The uniform rectangular door panel has a mass of 25 kg and is held in equilibrium above the horizontal at the position θ = 60° by rod BC. Determine the required stiffness of the torsional spring at A, so that the door's angular velocity becomes zero when the door reaches the closed position (θ = 0°) once the supporting rod BC is removed. The spring is undeformed when θ = 60°.
Given data:
Mass of uniform rectangular door
Initial position,
Let required torsional stiffness of spring be k
The diagram in initial and final positions is as shown below:
\(y_{1}=\frac{1.2 \sin 60^{\circ}}{2}\)
\(=0.5196 \mathrm{~m}\)
\(\theta_{e}=60^{\circ}=\pi / 3\) radians
POTENTIAL ENERGY:-
Initial position:
The gravitational potential energy of door panel
\(\left(V_{\text {door }}\right)_{1}=+W y_{1}\)
\(=+25 \times 9.81 \times 0.5196\)
\(=+127.4319 \mathrm{~J}\)
As springs are un stretched in initial position
So, potential energy becomes
\(V_{1}=\left(V_{\text {door }}\right)_{1}+\left(V_{\text {spring }}\right)_{1}\)
\(=127.4319+0\)
\(=127.43 \mathrm{~J}\)
Final position:
As door panel \(\mathrm{AB}\) has center of gravity on datum line, So it has zero gravitational potential energy.
The spring is stretched by angle \(=60^{\circ}\) in final position
So elastic potential energy of torsional spring,
\(\left(V_{\text {spring }}\right)_{2}=\frac{1}{2} k \theta_{e}^{2}\)
\(=\frac{1}{2} \times k \times\left(\frac{60^{\circ}}{180} \times \pi\right)^{2}\)
\(=\frac{1}{2} k\left(\frac{\pi}{3}\right)^{2}\)
Therefore the final potential energy is
\(V_{2}=\left(V_{\text {door }}\right)_{2}+\left(V_{\text {spring }}\right)_{2}\)
\(=0+\frac{1}{2} \times k(\pi / 3)^{2}\)
\(=\frac{\pi^{2}}{18} k\)
KINETIC ENERGY:
As system is at rest initially and also angular velocity of door reaches zero in final position. So in initial and final kinetic energy \(=0\)
\(T_{1}=0\)
\(T_{2}=0\)
Using conservation of energy principle,
\(V_{1}+T_{1}=V_{2}+T_{2}\)
\(127.43+0=\frac{\pi^{2}}{18} k+0\)
\(k=\frac{18 \times 127.43}{\pi^{2}}\)
\(k=232.4 \mathrm{~N} \cdot \mathrm{m} / \mathrm{rad}\)
Torsional stiffness of spring, \(k=232.4 \mathrm{~N} \cdot \mathrm{m} / \mathrm{rad}\)