A 47.3-cm diameter disk rotates with a constant angular acceleration of 2.3 rad/

Question

A 47.3-cm diameter disk rotates with a constant angular acceleration of 2.3 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/s^2

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

Explanation / Answer

(a) ? = a*t ? = (2.3 rad/s^2)*(2.3 s) ? = 5.29 rad/s (b) v = ?*r v = (5.29 rad/s)*(23.65cm) v = 125.1 cm/s v = 1.25 m/s a = a*r a = (2.3 rad/s^2)*( 23.65cm) a = 54.395 cm/s^2 a = .544 m/s^2 (c) ?f = ?i + ½*a*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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