Question

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The spring is initially unstrained and has a spring constant of k = 27.0 N/m. Find the tangential speed with which end A strikes the horizontal surface.

Answer 1

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 $m/s^{2}$

           L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where,    I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also,    $\omega = \frac{\nu}{L}$

where,    $\nu$ = speed of the tip of the rod

              x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

              x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then,  x = $x' - x_{o}$

          x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       $\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

      $\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

   $\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

  $\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

Answer 2

The tangential speed with which end A strikes the horizontal surface is 2.08 m/s.

The given parameters:

• Length of the bar, L = 0.2 m
• Mass of the bar, m = 0.795 kg
• Spring constant, k = 27 N/m
• Unstrained length = 0.1 m

The final length of the spring is calculated as follows;

$x = \sqrt{(0.2)^2 \ + \ (0.1)^2} \\\\x = 0.224 \ m$

The extension of the spring is calculated as follows;

$\Delta x = x_1 - x_0\\\\\Delta x = 0.224 \ m \ - \ 0.1\ m\\\\\Delta x = 0.124 \ m$

Apply the principle of conservation of energy;

$P.E_i = P.E_k + K.E_{rot}\\\\\frac{mgL}{2} = \frac{1}{2} k \Delta x^2 \ + \ \frac{1}{2} I \omega\\\\\frac{mgL}{2} = \frac{1}{2} k \Delta x^2 \ + \ \frac{1}{2} (\frac{mL^2}{3} )(\frac{V}{L} )^2\\\\\frac{mgL}{2} = \frac{1}{2} k \Delta x^2 \ + \ \frac{1}{6} (mV^2)\\\\mgL = k \Delta x^2 + \frac{1}{3} (mV^2)\\\\\frac{1}{3} mV^2 = mgL - k \Delta x^2\\\\V^2 = \frac{3(mgL - k \Delta x^2)}{m} \\\\V = \sqrt{\frac{3(mgL - k \Delta x^2)}{m}}$

$V = \sqrt{\frac{3(0.795 \times 9.8 \times 0.2 \ - \ 27 (0.124)^2)}{0.795}} \\\\V = 2.08 \ m/s$

Thus, the tangential speed with which end A strikes the horizontal surface is 2.08 m/s.

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