A 52.8-cm diameter disk rotates with a constant angular acceleration of 3.00 rad

Question

A 52.8-cm diameter disk rotates with a constant angular acceleration of 3.00 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

angular speed = angular acceleration * time =3*2.3=6.9 b). linear velocity = angular speed*r=6.9*.328=2.2632 tangential acceleration = angular acceleration * r=3*.328=.984 c. 1/2*alpha*t^2+omega*t+theta 0 = theta final convert radians to degrees 3=171.887339, 6.9=395.340879 1/2*171.887339*2.3^2+395.340879*2.3+57.3=1 421.22603 degrees 1421.22/360=3 + remainder, so it made three revolutions 1421.22-360*3=341.22 degrees

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