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28 January, 17:24

A hollow spherical shell with mass 1.65 kg rolls without slipping down a slope that makes an angle of 38.0 ∘ with the horizontal. Part A Find the magnitude of the acceleration acm of the center of mass of the spherical shell. Take the free-fall acceleration to be g = 9.80 m/s2.

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  1. 28 January, 17:50
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    The acceleration is 3.62 m/s²

    Explanation:

    Step 1: Data given

    mass of the shell = 1.65 kg

    angle = 38.0 °

    Step 2: Calculate the acceleration

    We have 2 forces working on the line of motion:

    ⇒ gravity down the slope = m*g*sinα

    ⇒ provides the linear acceleration

    ⇒ friction up the slope = F

    ⇒ provides the linear acceleration and also the torque about the CoM.

    ∑F = m*a = m*g*sin (α) - F

    I*dω/dt = F*R

    The spherical shell with mass m has moment of inertia I=2/3*m*R² Furthermore a pure rolling relates dω/dt and a through a = R dω/dt. So the two equations become

    m*a = m*g sin (α) - F

    2/3*m*a = F

    IF we combine both:

    m*a = m*g*sin (α) - 2/3*m*a

    1.65a = 1.65*9.81 * sin (38.0) - 2/3 * 1.65a

    1.65a + 1.1a = 9.9654

    2.75a = 9.9654

    a = 3.62 m/s²

    The acceleration is 3.62 m/s²
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