Answer:
a.[tex]-1.8rad/s^2[/tex]
b.[tex]332.5 rad[/tex]
c.52.9 rev
Explanation:
We are given that
Initial angular speed of wheel=[tex]\omega_0=35 rad/s[/tex]
Time=t=19 s
Final angular speed=[tex]\omega=0 rad/s[/tex]
a.We have to find the angular acceleration of the wheel.
We know that
Angular acceleration[tex],\alpha=-\frac{\omega_0}{t}[/tex]
Using the formula
[tex]\alpha=-\frac{35}{19}=-1.8rad/s^2[/tex]
b.[tex]\theta=\frac{1}{2}\omega_0t[/tex]
Using the formula
[tex]\theta=\frac{1}{2}(35)\times 19=332.5 rad[/tex]
[tex]\theta=332.5 rad[/tex]
c.[tex]2\pi[/tex] rad=1 rev
1 rad=[tex]\frac{1}{2\pi}[/tex] rev
332.5 rad=[tex]\frac{332.5}{2\pi}=52.9 rev[/tex]
Number of revolutions made by the flywheel in stopping=52.9 rev
