The 20kg block is constrained to move along a smooth circula

The 20-kg block is constrained to move along a smooth circular path which lies in a vertical plane as shown. The particle is released from rest at point C. The spring tension is 100 N when in the position shown. Determine the stiffness of the spring if the velocity of the particle is 2.5 m/s to the left when it passes through point B. Answers: K = 627 N/m

Solution

As it is smooth circular path, assume that there is no frictional force is involved here.

Initial force in the spring = 100 N.

Assume initial length of the spring = l₀ and stiffness of the spring = K.

At location C, length of the spring = (1200^ + 900^2)^0.5 = 1500 mm or 1.5 m.

So, force balnce along the length is : K x (1.5 - l₀ ) = 100 .................i) (As the block is at rest and it is released here, so no weight force is considered)

Use energy conservation equation between location B and C:

20 x g x 1.2 + 0.5 x K x (1.5 -  l₀ )^2 = 0.5 x 20 x 2.5^2 + 0.5 x K x (2.1 -  l₀ )^2

Or, 235.44 - 62.5 = 0.5 x K x (2.1 -  l₀ + 1.5 -  l₀ )(2.1 -  l₀ - 1.5 +  l₀ )

Or, 172.94 = 0.3 x K x (3.6 - 2 l₀ )

Or, 288.23 = K x (3.6 - 2 l₀ ) ......................ii)

Now, divide equation ii) by i) we get,

2.88 = (1.8 - l₀) /(1.5 - l₀ )

Or, l₀ = 1.342 m.

Put this value in equation i) : K x (1.5 - 1.342) = 100

Or, K = 634 N/m.