Is torque dependent on the moment of inertia? I think this is why more force is required for a shorter distance; the inertia causes the object to stop, and greater force is required to cover the angular displacement.
 A: The torque needed to achieve a certain angular acceleration depends on the moment of inertia similar to the force needed to achieve a certain acceleration depends on the mass. The force that needs to be applied to achieve a certain torque depends on the distance from the rotational axis of the object. Inertia does not cause on object to stop. It's quite opposite: An moving or rotating object of larger inertia will actually require a larger force to make it stop moving or rotating.
A: TL;DR For the same angular acceleration, it takes more torque for larger moment of inertia. Since torque and force are proportional for a constant radius, more torque means more force. The inertia does not cause object to stop rotating, just as mass does not cause object to stop moving.

Moment of inertia $I$ to angular acceleration $\vec{\alpha}$ has the same relationship as mass $m$ to linear acceleration $\vec{a}$
$$\boxed{\vec{\tau}_\text{net} = I \vec{\alpha}} \tag 1$$

the inertia causes the object to stop, and greater force is required to cover the angular displacement

This is not true - inertia does not cause object to stop. Once object is rotating at certain angular velocity $\omega$, if there is no (net) torque the object will keep rotating at the same angular velocity! If you want to stop the object at certain angular acceleration, more torque is required for larger moment of inertia.

Notice similarity between the Eq. (1) and the second Newton's law for translational motion
$$\vec{F}_\text{net} = m \vec{a}$$
Link between the force $\vec{F}$ and the torque $\vec{\tau}$ is
$$\vec{\tau} = \vec{r} \times \vec{F} \qquad \text{and} \qquad |\vec{\tau}| = |\vec{r}| |\vec{F}| \sin \phi$$
where $\vec{r}$ is radius of rotation, i.e. a vector from axis of rotation to where the force is applied, and $\phi$ is angle between $\vec{F}$ and $\vec{r}$. Symbol $\times$ does not mean "regular multiplication", but a vector product
Note about vectors: The overhead arrow indicates that a variable is a vector, which means it has both magnitude and direction. If you still have not learned about vectors, just ignore the arrow!
