A 40.4-cm diameter disk rotates with a constant angular acceleration of 2.60 rad

Question

A 40.4-cm diameter disk rotates with a constant angular acceleration of 2.60 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. 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

a) w = wo + alfa*t

= 0 + 2.6*2.3

= 5.98 rad/s

b) r = d/2 = 0.404/2 = 0.202 m

v = r*w

= 0.202*5.08

= 1.208 m/s

a_tan = r*alfa

= 0.202*2.6

= 0.525 m/s^2

c) angular displacement = wo*t + 0.5*alfa*t^2

= 0 + 0.5*2.6*2.3^2

= 6.0835 rad

= 6.0835*(360/2*pi) degrees

= 348.56 degrees

angle with +x axis = 57.3 - (360-348.56)

= 45.86 degrees

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