A 43.0-cm diameter disk rotates with a constant angular acceleration of 2.30 rad

Question

A 43.0-cm diameter disk rotates with a constant angular acceleration of 2.30 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

(a) = *t
= (2.3 rad/s^2)*(2.3 s)
= 5.29 rad/s

(b) v = *r
v = (5.29 rad/s)*(43cm)
v = 227.47 cm/s
v = 2.27 m/s

a = *r
a = (2.3 rad/s^2)*( 43cm)
a = 98.9 cm/s^2
a = 0.98 m/s^2

(c) f = i + ½**t^2
In this case, i equals 57.3°, which is 1 rad, so:
f = 1 + ½ *(2.3 rad/s^2)*(2.3 s)^2
f = 1 rad + 6.08 rad
f = 7.08 rad
f = 405.65°
f = 45.65° with respect to the positive x-axis.

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