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23 August, 15:31

A heavy turntable, used for rotating large objects, is a solid cylindrical wheel that can rotate about its central axle with negligible friction. The radius of the wheel is 0.330 m. A constant tangential force of 300 N applied to its edge causes the wheel to have an angular acceleration of 0.876 rad/s2.

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  1. 23 August, 15:32
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    I = 113.014 kg. m^2

    m = 2075.56 kg

    wf = 3.942 rad/s

    Explanation:

    Given:

    - The constant Force applied F = 300 N

    - The radius of the wheel r = 0.33 m

    - The angular acceleration α = 0.876 rad / s^2

    Find:

    (a) What is the moment of inertia of the wheel (in kg · m2) ?

    (b) What is the mass (in kg) of the wheel?

    (c) The wheel starts from rest and the tangential force remains constant over a time period of t = 4.50 s. What is the angular speed (in rad/s) of the wheel at the end of this time period?

    Solution:

    - We will apply Newton's second law for the rotational motion of the disc given by:

    F*r = I*α

    Where, I: The moment of inertia of the cylindrical wheel.

    I = F*r / α

    I = 300*0.33 / 0.876

    I = 113.014 kg. m^2

    - Assuming the cylindrical wheel as cylindrical disc with moment inertia given as:

    I = 0.5*m*r^2

    m = 2*I / r^2

    Where, m is the mass of the wheel in kg.

    m = 2*113.014 / 0.33^2

    m = 2075.56 kg

    - The initial angular velocity wi = 0, after time t sec the final angular speed wf can be determined by rotational kinematics equation 1:

    wf = wi + α*t

    wf = 0 + 0.876 * (4.5)

    wf = 3.942 rad/s
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