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

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.