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 2318

More general.

=IMoment of inertia I:I=(miri

2)

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 23111

Extended objects (like the stick)

I=(miri2)

=(m1+m2+…+mn)R2

=MR2

M

PHY 23112

Some common cases

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!

PHY 23117

Lon-capa

You can now do all problems up to and including 08-42.

Next lecture: Last part of chapter 8 and examples.