A 42.5-cm diameter disk rotates with a constant angular acceleration of 2.6 rad/

Question

A 42.5-cm diameter disk rotates with a constant angular acceleration of 2.6 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. = 5.98 rad/s (correct)
(b) Find the linear velocity and tangential acceleration of P at t = 2.30 s.
???? m/s
???? m/s^2

c) Find the position of P (in degrees, with respect to the positive x-axis) at t = 2.30s.

Explanation / Answer

(a) = *t
= (2.6 rad/s^2)*(2.3 s)
= 5.98 rad/s

(b) v = *r
v = (5.98 rad/s)*(42.5cm)
v = 254 cm/s
v = 2.54 m/s

a = *r
a = (2.6 rad/s^2)*( 42.5cm)
a = 110.5 cm/s^2
a = 1.10 m/s^2

(c) f = i + ½**t^2
In this case, i equals 57.3°, which is 1 rad, so:
f = 1 + ½ *(2.6 rad/s^2)*(2.3 s)^2
f = 1 rad + 6.88 rad
f = 7.88 rad
f = 451.3°
f = 91.3° with respect to the positive x-axis.

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