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3 December, 06:09

Starting from rest, a 12.32-cm-diameter compact disk takes 7.9 s to reach its operating angular velocity of 2298 rpm. Assume that the angular acceleration is constant. The disk's moment of inertia is 1.00*10-5 kg·m2. How much torque is applied to the disk?

Answers (1)
  1. M
    3 December, 06:16
    0
    T = 0.00030462 Nm

    Explanation:

    Given:-

    - The initial angular velocity, wi = 0 rad/s (rest)

    - The final angular velocity, wf = 2298 rpm = 240.646 rad/s

    - The time taken to reach wf, t = 7.9 s

    - The moment of inertia of disc, I = 1.00 * 10^-5 kg. m^2

    Find:-

    How much torque is applied to the disk?

    Solution:-

    - We will first determine the constant angular acceleration (α) of the disc when it starts from rest and reaches the final angular velocity in time t = 7.9 s.

    - We will use the first rotational kinematic equation of motion as follows:

    wf = wi + α*t

    α = (wf - wi) / t

    α = (240.646 - 0) / 7.9

    α = 30.462 rad/s^2

    - The amount of torque T required to accelerate (α) the disc uniformly in time t = 7.9s is given by the relationship as follows:

    T = I*α

    T = (10^-5) * (30.462)

    T = 0.00030462 Nm
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