Question

A wheel 2.25 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.30 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

Answer 1

Let r = 2.25 m; alpha = 4.30 rad/sec^2, omega0 = 57.3 deg = 1 rad at time t0 = 0.

a) Presuming w0 = 0 rad/sec, the initial angular velocity at t0 = 0, then w1 = alpha(dt) = 4.3*2 = 8.6 rad/sec when dt = t1 - t0 = 2 sec.

b) At t1 = 2 sec, w1 = 8.6 rad/sec so that v1 = w1*r = 8.6*2.25 =19.35 the tangential velocity.

c) Total acceleration A^2 = ar^2 + at^2; where ar is the radial acceleration ar = v1^2/r and at is the tangentail acceleration = alpha*r = 4.3*2.25 =9.675. Thus, A = sqrt((v1^2/r)^2 + (alpha*r)^2) = ? and everything on the RHS is known, you can do the math

d) d(omega) = omega1 - omega0 which is the angular amount the wheel moves in dt = 2 seconds starting at omega0; so that omega1 = d(omega) + omega0; where omega0 = 1 rad wrt the horizontal at t0 = 0 and d(omega) = 1/2 alpha(dt)^2 = (1/2)*4.3*2^25 = 8.6 rad which is how much the wheel rotates in dt = 2 sec. Therefore, omega1 = 8.6 rad + 1 rad = 9.6 rad, wrt the horizontal.