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.

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

it amazes me how much people are willing to write and give up time to answer anothers questions online.

Cheers man.

To find the speed of the end of the stick as it hits the floor, we can start by considering the conservation of energy. The stick starts from rest in an upright position, so it has only potential energy at that point, which is converted into kinetic energy as it falls.

To solve this problem, we need to know the length of the stick (meter-long in this case) and any information about the mass of the stick. Assuming the stick has a mass of M kg, we can proceed with the calculations.

The potential energy of the stick at the initial position is given by:

PE_initial = M * g * h

Where g is the acceleration due to gravity (approximately 9.8 m/s²) and h is the height of the center of mass of the stick above the floor. In this case, h is equal to half of the length of the stick.

PE_initial = M * g * (length of stick / 2)

As the stick falls, potential energy is converted into kinetic energy. At the moment the stick hits the floor, all of its potential energy is converted into kinetic energy.

KE_final = PE_initial

The formula for kinetic energy is:

KE_final = (1/2) * M * v²

Where v is the velocity of the end of the stick as it hits the floor.

Now, equating the two equations above, we have:

(1/2) * M * v² = M * g * (length of stick / 2)

We can simplify the equation by canceling M from both sides and rearranging the terms:

v² = 2 * g * (length of stick / 2)

v² = g * length of stick

Finally, taking the square root of both sides, we get the speed of the end of the stick as it hits the floor:

v = √(g * length of stick)

Therefore, the speed of the end of the stick as it hits the floor is equal to the square root of the product of the acceleration due to gravity and the length of the stick.