Find the Speed of a Falling Meter Stick

[nchin]

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 just before 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)

Solution on page 6 # HRW 10.63
http://www.nvcc.edu/home/tstantcheva/231files/hrwch10hw.pdf

Can someone explain the solution to me please??

I do not understand why PE = 1/2 mgl? where did 1/2 come from?
How did the first step become ω^(2) equal 3g/L?
does v = ωL?
how did ωL = sqrt (3gL)?

thanks!

 

[ehild]

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 just before 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)

Solution on page 6 # HRW 10.63
http://www.nvcc.edu/home/tstantcheva/231files/hrwch10hw.pdf

Can someone explain the solution to me please??

I do not understand why PE = 1/2 mgl? where did 1/2 come from?
How did the first step become ω^(2) equal 3g/L?
does v = ωL?
how did ωL = sqrt (3gL)?

thanks!

The potential energy of the rod is the sum of the potential energies of its pieces. A little piece at the top has potential energy mgL while the bottom piece has zero PE. The whole potential energy is the same as if all the mass concentrated into the centre of mass, at half the height of the rod.

ehild.

 

[nchin]

The potential energy of the rod is the sum of the potential energies of its pieces. A little piece at the top has potential energy mgL while the bottom piece has zero PE. The whole potential energy is the same as if all the mass concentrated into the centre of mass, at half the height of the rod.

ehild.

so how did 1/2mgL = 1/2(mL^(2)/3)ω^(2) become ω^(2) = 3g/L?

 

[ehild]

$$\frac {1}{2}mgL=\frac {1}{2}\frac{mL^2}{3}\omega ^2$$

Isolate ω².

ehild

 

[nchin]

$$\frac {1}{2}mgL=\frac {1}{2}\frac{mL^2}{3}\omega ^2$$

Isolate ω².

ehild

im sorry, i don't really understand. how would you isolate ω^2?

 

[ehild]

Divide both sides of the equation by mL^2, and multiply by 6. What do you get?

ehild

 

[nchin]

Divide both sides of the equation by mL^2, and multiply by 6. What do you get?

ehild

oooohhhh i got it! ω = sqrt (3g/L)

so can you explain this part to me

v = ωL = sqrt (3gL)

so i know they multiplied L to both side

but why do we need to find the linear velocity instead of the angular velocity?

 

[ehild]

oooohhhh i got it! ω = sqrt (3g/L)

so can you explain this part to me

v = ωL = sqrt (3gL)

so i know they multiplied L to both side

but why do we need to find the linear velocity instead of the angular velocity?

You need it because that was the question in the problem.

ehild

 

[nchin]

You need it because that was the question in the problem.

ehild

haha ok thanks for your help!

 

[ehild]

ω was multiplied by L as the rod rotated about one end, the other end was at distance L from the centre of rotation. L is the radius of the circle the endpoint moves. You know that a point of a roting body at distance R from the centre has the linear velocity Rω.

ehild