Answer:
[tex]\Delta t = 8 s[/tex]
Explanation:
As we know that the angular acceleration of the wheel due to friction is constant
so we can use kinematics
[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2[/tex]
so we have
[tex](65 \times 2\pi) = (2\pi \times 9)(10) + \frac{1}{2}(\alpha)(10^2)[/tex]
[tex]130\pi = 180\pi + 50 \alpha[/tex]
[tex]\alpha = -\pi rad/s^2[/tex]
now time required to completely stop the wheel is given as
[tex]\omega_f = \omega_i + \alpha t[/tex]
[tex]0 = (2\pi \times 9) + (-\pi) t[/tex]
[tex]t = 18 s[/tex]
now time required to stop the wheel is given as
[tex]\Delta t = 18 - 10[/tex]
[tex]\Delta t = 8 s[/tex]