a meter stick is pivoted on a physical pendulum at a distance x from a 50cm mark.period is 2.5s. what is x?

Answer 1

As you know a period T for any pendulum is
T=2Pi Sqrt(I/(mgl))
Where Pi=3.14…
Sqrt - square root
I – second moment or moment of inertia
m – mass of the pendulum (?)
g - gravitational acceleration
l – is the distance from the axis of oscillation to its center of gravity.
Then l=x+50cm


Since we have a bar pendulum of length L and width B (which you did not specify) we need to express the moment of inertia as
I = m(k^2 + l^2) where
k = Sqrt((L^2+B^2)/12)

Then T= 2Pi Sqrt((k^2 + l^2)/(gl))

And if my algebra is correct we have
l^2 - (g(T/(2Pi))^2) l – k^2=0 (a quadratic equation)

solve for l

then x=l - 50cm

Answer 2

Yea...what Edward said

Answer 3

why should i tell you?