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

step-by-step explanation:

we are given

we have to simplify it

firstly, we will factor it

now, we can cancel like terms

now, we can combine it

so, we get

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .