Answer:
0.231 N
Explanation:
To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be
[tex]\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2[/tex]
If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:
[tex] I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2[/tex]
According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is
[tex] T = I\alpha = 0.303*0.6454 = 0.196 Nm[/tex]
So the force acting on the other end to generate this torque mush be:
[tex] F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N[/tex]
