Question

A wheel is being held in the air and rotates about a fixed axle. At time t = 0 the wheel is spinning with an angular velocity LaTeX: \omega_i = ? i = 285 rad/sec in the counter-clockwise direction. The position of a spot on the wheel is at an angle LaTeX: \theta_i = 0 ? i = 0 at the time t = 0. At t = 0 a constant clockwise torque is applied to the wheel such that there is a constant clockwise angular acceleration LaTeX: \alpha = - ? = ? 5.14 rad/s2. How long will it take for the wheel to turn back to where the spot on the wheel is at LaTeX: \theta_f = 0 ? f = 0 ? Give your answer in Seconds to at least three digits to avoid being counted incorrect due to rounding. NOTE: The angle in this problem is not limited to the values LaTeX: 0 \leq \theta \leq 2 \pi 0 ? ? ? 2 ? . The angle can be larger than LaTeX: 2 \pi 2 ? , indicating that the wheel has gone around more than one revolution. For example, if the wheel turns through three revolutions, LaTeX: \theta = 6 \pi ? = 6 ? . Thus LaTeX: \theta_f = 0 ? f = 0 implies that wheel may have turned several revolutions counter-clockwise and then rotated BACK the same number of revolutions clockwise.

Answer 1

Given : Initial angular speed (?_i) = 285 rad/sec at t = 0 and angular acceleration(?) = ? 5.14 rad/sec²

using law of motion : ?_{f =} ?_{i +} ?t ----------------------- 1

time taken for wheel to stop due to negative acceleration

0 = 285 + (-5.14)t

t = 285/5.14 = 55.44 sec

angle rotated during deaccleration :  ? = ?_it + 1/2(?)t² --------------------------- 2

? = 285x55.44 - (0.5)x(5.14)x(55.44)²

  ? = 7901.264 rad

? = 1257.525 revolutions

   for the wheel to turn back to where the spot on the wheel ? should be integer ? = 1257 rev

  ? = 2 x pi x1257 = 7897.96 rad

using  ? = ?_it + 1/2(?)t² to calculate time

7897.96 = 285t - 0.5x5.14t²

on sloving t =54.31 sec

  time for the wheel to turn back to where the spot on the wheel is: t = 54.31sec