<p>A 50.2-cm diameter disk rotates with a constant angular acceleration of 2.9 r

Question

<p>A 50.2-cm diameter disk rotates with a constant angular acceleration of 2.9 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&#176; with the positive x-axis at this time.</p>
<p>The angular speed of the wheel at <em>t</em> = 2.30 s is <strong>6.67 rad/s</strong></p>
<p>a) Find the tangential acceleration of P at t = 2.30 s. (the linear velocity is <strong>1.67 m/s</strong>)</p>
<p>b) Find the position of P (in degrees, with respect to the positive x-axis) at <em>t</em> = 2.30s.</p>

Explanation / Answer

a) = o + t = 0+2.9*2.3 = 6.67 rad/s

linear velocity = v = r = (0.502/2) * 6.67 = 1.67 m/s

tangential acceleration = v2/r = 1.67*1.67/0.251 = 11.16 m/s2

b) -o = ot+0.5*t2 = 0.5*2.9*2.3*2.3 = 7.6705 rad = 879.42o => = 936.72

=> position of P = 936.72-n*360 = 216.72o from positive x-axis

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