Answer:
F = 345.45 N
Explanation:
Angular acceleration of the disc is given as rate of change in angular speed
it is given by formula
[tex]\alpha = \frac{d\omega}{dt}[/tex]
[tex]\alpha = \frac{2\pi(0.600)}{2}[/tex]
[tex]\alpha = 1.88 rad/s^2[/tex]
now we know that moment of inertia of the solid uniform disc is given as
[tex]I = \frac{1}{2}mR^2[/tex]
[tex]I = \frac{1}{2}245(1.50)^2[/tex]
[tex]I = 275.625 kg m^2[/tex]
now we have an equation for torque as
[tex]\Tau = I\alpha[/tex]
[tex]r F = 275.625(1.88)[/tex]
[tex]F = \frac{275.625(1.88)}{1.50}[/tex]
[tex]F = 345.45 N[/tex]
