A thin rod of length 0.75 m and mass 120 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed 2.76 rad/s. Neglecting friction and air resistance, find (a) the rod’s kinetic energy at its lowest position and (b) how far above that position the center of mass rises.

a) KE = 0.0857Joules

b) height = 0.073m

Explanation:

The rotational inertia of the rod at an axis passing through the fixed end is given by:

I = Icm + my

I= 1/12 mL^2 + m(L/2)

I= 1/12 ×0.12×(0.75)^2 + 0.12× (0.75/2)^2

I = 5.625×10^-3 + 0.01688

I = 0.0225Kg/m^-2

a) KE = 1/2 Iw^2

Where I= rotational inertia

w= angular speed at lowest point

KE = 1/2 × 0.0225 ×(2.76)^2

KE= 0.0857 Joules

b) Ki + Ui = Kf + Uf

Ki = Uf - Ui = 0

mgh = K

0.0857= mgh

h = 0.0857/( 0.120× 9.8)

h = 0.0857/ 1.176

h = 0.073m

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.