Physic C.................help <span title="me!!!!!!!!!!!!!!!!!!!!!!!!?">me!!!!!!!!!!!!!!!!!!!!!!!...</span>

Question

A thin rod of length l and mass m 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 w. Neglect friction and air resistance.

(a) Find the rod's kinetic energy at its lowest position

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(b) Find how far above that position the center of mass rises.

Answer 1

For a physical pendulum its period T is given in terms of  its length L and acceleration due to gravity g as

T= 2pi sqrt(2L/3g)

or

w= 2pi f= 2pi / T

w= sqrt(3g/2L)

L= 3g/(2w^2)

Speed V is then

v= w r or since r=0.5 L

v= 0.5w L

v= 3g/4w

Now

Ke= 0.5 m v^2

Ke= 0.5 m (3g/4w)^2

b) Kinetic energy Ke on the bottom = potential energy Pe at the top of the swing

Ke(max)=Pe(max)=mg h(max)

h(max)= Ke(max)/mg

Answer 2

a. KE=1/2(mVf2)

b. PE=mgh

    h=PE/W

Answer 3

(a) Moment of inertia I of a rod about one end is 1/3 ml^2.

K = 1/2 Iw^2 = 1/2(1/3ml^2)w^2.

(b) Let the initial position i be at the lowest position and the final position f be at the highest position. Take the gravitational potential energy of the center of the rod to be zero at the lowest position, that is, Ui = 0. The rod comes momentarily at rest at the highest position, that is, Kf = 0. From conservation of energy, Uf - Ui = (mg)change in height of center of mass = Kf - Ki = 1/6 (ml^2)w^2. Change in height of center of mass = (l2w^2/6g).