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.