A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it hits the floor, assuming that the end on the floor does not slip. (Hint: Consider the stick to be a thin rod and use the conservation of energy principle.)
mgh=.5mv^2
1/gh=2/v^2
v=sqrt(2gh)
where g is gravity (9.8)
h is hieght of meter stick (1 meter)
m is mass
why is this wrong??
mgh=1/2 mv^2
h (cg height)=1/2
g=v^2
the energy is due to the mass falling, assume the mass is at the center of gravity (h=1/2)
You forgot that the stick is rotating. Add in (1/2) I w^2
if the center of mass is moving at speed v at the bottom when the stick is horizontal then the tip is moving at 2v
the tip velocity = 2v = rw = 1w
so w = 2v
total Ke = (1/2) m v^2 + (1/2)I w^2 = m g h
but w^2 = 4 v^2
remember you want the speed of the tip, not the center so the answer is 2v
correct, thanks Professor.
so what is the answer???
√24÷7L
By energy conservation
Mgl/2 =1/2Iw^2
Solving this equation by putting the value of I= ml^2/3
You get speed=sqrt 3g