A wheel 2.10 m in diameter lies in a vertical plane and rotates about its centra

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 4.45 rad/s^2. 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 degree 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/s direction compositefunction with respect to the radius to point P (d) the angular position of the point P A car accelerates uniformly from rest and reaches a speed of 22.3 m/s in 8.99 s. Assume the diameter of a tire is 58.5 cm. (a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. (b) What is the final angular speed of a tire in revolutions per second?

Explanation / Answer

here,

diameter , d = 2.1 m

radius , r = 1.05 m

angular accelration , alpha = 4.45 rad/s^2

theta0 = 57.3 degree = 0.9996 rad

at t = 2 s

a)

the angular speed of the wheel , w = 0 + alpha * t

w = 0 + 4.45 * 2 = 8.9 rad/s

b)

the tangential speed of point P , v = r * w

v = 1.05 * 8.9 = 9.35 m/s

c)

centripital accelration , ac = v^2/r = 88.17 m/s^2

tangential accelration , at = alpha * r = 4.67 m/s^2

the magnitude of total accelration , |a| = sqrt(88.17^2 + 4.67^3) = 88.3 m/s^2

direction = arctan(4.67/88.17) = 3.03 degree from the radius

d)

the angular position of P be theta

theta - theta0 = 0 + 0.5 * alpha * t^2

theta - 0.9996 = 0.5 * 4.45 * 2^2

theta = 9.9 rad

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