A 46.8-cm diameter disk rotates with a constant angular acceleration of 2.8 rad/

Question

A 46.8-cm diameter disk rotates with a constant angular acceleration of 2.8 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.
rad/s

(b) Find the linear velocity and tangential acceleration of P at t = 2.30 s.
linear velocity m/s
tangential acceleration m/s2

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

Explanation / Answer

The diameter of the disk is d = 46.8 cm The radius of the disk is r = 23.4 cm                                           = 0.234 m The angular acceleration = 2.8 rad/s2 The angular displacement at t = 0 is 0 = 57.3o                                                               = 1 rad a) The initial angular velocity of the disk is 0 = 0 The equation of motion of the disk will be = 0 + t = 0 + (2.8 rad/s2)(2.3 s) = 6.44 rad/s b) The linear velocity is v = r                                       = (0.234 m)(6.44 rad/s)                                       = 1.51 m/s The tangential acceleration is at = v                                                    = (1.51 m/s)(6.44 rad/s)                                                    = 9.7 m/s2 c) The position of the point is = 0 + 0t + 0.5t2    = 1 rad + 0 + 0.5(2.8 rad/s2)(2.3 s)2    = 8.4 rad    = 481.63o    = 121.63o with resect to the positive X axis in counter clockwise direction                                                               = 1 rad a) The initial angular velocity of the disk is 0 = 0 The equation of motion of the disk will be = 0 + t = 0 + (2.8 rad/s2)(2.3 s) = 6.44 rad/s b) The linear velocity is v = r                                       = (0.234 m)(6.44 rad/s)                                       = 1.51 m/s The tangential acceleration is at = v                                                    = (1.51 m/s)(6.44 rad/s)                                                    = 9.7 m/s2 c) The position of the point is = 0 + 0t + 0.5t2    = 1 rad + 0 + 0.5(2.8 rad/s2)(2.3 s)2    = 8.4 rad    = 481.63o    = 121.63o with resect to the positive X axis in counter clockwise direction    = 121.63o with resect to the positive X axis in counter clockwise direction

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