A disk with a rotational inertia of 2.0 kg·m2 and a radius of 0.40 m rotates on a frictionless fixed axis perpendicular to the disk faces an

Question

A disk with a rotational inertia of 2.0 kg·m2 and a radius of 0.40 m rotates on a frictionless fixed axis perpendicular to the disk faces and through its center. A force of 5.0 N is applied tangentially to the rim. The angular acceleration of the disk is?

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Mít Mít 2 days 2021-07-22T11:12:05+00:00 1 Answers 1 views 0

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    2021-07-22T11:13:15+00:00

    To solve this problem we must apply the concepts related to the torque expressed as a function of the angular acceleration and the moment of inertia as well as the radius and the force. From these two definitions we will seek to find the angular acceleration of the body:

    We know that Torque is defined as,

    \tau = I \alpha

    Here,

    I = Moment of Inertia

    \alpha = Angular acceleration

    And at the same time, the torque is the product between the force and the radius, then we have

    \tau = rF

    Here,

    r = Radius

    F = Force

    Equation we have,

    rF = I\alpha

    Rearranging to find the acceleration

    \alpha = \frac{rF}{I}

    Our values are,

    \text{Rotational Inertia of Disk} = I = 2.0 kg\cdot m^2

    \text{Radius of disk} = r = 0.40m

    \text{Force} = F = 5.0N

    Replacing this value at the previous equation

    \alpha = \frac{(0.40m)(5.0N)}{2.0kg\cdot m^2}

    \alpha = 1 rad/s^2

    Therefore the angular acceleration of the disk is 1rad/s^2

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