Problem

The uniform rectangular door panel has a mass of 25 kg and is held in equilibrium above...

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°.

Step-by-Step Solution

Solution 1

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, \(k=232.4 \mathrm{~N} \cdot \mathrm{m} / \mathrm{rad}\)

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