The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in the horizontal position θ = 0 where the spring is unstretched. If the bar is observed to momentarily stop in the position θ = 64°, determine the spring constant k. For your computed value of k, what is magnitude of the angular velocity of the bar when θ = 32°.

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

explanation:

newton's third law, for every action there is an equal and opposite reaction.