A machine part has the shape of a solid uniform sphere of mass 240 g and diameter 2.50 cm . It is spinning about a frictionless axle through

Question

A machine part has the shape of a solid uniform sphere of mass 240 g and diameter 2.50 cm . It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of 0.0200 N at that point. Part A Find its angular acceleration. Let the direction the sphere is spinning be the positive sense of rotation.

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Thanh Hà 4 years 2021-07-31T17:44:58+00:00 1 Answers 23 views 0

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  1. Nem
    0
    2021-07-31T17:46:53+00:00
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    Answer:

    -16.6 rad/s^2

    Explanation:

    The torque exerted on a rigid body is related to the angular acceleration by the equation

    \tau = I \alpha (1)

    where

    \tau is the torque

    I is the moment  of inertia of the body

    \alpha is the angular acceleration

    Here we have a solid sphere: the moment of inertia of a sphere rotating about is centre is

    I=\frac{2}{5}MR^2

    where

    M = 240 g = 0.240 kg is the mass of the sphere

    R=\frac{2.50}{2}=1.25 cm = 0.0125 m is the radius of the sphere

    Substituting,

    I=\frac{2}{5}(0.240)(0.0125)^2=1.5\cdot 10^{-5} kg m^2

    The torque exerted on the sphere is

    \tau = Fr

    where

    F = -0.0200 N is the force of friction

    r = 0.0125 m is the radius of the sphere

    So

    \tau=(-0.0200)(0.0125)=-2.5\cdot 10^{-4} Nm

    Substituting into (1), we find the angular acceleration:

    \alpha = \frac{\tau}{I}=\frac{-2.5\cdot 10^{-4}}{1.5\cdot 10^{-5}}=-16.6 rad/s^2

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