Question

A 2-kg disk has a radius of 18 cm and rotates with an angular acceleration of 12 rad/s2

about

an axis through its center and perpendicular to the plane of the disk. Determine the tangential

force at the rim of the disk

Answer 1

By the definition of the torque, we have:

$$\tau=FR,$$

here, $F$ is the tangential force at the rim of the disk, $r$ is the radius of the disk.

From the other hand,

$$\tau=I\alpha.$$

The moment of inertia of the disk can be written as follows:

$$I=\dfrac{1}{2}MR^2.$$

Finally, we have:

$$FR=\dfrac{1}{2}MR^2\alpha,$$ $$F=\dfrac{1}{2}MR\alpha,$$ $$F=\dfrac{1}{2}\cdot2\ kg\cdot0.18\ m\cdot12\ \dfrac{rad}{s^2}=2.16\ N.$$