Torque — Question 5

Question 5

Gravity torque depends on the horizontal offset of the center

A uniform rod of mass $4\,\mathrm{kg}$ and length $1.2\,\mathrm m$ is hinged at one end. In a vertical plane it makes angle $\theta$ counterclockwise from the horizontal rightward direction. Use $g=9.8\,\mathrm{m/s^2}$.

1. Find gravity torque about the hinge as a function of angle and evaluate it at $60^\circ$.

2. Identify the zero-torque orientations and explain the sign change across the upward vertical.

Question 5 - Solution

Gravity torque depends on the horizontal offset of the center

(a) Derivation and calculation

The center position and gravity force are $$\vec r_c=\frac L2(\cos\theta,\sin\theta),\qquad\vec F_g=(0,-Mg).$$ $$\boxed{\tau_z=-\frac{MgL}{2}\cos\theta},\qquad \boxed{\tau_z(60^\circ)=-4(9.8)(0.60)(0.50)=-11.76\ \mathrm{N\,m}}.$$

(b) Result and interpretation

The moment vanishes when the force line passes through the hinge: $$\cos\theta=0\Rightarrow\boxed{\theta=90^\circ,270^\circ\pmod{360^\circ}}.$$ Near the upward vertical, $\theta=\pi/2+\delta$ gives $$\tau_z\simeq\frac{MgL}{2}\delta.$$ The sign changes as the center crosses the hinge’s vertical line; this torque drives small deviations away from the upward orientation.

Physical check

The downward vertical instead has a restoring torque for small angular deviations.