Question

A wheel 2.10 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 3.50 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

(b) the tangential speed of the point P

(c) the total acceleration of the point P

magnitude-

direction (degrees with respect to the radius point p)-

(d) the angular position of the point P

Answer 1

a) from the relation

wf = wi+alpha*t

here initial angular speed wi = 0

alpha = angular acceleration = 3.5 rad/s^2

t = 2 s

wf = 3.5*2 = 7 rad/s

b) tangential speed v = r*w

v = (2.1/1)*7 = 7.35 m/s

c) tangential acceleration at = r*alpha = (2.1/2)*3.5 = 3.675 m/s^2

radial acceleration ar = v^2/r = rw^2 = (2.1/2)*7^2 = 51.45 m/s^2

total acceleration a tot = sqrt(ar^2+at^2)

magnitude: atot = sqrt(51.45^2+3.675^2) = 51.58 m/s^2

direction = theta = tan^-1(at/ar) = tan^-1(3.675/51.45) = 4.09 degrees

d) from the relation

angular position = f = i+wi?*t+1/2*alpha*t^2

i = 57.3 degree = 1 rad

f = 1+0+1/2*3.5*2^2 = 8 rad