the value of the angular velocity:

    ½ * v² + ½ I * (v/r) ² = g*h (1)

    In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

    Solid Sphere I = 2/5 * m * r²

    Replacing in (1):

    ½ * v² + ½ (2/5 * m*r²) * (v/r) ² = g*h

    Replacing by the value given for h, and solving for v:

    v = √ (10/7*9.8 m/s2*14.7 m) = 14. 4 m/s

    Spherical shell I=2/3*m*r²

    Replacing in (1):

    ½ * v² + ½ (2/3 * m*r²) * (v/r) ² = g*h

    Replacing by the value given for h, and solving for v:

    v = √ (6/5*9.8 m/s2*14.7 m) = 13.2 m/s

    Hoop I = m*r²

    Replacing in (1):

    ½ * v² + ½ (m*r²) * (v/r) ² = g*h

    Replacing by the value given for h, and solving for v:

    v = √ (9.8 m/s2*14.7 m) = 12.0 m/s

    Cylinder I = 1/2 * m * r²

    Replacing in (1):

    ½ * v² + ½ (1/2 * m*r²) * (v/r) ² = g*h

    Replacing by the value given for h, and solving for v:

    v = 2*√ (1/3*9.8 m/s2*14.7 m) = 13.9 m/s
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