# Rotational momentum (previous problem)

#### anelmarx

1. Homework Statement
A thin rod of length 1.50m is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge. The mass of the rod may be ignored compared to the mass of an object fixed to the top of the rod. The rod, starting from rest, tips over and rotates downward.
Find the angular speed of the rod just before it strikes the floor.

2. Homework Equations

KE=PE
Iw*w/2=mgh where h=L/2=CG
I=mr*r

3. The Attempt at a Solution
w=?
r=1.5m=L
g=6.50 rad/s*s
h=0.75m (CG=L/2)
s=2.36m
angle=90degrees=pi/2 rad
T=14.7

mr*r*w*w/2=mgL/2
Cancel out m
r*r*w*w/2=gL/2
now I have calcultated g = 6.50 rad/s*s
so
1.5*1.5*w*w/2=6.5*1.5/2
1.125*w*w=4.875
w*w=4.875/1.125
w=2.00

But the correct answer is 3.61 according to textbook.

What am I doing wrong?
Please help, I am so frustrated.

Related Advanced Physics Homework Help News on Phys.org

#### tiny-tim

Science Advisor
Homework Helper
Welcome to PF!

Hi anelmarx! Welcome to PF! A thin rod of length 1.50m is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge. The mass of the rod may be ignored compared to the mass of an object fixed to the top of the rod. The rod, starting from rest, tips over and rotates downward.
Find the angular speed of the rod just before it strikes the floor.

KE=PE
Iw*w/2=mgh where h=L/2=CG
I=mr*r

now I have calcultated g = 6.50 rad/s*s
Why are you using the centre of the rod (L/2)? The mass is all at the end of the rod.

And how do you get g = 6.50 rad/s*s? g is length/time2 … radian is not a length.

ok, that's tactics … now overall strategy …

this is not a moment of inertia case, it's a perfectly simple ordinary conservation of energy case with a point mass …

it's exactly the same as as a mass on a string (except upside down, of course) …

just use mgh and 1/2 mv2

the only relevance of the angular speed is that for some reason they've asked you to convert the ordinary speed into angular speed at the end #### anelmarx

Ok now I am really lost.

I am studying online and do not have anyone to ask - lecturer not resonding .

If I use v*v=v_o*v_o+2ax
and a=-9.81m/s*s
and x=1.5m
then
v*v=0+2*-9.81*1.5
but then you cannot take the root of a negative?

OMG I am just so frustrated

#### tiny-tim

Science Advisor
Homework Helper
If I use v*v=v_o*v_o+2ax
uhh? that's a constant acceleration equation … why are you using it? this isn't constant acceleration

use mgh and 1/2 mv2 and conservation of energy

#### anelmarx

Ok, think I might have figured it out:

PE=KE
mgh=mv*v/2
m falls away so
gh=v*v/2
and g = 9.81
so
9.81*1.5*2=v*v
v=5.43m/s
then ang speed=v/r=5.43/1.5=3.61rad/s

and to find ang acc:

90degrees = pi/2 and s=r*pi/2=1.5*pi/2=2.36m
so
9.81m/s*((pi/2)rad/2.36m)=6.53rad/s

#### tiny-tim

Science Advisor
Homework Helper
PE=KE

then ang speed=v/r=5.43/1.5=3.61rad/s
Fine, except better if you say PE lost = KE gained and to find ang acc:

90degrees = pi/2 and s=r*pi/2=1.5*pi/2=2.36m
so
9.81m/s*((pi/2)rad/2.36m)=6.53rad/s
oh, you didn't say anything about angular acceleration in the original question …

hmm … i'm not following your reasoning …

where does π/2 come into it? either use general principles, or use torque (= moment of force) = moment of inertia times angular acceleration #### anelmarx

Wow! thank you so much for your help, I have been struggling with this for hours and hours.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving