Answer
(a) $11N.m$
(b) $12N.m$
(c) $23N.m$
Work Step by Step
(a) The torque about the x-axis can be determined as follows:
$I_x=m_1r_1^2+m_2r_2^2+m_3r_3^2$
$I_x=(9.0)(1.0)^2+0+0$
$I_x=9.0Kg.m^2$
Now $\tau_x=I_x\alpha$
We plug in the known values to obtain:
$\tau_x=(9.0Kg.m^2)(1.20rad/s^2)$
$\tau_x=11N.m$
(b) The torque about the y axis can be determined as
$I_y=0+0+(2.5)(2.0)^2=10Kgm^2$
Now$\tau_y=I_y\alpha$
We plug in the known values to obtain:
$\tau=(10)(1.20)$
$\tau=12N.m$
(c) The torque about the z axis can be determined as follows:
$I_z=(9.0)(1.0)^2+0+(2.5)(2.0)^2=19Kg.m^2$
Now $\tau_z=I_z\alpha$
$\implies \tau_z=(19Kg.m^2)(1.20rad/s^2)=23N.m$
