A wheel 2.30 m in diameter lies in a vertical plane and rotates with a constant angular acceleration of 4.20 r?

4.20 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掳 with the horizontal at this time. At t = 2.00 s find the following.


(a) the angular speed of the wheel


correct check mark rad/s





(b) the tangential speed and the total acceleration of the point P


wrong check mark


Your answer differs from the correct answer by 10% to 100%. m/s


m/s2





(c) the angular position of the point P


rad





How do you do parts b and c? I thought you have to take the distance formula to find time and then you can find angular velocity. Then I can use the equation v=rw .??? Some help would be greatly appreciated. |||Let r = 2.3 m; alpha = 4.2 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) = 4.2*2 = 8.4 rad/sec when dt = t1 - t0 = 2 sec.





b) At t1 = 2 sec, w1 = 8.4 rad/sec so that v1 = w1*r = 8.4*2.3 = ? the tangential velocity. 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 = 4.2*2.3. Thus, A = sqrt((v1^2/r)^2 + (alpha*r)^2) = ? and everything on the RHS is known, you can do the math.





c) 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)*4.2*2^2 = 8.4 rad which is how much the wheel rotates in dt = 2 sec. Therefore, omega1 = 8.4 rad + 1 rad = 9.4 rad, wrt the horizontal. [Note. Recognize that once around is 2 pi ~ 6 rads to get to the starting point at 1 rad. So the point P will be about 2.4 rads beyond the 1 rad position is started at. That results because 6 rads brings it back to 57.3 deg (1 rad) and then it moves on another 2.4 rads beyond the starting point that was at omeg0 = 1 rad.]





The distance formulae are angular distances (i.e., omega) and angular velocities (i.e., w) But they are similar to the linear ones. For example, s = 1/2 at^2 is the linear equation for traveling s distance with average acceleration a in time t and starting with v0 = 0 initial velocity. In angular math talk, that would be omega = 1/2 alpha t^2; where alpha = rad/sec^2 average angular acceleration and starting with w0 = 0 angular velocity.