We start from the definition of Torque,
[tex] T = I \alpha [/tex]
Where ,
I = moment of inertia
[tex] \alpha = [/tex] Angular acceleration.
The torque given in the problem is 1.96mN.
We look for the moment of inertia of a solid cylinder,
[tex] I = \frac {1} {2} mR ^ 2 [/tex]
Where m is the mass of 4Kg and R the radius 0.035m
[tex] I = \frac {1} {2} (4) (0.035) ^ 2 [/tex]
[tex] I = 2.45 * 10 ^ - 3 Kgm ^ 2 [/tex]
Replacing,
[tex] -1.96 = 2.45 * 10 ^{-3} \alpha \\\alpha = -800rad / s ^ 2 [/tex]
A) With angular acceleration we can find the number of revolutions, the given equation would be,
[tex] w_f ^ 2-w_i ^ 2 = 2 \alpha \theta [/tex]
[tex] 0 ^ 2- 9700rpm (2 \pi / 60rpm) ^ 2 = -2 * 800 \ theta [/tex]
[tex] \theta = \frac {1031812.3} {1600} [/tex]
[tex] \theta = 644.88 revolutions. [/tex]
B) We apply the rotational dynamics formula and we can find the time,
[tex] w_f = w_i + \alpha t [/tex]
[tex] 0 = 9700 rpm (2 \pi / 60 rpm) -800t [/tex]
[tex] t = \frac {1015.78} {800} [/tex]
[tex] t = 1.26s [/tex]
