A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of −5.40 rad/s2. during a 3.40-s time interval, the wheel rotates through 50.4 rad. what is the angular speed of the wheel at the end of the 3.40-s interval?
The law of motion for the wheel is [tex]\theta(t)= \omega_0 t + \frac{1}{2} \alpha t^2[/tex] where [tex]\theta(t)[/tex] is the angle covered after a time interval t [tex]\omega _0 [/tex] is the initial angular speed of the wheel [tex]\alpha[/tex] is the angular acceleration of the wheel
We know that after a time t=3.40 s the angle covered by the wheel is 50.4 rad, so we can use these data and re-arrange the previous formula to find the initial angular speed: [tex]\omega_0 = \frac{\theta - \frac{1}{2}\alpha t^2 }{t}= \frac{50.4 rad - \frac{1}{2}(-5.40 rad/s^2)(3.40 s)^2 }{3.40 s}= 24.0 rad/s[/tex]
And then, the final angular speed will be given by [tex]\omega_f = \omega_0 + \alpha t=24.0 rad/s +(-5.4 rad/s^2)(3.40 s)=5.64 rad/s[/tex]
