A machinist turns the power on to a grinding wheel, at rest, at time t = 0 s. The wheelaccelerates uniformly for 10 s and reaches the operating angular velocity of 38 rad/s.The wheel is run at that angular velocity for 30 s and then power is shut off. The wheelslows down uniformly at 2.1 rad/s2until the wheel stops. In this situation, the angularacceleration of the wheel between t = 0 s and t = 10 s is closest to:

A) 6.1 rad/s2
B) 6.8 rad/s2
C) 3.8 rad/s2
D) 4.6 rad/s2
E) 5.3 rad/s2

Since, between the time period from t=0 s to t=10 s the wheel is in uniformly accelerated motion . Therefore angular acceleration is constant at that time period.

Using equation of angular motion.

[tex]\omega=\omega_0+\alpha t[/tex]

Here, [tex]\omega=angular \ velocity\ at\ any\ time\ t\\\omega_0=angular \ velocity\ at\ time\ t=0\\\alpha=angular \ acceleration [/tex]

putting , [tex]\omega=38 \ rad/s\\t=10\ s\\\omega_0=0 \ rad/s[/tex]

[tex]38=0+\alpha\times 10\\\alpha=3.8\ rad/s^2.[/tex]

Hence, this is the required solution.