A wheel 1.72 m in diameter lies in a vertical plane and rotates about its centra

Question

A wheel 1.72 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 3.97 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

(d) the angular position of the point P

Visualize a large carnival roulette wheel. It spins at several radians per second. A point on the rim will have a rather large centripetal acceleration and a smaller tangential acceleration.

Parts (a) and (d) are about the kinematics of rotation, describing rotation with constant . In parts (b) and (c), we relate rotation at one instant to the linear motion of a point on the object, through angular position measured in radians.

We are given

= 3.97 rad/s2,   i = 0,   i = 57.3° = 1 rad,  and t = 2 s.

(a) We choose the equation

f = i + t = 0 + t

Explanation / Answer

a) the radius of the wheel is 0.86 m
angular velocity = initial angular velocity + A t where A is angular accel and t is time
ang vel = 0 + 3.97rad/s/s x 2s = 7.94 rad/s

b) tangential vel = r w = 0.86 m x 7.94 rad/s = 6.8284 m/s

c) total accel = Sqrt[a tan^2 + a radial^2]
a tan = 3.97 rad/s/s, a radial = v^2/r = (6.8284m/s)^2/0.86 m = 54.22 m/s/s
total accel = 54.36
d) total angular displacement of the point P = w0 t + 1/2 A t^2 = 0 + 1/2 (3.97rad/s/s)(2s)^2 = 7.94 rad
the initial position, 57.3 deg = 1 rad, so final position = 8.94 rad = 127 deg with respect to the horizontal

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