L = 1 meter
Well, there is an easy way and a hard way
The easy way is that the potential energy at start is the kinetic energy when it hits
PE = m g * height of center of mass
= (1/2) m g L
when it hits its motion is all rotation about the end on the floor
rotational KE = (1/2) I omega^2
I about end = (1/3) m L^2
so
Ke = (1/6) m L^2 omega^2
so
(1/6) m L^2 omega^2 = (1/2) m g L
omega^2 = 3 g /L
omega = sqrt( 3g/L)
v = L omega at the end of the rod so
v = sqrt (3 g L)
here of course L = 1 meter
so v = sqrt(3g)
=============================
then I will just outline a hard way
There is a force up from the floor on the rod, call it F
there is a force down on the rod due to gravity, call it m g
Those are all the vertical forces so
m g - F = m a
where a is the acceleration of the center of the rod downward.
Now there is a moment about the center of mass = F(L/2)
which results in an anglar acceleration about the center of alpha
F L/2 = I alpha = (1/12) m L^2 alpha
but a = (L/2) alpha
so
F /2 = (1/12) m L (2 a/L)
F = (1/3) m a (note g is gone because we took moments about CG
so use that for F in our force problem
m g - F = m a
m g - (1/3) m a = m a
g = (4/3) a
a = (2/3)g
and alpha = 2 a/L = (4/3)g/L radians/s^2
how long to go pi/2 radians?
omega = alpha t
theta = (1/2) alpha t^2
pi = alpha t^2
t^2 = pi (3/4)L/g
t = sqrt [ pi (3/4)L/g ]
omega = alpha * t = (4/3) g/L *sqrt [ pi (3/4)L/g ]
velocity of end = omega L
= (4/3) g *sqrt [ pi (3/4)L/g ]
= sqrt [ pi (4/3) g L ]
THE trouble is that a and alpha are not constant, because the moment and force are not constant, so a far more detailed solution is required with F and moment a function of theta
A meter-long stick is balanced vertically on the ground. It is then made to fall with a gentle nudge. What is the speed of the end of the stick as it hits the floor? Assume the end on the floor does not slip. Treat the stick as a “thin rod.”
Really need help with this one.... All i've come to think of is the fact that i might be able to use t=rFsin(0) or a=t(toque)/I somehow.
2 answers
it amazes me how much people are willing to write and give up time to answer anothers questions online.
Cheers man.
Cheers man.
1 answer