A 41.0-cm diameter disk rotateswith a constant angular acceleration of 2.8 rad/s

Question

A 41.0-cm diameter disk rotateswith a constant angular acceleration of 2.8 rad/s2. It starts from restat t = 0, and a line drawn from the center ofthe disk to a point P on the rim of the diskmakes an angle of 57.3° with the positivex-axis atthis 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 Pat t = 2.30 s.
linear velocity m/s tangential acceleration m/s2
c) Find the position of P (in degrees, with respect to the positivex-axis) at t = 2.30s.
°
(a) Find the angular speed of the wheel at t =2.30 s.
rad/s

(b) Find the linear velocity and tangential acceleration of Pat t = 2.30 s.
linear velocity m/s tangential acceleration m/s2
c) Find the position of P (in degrees, with respect to the positivex-axis) at t = 2.30s.
°
linear velocity m/s tangential acceleration m/s2

Explanation / Answer

The diameter of the disk is d = 41.0 cm = 41.0 *10-2 m The angular acceleration of the disk is = 2.8rad/s2 The disk starts from rest at t = 0 The line drawn from the center of the disk makes an angle of = 57.3o with the positive x-axis at thistime. (a)The angular speed of the wheel at t = 2.30 sis w = wo + t Here,wo= 0 rad/s or w = 0 + 2.8 * 2.30 or w = 6.44 rad/s (b)The linear velocity of P at t = 2.30 s is v = r * w Here,r = (d/2) = (41.0 * 10-2/2) = 20.5 *10-2 m or v = 20.5 * 10-2 * 6.44 = 132.02 m/s The tangential acceleration of P at t = 2.30 sis T= r * or T= 20.5 * 10-2 * 2.8 = 0.574rad/s2 (c)The position of P (in degrees, with respect to the positivex-axis) at t = 2.30s is = wot + (1/2)t2 Substituting the values in the above equation,we get = 0 * 2.30 + (1/2) * 2.8 * (2.30)2 or = 7.406 rad or = 7.406 rad

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