To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.
Torque in a body is defined as,
[tex]\tau_l = F*d[/tex]
And in angular movement like
[tex]\tau_a = I*\alpha[/tex]
Where,
F= Force
d= Distance
I = Inertia
[tex]\alpha =[/tex] Acceleration Angular
PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.
[tex]\tau= F*cos(19)*d[/tex]
On the other hand we have the speed data expressed in RPM, as well
[tex]\omega_f = 10rpm = 10\frac{1rev}{1min}(\frac{1min}{60s})(\frac{2\pi rad}{1rev})[/tex]
[tex]\omega_f = 1.0471rad/s[/tex]
Acceleration can be calculated by
[tex]\alpha = \frac{\omega_f}{t}[/tex]
[tex]\alpha = \frac{1.0471}{9}[/tex]
[tex]\alpha = 0.11rad/s^2[/tex]
In the case of Inertia we know that it is equivalent to
[tex]I = \frac{1}{2}mr^2 = \frac{1}{2}(750)*(2.3)^2[/tex]
[tex]I = 1983.75kg.m^2[/tex]
Matching the two types of torque we have to,
[tex]\tau_l=\tau_a[/tex]
[tex]Fd=I\alpha[/tex]
[tex]Fcos(19)*2.3=1983.75(0.11)[/tex]
[tex]F=100.34N[/tex]
PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,
[tex]W = \frac{1}{2}I\omega_f^2[/tex]
[tex]W= \frac{1}{2}(1983.75)(1.0471)^2[/tex]
[tex]W=1087.51J[/tex]
