A 41.9-cm diameter disk rotates with a constant angular acceleration of 2.2 rad/

Question

A 41.9-cm diameter disk rotates with a constant angular acceleration of 2.2 rad/s2. It starts from rest at t = 0, and a line drawn from the center of the disk to a point P on the rim of the disk makes an angle of 57.3° with the positive x-axis at this time.
(a) Find the angular speed of the wheel at t = 2.30 s.
Correct: Your answer is correct. rad/s

(b) Find the linear velocity and tangential acceleration of P at t = 2.30 s.
linear velocity Correct: Your answer is correct. m/s
tangential acceleration Correct: Your answer is correct. m/s2

c) Find the position of P (in degrees, with respect to the positive x-axis) at t = 2.30s.
°

Explanation / Answer

Given that The diameter of the disk = 41.9 cm                                     = 0.419 m Radius of the disk, r = 0.2095 m Angular acceleration, a = 2.2 rad/s^2 initial angular velocity, 1 = 0,
initial position of P= o = 57.3 = 1 rad
At t = 2.30 s ---------------------------------------------------------

a) the angular speed of the wheel
= at
    = (2.2 rad/s^2)(2.30 s)     = 5.06 rad/s ---------------------------------------------------------- ---------------------------------------------------------- b) the linear velocity of P:
               v = r
                  = 1.06 m/s
tangentialacceleration of P:
            at = a * r
                = 0.461 m/s2 ----------------------------------------------------------
c) = o +0.5at^2       = 1 rad + 0.5 * (0.461 m/s2)(2.3)^2       = 1 rad + 0.5 * (0.461 m/s2)(2.3)^2
       = 6.819 rad
       = 390.7 deg
Finally It appears as 390.7 - 360 =30.7 deg

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