Question

 A 45.0-cm diameter disk rotates with a constant angular acceleration of 2.50 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. At t = 2.30 s, find (a) the angular speed of the wheel, (b) the linear speed and tangential acceleration of P, and (c) the position of P (in degrees, with respect to the positive x-axis).

Answer 1

here,

the angular acceleration , alpha = 2.5 rad/s^2

theta0 = 57.3 degree = 0.9997 rad

diameter , d = 45 cm

radius , r = d/1 = 22.5 cm = 0.225 m

at t = 2.3 s

a)

the angular speed of the wheel , w = w0 + alpha * t

w = 0 + 2.5 * 2.3 rad/s

w = 5.75 rad/s

b)

the linear speed , v = r * w

v = 0.225 * 5.75 m/s = 1.29 m/s

the tangential acceleration , a = alpha * r

alpha = 2.5 * 0.225 rad/s^2 = 0.56 m/s^2

c)

the angle covered, theta = theta0 + (w0 * t + 0.5 * alpha * t^2)

theta = 0.9997 + ( 0 + 0.5 * 2.5 * 2.3^2) rad

theta = 7.61 rad = 436.4 degree

the angle with positive x-axis , phi = (436.4 - 360) degree

phi = 76.4 degree