Answer:
a. [tex]-12.7 Nm[/tex]
b. [tex]-7.9 rad/s^2[/tex]
Explanation:
I have attached an illustration of a solid disk with the respective forces applied, as stated in this question.
Forces applied to the solid disk include:
[tex]F_1 = 90.0N\\F_2 = 125N[/tex]
Other parameters given include:
Mass of solid disk, [tex]M = 24.3kg[/tex]
and radius of solid disk, [tex]r = 0.364m[/tex]
a.) The formula for determining torque ([tex]T[/tex]), is [tex]T = r * F[/tex]
Hence the net torque produced by the two forces is given as a summation of both forces:
[tex]T = T_{125} + T_{90}\\= -r(125)sin90 + r(90)sin90\\= 0.364(-125 + 90)\\= -12.7 Nm[/tex]
b.) Â The angular acceleration of the disk can be found thus:
using the formula for the Moment of Inertia of a solid disk;
[tex]I_{disk} = {\frac{1}{2}}Mr^2[/tex]
where [tex]M[/tex] = Mass of solid disk
and [tex]r[/tex] = radius of solid disk
We then relate the torque and angular acceleration ([tex]\alpha[/tex]) with the formula:
[tex]T = I\alpha \\-12.7 = ({\frac{1}{2}}Mr^2)\alpha \\\alpha = -{\frac{12.7}{1.61}} = -7.9 rad/s^2[/tex]
