A 42.0-cm diameter disk rotates with a constant angular acceleration of 2.60 rad

Question

A 42.0-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) At t = 2.35 s, find the angular speed of the wheel.
rad/s

(b) At t = 2.35 s, find the magnitude of the linear velocity and tangential acceleration of P.

(c) At t = 2.35 s, find the position of P (in degrees, with respect to the positive x-axis).
° counterclockwise from the +x-axis

Explanation / Answer

here,

diameter , d = 0.42 m

d = 0.21 m

constant angular accelration , alpha = 2.6 rad/s^2

theta0 = 57.3 degree

a)

the final angular speed , w = w0 + alpha * t

w = 0 + 2.6 * 2.35

w = 6.11 rad/s

b)

the magnitude of linear velocity , v = r * w

v = 1.28 m/s

the tangential accelration , at = w^2 * r

at = 6.11^2 * 0.21 = 7.84 m/s^2

c)

the position of P , theta = theta0 + 0.5 * alpha * t^2 * 360/6.28

theta = 57.3 + 0.5 * 2.6 * 2.35^2 * 360 /6.28

theta = 468.84 degree = 108.85 degree counterclockwise from +x axis

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