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PHY 2311
PHYSICS 231Lecture 19: More about rotations
Remco ZegersWalk-in hour: Thursday 11:30-13:30 am
Helproom
Demo: fighting sticks
PHY 2312
What we did so far
Translational equilibrium: F=ma=0 The center of gravitydoes not move!
Rotational equilibrium: =0 The object does notrotate
Mechanical equilibrium: F=ma=0 & =0 No movement!
Torque: =Fd
ii
iii
CG m
xmx
ii
iii
CG m
ymyCenter of
Gravity:
Demo: Leaning tower
Demo: turning screws
PHY 2313
rFt=mat
Torque and angular acceleration
m
FNewton 2nd law: F=ma
Ftr=mrat
Ftr=mr2 Used at=r
=mr2 Used =Ftr
The angular acceleration goes linear with the torque.
Mr2=moment of inertia
PHY 2314
Two masses
r
m
F
m r
=mr2=(m1r1
2+m2r22
)
If m1=m2 and r1=r2
=2mr2Compared to the case with only one mass, the angularacceleration 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.
PHY 2315
Two masses at different radii
r
m
F
mr
=mr2=(m1r1
2+m2r22
)
If m1=m2 and r2=2r1
=5mr2When increasing the distance between a mass and therotation axis, the moment of inertia increases quadraticly.So, for the same torque, you will get a much smallerangular acceleration.
PHY 2316
A homogeneous stick
Rotation point
mmmm
mmmm
m
m
F =mr2=(m1r1
2+m2r22+…+mnrn
2)=(miri
2)=I
Moment of inertia I:
I=(miri2)
PHY 2317
Two inhomogeneous sticks
mmmm
mmmm
5m
5m
F5mmmm
mmm5m
m
m
F
18m 18 m
=(miri2)
118mr2=(miri
2) 310mr2
r
Easy to rotate! Difficult to rotate
PHY 2319
A simple example
A and B have the same total mass. If the sametorque is applied, which one accelerates faster?
FF
r r
Answer: A=IMoment of inertia I:I=(miri
2)
PHY 23110
The rotation axis matters!
I=(miri2)
=0.2*0.5+0.3*0.5+ 0.2*0.5+0.3*0.5=0.5 kgm2
I=(miri2)
=0.2*0.+0.3*0.5+ 0.2*0+0.3*0.5=0.3 kgm2
PHY 23113
ExampleA monocycle (bicycle with one wheel) has a wheel thathas a diameter of 1 meter. The mass of the wheel is 5kg (assume all mass is sitting at the outside of the wheel).The friction force from the road is 25 N. If the cycleis 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?
25N
0.5m
0.3mF
=I=a/r so =0.3/0.5=0.6 rad/sI=(miri
2)=MR2=5(0.52)=1.25 kgm2
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
PHY 23114
Lon-capa
You can now do all problems up to and including 08-34.Look at the lecture sheets of last 2 lecturesand this one for help!!
PHY 23115
Rotational kinetic energy
Consider a object rotatingwith constant velocity. Each pointmoves with velocity vi. The totalkinetic energy is:
Irmrmvmi i
iiiiii
ii222222
21
21
21
21
KEr=½I2
Conservation of energy for rotating object:
[PE+KEt+KEr]initial= [PE+KEt+KEr]final
PHY 23116
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 bothhave 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/sBall: [½mv2+mgh+½I2]initial= [½mv2+mgh+½I2]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!