Answer:
By the definition of the torque, we have:
[tex]\tau=FR,τ=FR,[/tex]
here, FF is the tangential force at the rim of the disk, rr is the radius of the disk.
From the other hand,
[tex]\tau=I\alpha.τ=Iα.[/tex]
The moment of inertia of the disk can be written as follows:
[tex]I=\dfrac{1}{2}MR^2.I= [/tex]
2
1
MR
2
.
Finally, we have:
[tex]FR=\dfrac{1}{2}MR^2\alpha,FR=
2
1
MR
2
α,
F=\dfrac{1}{2}MR\alpha,F=
2
1
MRα,
F=\dfrac{1}{2}\cdot2\ kg\cdot0.18\ m\cdot12\ \dfrac{rad}{s^2}=2.16\ N.F=
2
1
⋅2 kg⋅0.18 m⋅12
s
2
rad
=2.16 N.[/tex]
Explanation:
Hope it helps bye
