# Homework Help: Uniform rod pivoted off-center

1. Jul 31, 2013

### natasha13100

√ω1. The problem statement, all variables and given/known data
A uniform rod of mass 1.2 kg and length 1.2 m is pivoted in the horizontal position as shown (black point). The rod is at rest and then released. The acceleration due to gravity is g = 9.8 m/s2.
Part 1. Find the rotational inertia of the rod relative to the axis perpendicular to the screen and passing through the pivot point (in kgm2).
Part 2. As the rod swings downward, its angular speed velocity and the magnitude of its angular acceleration, respectively are (increasing, decreasing, staying the same) and (increasing, decreasing, staying the same).
Part 3. What is the angular speed (in rad/s) of the rod as it passes through the vertical position (when end marked B is at the bottom)?
Part 4. What is the linear speed of the bottom end of the rod (marked B) when it passes the vertical position?(in m/s)

2. Relevant equations
Icm=1/12ml2
Ip=Icm+md2
Ug=mgh
Ui+Ki+Wnoncon=Uf+Kf
Ktrans=1/2mv2
Krot=1/2Iω2
t=Frsinθ
K=Ktrans+Krot
v=ωR

3. The attempt at a solution
Picture/FBD is attached.
Part 1
Ip=1/12m(l)2+m(l/4)2=1/12*1.2*1.22+1.2*.32=.252
Part 2
increasing and decreasing
Part 3
0=1/2mv2+1/2Iω2+mgh where h=-l/4(distance from pivot to cm which will be directly below pivot when the rod is at the bottom of its swing)
mgh=1/2mv2+1/2Iω2=mgh=1/2mω2R2+1/2Iω2=1/2mω2R2+1/2Iω2
1.2(9.8)(.3)=(1/2(1.2)(.3)2+1/2*.252)ω2=(.18)ω2
ω=√(3.528/.18)=4.42718872424
I am not sure if this is correct, but I used l/4 (distance from pivot to center of mass) as R.
Part 4
mgh=1/2mv2+1/2Iω2
2mgh=mv2+Iv2/R2
2(1.2)(9.8)(.3)=(1.2+.252/.32)v2
v=√(7.056/4)=1.32815661727
or v=ωR2=4.42718872424(.3)^2=0.398446985181
Since these two values for v don't match up, I think I'm doing something wrong.

#### Attached Files:

• ###### swinging rod.png
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2. Jul 31, 2013

### tiny-tim

hi natasha13100!

i think you're using the wrong formula for kinetic energy

it's 1/2mv2 of the c.o.m. plus 1/2Ic.o.mω2 about the c.o.m.

(which also equals 1/2Ic.o.rω2 about the instantaneous centre of rotation, ie the pivot)

3. Aug 1, 2013

### natasha13100

What does that mean?
So K=1/2mv^2+1/2(1/12ml^2)w^2=1/2mw^2/(l/2)^2+1/24ml^2w^2=w^2(ml^2/8+ml^2/24)
Uf+Kf=Ui+Ki
0=w^2(ml^2/8+ml^2/24)-mgl/4
mgl/4=w^2(ml^2/8+ml^2/24)
g=w^2(l/2+l/6)
w^2=g/(2l/3)
w=sqrt(3g/2l)
and
v=wl/2=sqrt(3g/2l)l/2
or
K=1/2mv^2+1/2(1/12ml^2)w^2=1/2mv^2+1/2(1/12ml^2)v^2/(l/2)^2=v^2(ml/2+1/6m)
gl/4=v^2(l/2+1/6)
v^2(3l+1)/6=gl/4
v^2=3/(6l+2)
v=sqrt(3/(6l+2))
but v is still off

Last edited: Aug 1, 2013
4. Aug 1, 2013

### tiny-tim

the v in 1/2 mv2 will be (l/4)ω

(because l24 is the distance from the centre of rotation to the centre of mass)

5. Aug 1, 2013

### natasha13100

0=w^2(1/2ml^2/16+ml^2/24)-mgl/4
mgl/4=w^2(ml^2/32+ml^2/24)
g=w^2(l/8+l/6)
w^2=24g/(3l+4l)
w=sqrt(24g/7l) which is correct
and
v=lsqrt(24g/7l)/4 but that's not right.

Last edited: Aug 1, 2013
6. Aug 1, 2013

### tiny-tim

looks ok down to here

(btw, you'd get the same result if you used the parallel axis theorem, and just 1/2 Ic.o.r.ω2 )

7. Aug 1, 2013

### natasha13100

1/2Ic.o.r.ω2=mgl/4
c.o.r.=Ic.o.m.+m(l/4)2
Ic.o.m.=1/12ml2
1/2(1/12ml2+m(l/4)22=mgl/4
ω2=g/(l(1/6+1/8))=g/l/(7/24)
ω=√(24g/7l)
okay, I see what you mean.
For v, wouldn't you just plug in for ω?
1/2mv2+1/2(1/12ml22=mgl/4
m/2*v2=mgl/4-1/24ml^2*24g/7l=mgl/4-mgl/7
v2=(7gl-4gl)/14=3gl/14
v=√(3gl/14)

8. Aug 1, 2013

### tiny-tim

yes

(but in your last post you used the wrong distance … the question asks for v for the bottom of the rod, which is not the same v as you used earlier)
i'm confused … what is this?

9. Aug 1, 2013

### natasha13100

Oh that was the v of the center of mass (I think). So for the bottom of the rod, r=3l/4 (distance from pivot to bottom of rod) and v=√(24g/7l)*3l/4

10. Aug 1, 2013

yup!