Physics - Rotation Question with Solution | TestHub

Given, diameter of the disc, $d=0.8$ m.
Radius of the disc, $r=\frac{d}{2}=\frac{0.8}{2}=0.4$ m.
Force applied, $F=120$ N.

The diagram shows a curved arrow indicating that the disc is rotating about its central axis. For the applied force to cause this specific rotation, it must act tangentially to the edge of the disc. In the given 2D perspective of the 3D disc, a tangential force at the rightmost edge is drawn vertically.

The magnitude of the torque $\tau$ is given by the product of the force and the perpendicular distance from the axis of rotation:
$\tau =r\times F$
$\tau =0.4\times 120=48$ Nm.

The direction of the torque vector is given by the cross product $\vec{\tau }=\vec{r}\times \vec{F}$.
Since both the position vector $\vec{r}$ (from the center to the edge) and the tangential force vector $\vec{F}$ lie in the plane of the disc, their cross product $\vec{\tau }$ must be perpendicular to both. Therefore, the torque vector is perpendicular to the plane of the disc.

Thus, the magnitude and direction of the torque is $48$ Nm perpendicular to the plane of the disc.

Answer: $48$ Nm perpendicular to plane of disc