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.

Explanation:

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

So, initial potential energy stored in the spring is = 0

And, initial potential gravitational potential energy of the rod is .

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8

           L = length of the rod = 0.2 m

Initial total energy T =

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

            T =

where,    I = moment of inertia of the rod about the end =

Also,    

where,     = speed of the tip of the rod

              x = spring extension

The initial unstrained length is = 0.1 m

Therefore, final length will be calculated as follows.

              x' = m

Then,  x =

          x = m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

Putting the given values into the above formula as follows.

          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.

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