PHYSICS 231
Lecture 19: More about rotations
Demo: fighting sticks
Remco Zegers
Walk-in hour: Thursday 11:30-13:30 am
Helproom
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What we did so far
Torque: =Fd Demo: turning screws
Center of
Gravity:
m x

m
i
xCG
i
i
i
m y

m
i
i
y CG
i
i
i
i
Demo: Leaning tower
Translational equilibrium: F=ma=0 The center of gravity
does not move!
Rotational equilibrium:
=0
The object does not
rotate
Mechanical equilibrium: F=ma=0 & =0 No movement!
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Torque and angular acceleration
F
r
m
Newton 2nd law: F=ma
Ft=mat
Ftr=mrat
Ftr=mr2
Used at=r
=mr2
Used =Ftr
The angular acceleration goes linear with the torque.
Mr2=moment of inertia
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Two masses
F
r
m
r
m
=mr2
=(m1r12+m2r22)

If m1=m2 and r1=r2
=2mr2
Compared to the case with only one mass, the angular
acceleration will be twice smaller when applying the
same torque, if the mass increases by a factor of two.
The moment of inertia has increased by a factor of 2.
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Two masses at different radii
F
r
m
r
m
=mr2
=(m1r12+m2r22)

If m1=m2 and r2=2r1
=
5mr2
When increasing the distance between a mass and the
rotation axis, the moment of inertia increases quadraticly.
So, for the same torque, you will get a much smaller
angular acceleration.
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m
m
m
m
m
m
m
m
m
m
F
A homogeneous stick
=mr2
=(m1r12+m2r22+…+mnrn2)
=(miri2)
=I
Moment of inertia I:
I=(miri2)
Rotation point
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Two inhomogeneous sticks
r
m
m
m
m
5m
5m
m
m
m
m
F
5m
m
m
m
m
m
m
m
m
5m
=(miri2)
118mr2
18m
Easy to rotate!
F
=(miri2)
310mr2
18 m
Difficult to rotate
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More general.
=I
Moment of inertia I:
I=(miri2)
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F
A simple example
F
r
r
A and B have the same total mass. If the same
torque is applied, which one accelerates faster?
Answer: A
=I
Moment of inertia I:
I=(miri2)
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The rotation axis matters!
I=(miri2)
I=(miri2)
=0.2*0.+0.3*0.5+
0.2*0+0.3*0.5
=0.3 kgm2
=0.2*0.5+0.3*0.5+
0.2*0.5+0.3*0.5
=0.5 kgm2
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Extended objects (like the stick)
M
I=(miri2)
=(m1+m2+…+mn)R2
=MR2
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Some common cases
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Example
A monocycle (bicycle with one wheel) has a wheel that
has a diameter of 1 meter. The mass of the wheel is 5
kg (assume all mass is sitting at the outside of the wheel).
The friction force from the road is 25 N. If the cycle
is accelerating with 0.3 m/s, what is the force applied on
each of the paddles if the paddles are 30 cm from the
center of the wheel?
=I
=a/r so =0.3/0.5=0.6 rad/s
I=(miri2)=MR2=5(0.52)=1.25 kgm2
0.5m
F
25N
0.3m
friction=-25*0.5=-12.5
paddles=F*0.3+F*0.3=0.6F
0.6F-12.5=1.25*0.6, so F=22.1 N
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Lon-capa
You can now do all problems up to and including 08-34.
Look at the lecture sheets of last 2 lectures
and this one for help!!
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Rotational kinetic energy
Consider a object rotating
with constant velocity. Each point
moves with velocity vi. The total
kinetic energy is:

1
1
1




m
v

m

r


mr

 2
 2

2 
2
i i
i
i
2 2
i i
i
2
i i
i
2
 1 2
  2 I

KEr=½I2
Conservation of energy for rotating object:
[PE+KEt+KEr]initial= [PE+KEt+KEr]final
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Example.
1m
Consider a ball and a block going down the same 1m-high slope.
The ball rolls and both objects do not feel friction. If both
have mass 1kg, what are their velocities at the bottom (I.e.
which one arrives first?). The diameter of the ball is 0.4 m.
Block: [½mv2+mgh]initial= [½mv2+mgh]final
1*9.8*1 = 0.5*1*v2 so v=4.4 m/s
Ball: [½mv2+mgh+½I2]initial= [½mv2+mgh+½I2]final
I=0.4*MR2=0.064 kgm2 and =v/R=2.5v
1*9.8*1 = 0.5*1*v2+0.5*0.064*(2.5v)2
so v=3.7 m/s Part of the energy goes to the rotation: slower!
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Lon-capa
You can now do all problems up to and including 08-42.
Next lecture: Last part of chapter 8 and examples.
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