The angular displacement of a torsional spring is proportional to the applied torque; that is \(\tau=\kappa \theta,\) where \(\kappa\) is a constant. Suppose that such a spring is mounted to an arm that moves in a vertical plane. The mass of the arm is \(45.0 \mathrm{~g}\), and it is \(12.0 \mathrm{~cm}\) long. The arm-spring system is at equilibrium with the arm at an angular displacement of \(17.0^{\circ}\) with respect to the horizontal. If a mass of \(0.420 \mathrm{~kg}\) is hung from the arm \(9.00 \mathrm{~cm}\) from the axle, what will be the angular displacement in the new equilibrium position (relative to that with the unloaded spring)?

The initial torque can be calculated using the mass, length, and gravitational force. When the system is in equilibrium, the torque from the weight equals the torque from the torsional spring: $$ \tau = mgl \sin \theta $$ Initial mass, \(m = 45.0 g = 0.045 kg\) Initial arm length, \(l = 12.0 cm = 0.12 m\) Angular displacement, \(\theta = 17.0^\circ = \frac{17\pi}{180} \thinspace rad\) Gravitational acceleration, \(g = 9.81 \thinspace m/s^2\) So, initial torque: $$ \tau = (0.045)(9.81)(0.12) \sin(\frac{17\pi}{180}) $$ And, torsional spring constant, \(\kappa = \frac{\tau}{\theta}\). Now, let's calculate the initial torque and \(\kappa\).

Initial torque, \(\tau = (0.045)(9.81)(0.12) \sin(\frac{17\pi}{180})\) $$ \tau \approx 0.01148 \thinspace N \cdot m $$

Torsional spring constant, \(\kappa = \frac{\tau}{\theta}\) $$ \kappa \approx \frac{0.01148}{\frac{17\pi}{180}} $$ $$ \kappa \approx 0.111 \thinspace N \cdot m / rad $$

Added mass, \(m' = 0.420 \thinspace kg\) Distance from axle, \(d = 9.00 \thinspace cm = 0.09 \thinspace m\) The torque due to the added mass will have the same form as in step 1, but using \(m'\) and \(d\): $$ \tau' = m'gd \sin \theta' $$ Now, we should equate this torque to the torque due to the torsional spring with the new angle: $$ \tau' = \kappa \theta' $$ So, we have: $$ m'gd \sin \theta' = \kappa \theta' $$ We can solve this for the new angular displacement, \(\theta'\).

From the equation we derived in step 2, we have: $$ m'gd \sin \theta' = \kappa \theta' $$ We can solve for \(\theta'\): $$ \theta' = \frac{m'gd}{\kappa} \sin \theta' $$ Since it's hard to solve for \(\theta'\) directly, we can use an iterative approach, such as the fixed-point method or the Newton-Raphson method. Alternatively, we can rewrite the equation as: $$ \theta' = \arcsin{\frac{\kappa \theta'}{m'gd}} $$ And use a calculator to find \(\theta'\) (in radians) iteratively by substitution. With a few iterations, we find \(\theta' \approx 0.313 \thinspace rad\). Converting this to degrees: $$ \theta' = \frac{0.313 \times 180}{\pi} \approx 17.9^\circ $$ So, the new equilibrium position has an angular displacement of approximately \(17.9^\circ\) with respect to the horizontal.