Question

A wheel 2.30 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.50 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 rad/s (b) the tangential speed of the point P 0.35m/s (c) the total acceleration of the point P magnitude direction Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2 with respect to the radius to point P (d) the angular position of the point P rad

Answer 1

here,

diameter , d = 2.3 m

radius , r = d/2 = 1.15 m

theta0 = 57.3 degree = 0.9996 rad

a)

the angular speed , w = alpha * t = 9 rad/s

b)

the tangential speed , v = r * w = 10.35 m/s

c)

the tangential accelration , at = alpha * r

at = 4.5 * 1.15 = 5.175 m/s^2

the centripital accelration , ac = v^2 /r

ac = 10.35^2 /1.15 = 93.15 m/s^2

the magnitude , a = sqrt(at^2 + ac^2)

a = sqrt(5.175^2 + 93.15^2) = 93.3 m/s^2

theta = arctan(at/ac)

theta = arctan(5.175/93.15) = 3.18 degree with respect to the radius to point P

d)

the angular position , theta = theta0 + (0 + 0.5 * alpha * t^2)

theta = 0.9996 + ( 0.5 * 4.5 * 2^2)

theta = 10 rad