A 6.0-cm-diameter gear rotates with angular velocity ω = ( 20 ─ ½ t² ) rad/s where t is in seconds. At t = 4.0 s, what are: The gear's angular acceleration?

In Problems 78, 79, and 80 you are given the equations that are used to solve a problem. For each of these, you are to write a realistic problem for which these are the correct equations. Be sure that the answer your problem requests is consistent with the equations given.

$\(\begin{aligned}\) 100 \, \(\text{m}\) &= 0 \, \(\text{m}\) + (50 \(\cos\) \(\theta\) \, \(\text{m/s}\)) t_1 \\ 0 \, \(\text{m}\) &= 0 \, \(\text{m}\) + (50 \(\sin\) \(\theta\) \, \(\text{m/s}\)) t_1 - \(\frac{1}{2}\) (9.80 \, \(\text{m/s}\)^2) t_1^2 \(\end{aligned}\)$

A 25 g steel ball is attached to the top of a 24-cm-diameter vertical wheel. Starting from rest, the wheel accelerates at 470 rad/s². The ball is released after ¾ of a revolution. How high does it go above the center of the wheel?

A 6.0-cm-diameter gear rotates with angular velocity ω = ( 20 ─ ½ t² ) rad/s where t is in seconds. At t = 4.0 s, what are: The tangential acceleration of a tooth on the gear?

A painted tooth on a spinning gear has angular position θ = (6.0 rad/s⁴)t⁴. What is the tooth's angular acceleration at the end of 10 revolutions?

The angular velocity of a process control motor is ω = ( 20 - ½ t² ) rad/s, where t is in seconds. At what time does the motor reverse direction?

The angular velocity of a process control motor is ω = ( 20 ─ ½ t² ) rad/s, where t is in seconds. Through what angle does the motor turn between t = 0 s and the instant at which it reverses direction?