Answer:
The spring constant = 104.82 N/m
The angular velocity of the bar when θ = 32° is 1.70 rad/s
Explanation:
From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:


Also;

Thus;

where;
= deflection in the spring
k = spring constant
b = remaining length in the rod
m = mass of the slender bar
g = acceleration due to gravity


Thus; the spring constant = 104.82 N/m
b
The angular velocity can be calculated by also using the conservation of energy;






Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s
Answer: 5.5m/s
Explanation:
vf=vi+at
vf= 4.0m/s + (0.50m/s^2)(3.0s)
To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.
PART A) We will begin by finding the two net distances.

And the distance 'd' is



Through the free-body diagram the tension components are given by


Here we can watch that,

Dividing both expression we have that,

Replacing the values,


PART B) Using the vertical component we can find the tension,



