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

(d) the angular position of the point P
rad

Explanation / Answer

a) given that alpha = 4.25 rad/s^2

radius is r = d/2 = 2.2/2 = 1.1 m

but alpha = (Wf-wi)/t

4.25 = (Wf-0)/2

Wf = 4.25*2 = 8.5 rad/s

b) v = r*w = 1.1*8.5 = 9.35 m/s

c) total accelaration is a = sqrt(a_tan^2 + a_rad^2)

a_tan = r*alpha = 1.1*4.25 = 4.675 m/s^2

a_rad = r*w^2 = 1.1*8.5^2 = 79.475 m/s^2

a = sqrt(4.675^2+79.475^2) = 79.62 m/s^2

angular displacement after t = 2 sec is theta = (0.5*alpha*t^2) = (0.5*4.5*2^2) = 9 rad

so the direction is 9 rad = 515.662 deg = 155.662 deg

d) angular position of the point P is 57.3 deg = 1 rad

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