A wheel 2.20 m in diameter lies in a vertical plane and rotates about its centra

Question

A wheel 2.20 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.25 rad/s2. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s, find the following. (a) the angular speed of the wheel rad/s (b) the tangential speed of the point P m/s (c) the total acceleration of the point P magnitude .m/s2 direction ° with respect to the radius to point P (d) the angular position of the point P rad

Explanation / Answer

a) angular speed = 4.25 * 2.00

= 8.5 rad/s

b) r = d/2 = 2.20/2 = 1.10 m

tangential speed = r w = 1.10 * 8.5

tangential speed = 9.35 m/s

c) total acceleration = v2 / r = 9.352 / 1.1 m

= 79.47 m/s2

d) P = w0 t + 1/2 a t2 = 0 + 1/2 * 4.25 * 22 = 8.5 rad

Initial position 57.3 deg = 1 rad

Final position = 8.5 + 1 rad

angular position of the point P = 9.5 rad

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