(a) [tex]50.4 N\cdot m[/tex]
The torque exerted on the solid disk is given by
[tex]\tau=Frsin \theta[/tex]
where
F is the magnitude of the force
r is the radius of the disk
[tex]\theta[/tex] is the angle between F and r
Here we have
F = 180 N
r = 0.280 m
[tex]\theta=90^{\circ}[/tex] (because the force is applied tangentially to the disk)
So the torque is
[tex]\tau = (180 N)(0.280 m)(sin 90^{\circ})=50.4 N\cdot m[/tex]
(b) [tex]17.2 rad/s^2[/tex]
First of all, we need to calculate the moment of inertia of the disk, which is given by
[tex]I=\frac{1}{2}mr^2[/tex]
where
m = 75.0 kg is the mass of the disk
r = 0.280 m is the radius
Substituting,
[tex]I=\frac{1}{2}(75.0)(0.280)^2=2.94 kg m^2[/tex]
And the angular acceleration can be found by using the equivalent of Newtons' second law for rotational motions:
[tex]\tau = I \alpha[/tex]
where
[tex]\tau = 50.4 N \cdot m[/tex] is the torque exerted
I is the moment of inertia
[tex]\alpha[/tex] is the angular acceleration
Solving for [tex]\alpha[/tex], we find:
[tex]\alpha = \frac{\tau}{I}=\frac{50.4 N \cdot m}{2.94 kg m^2}=17.2 rad/s^2[/tex]
