PHY 2311

PHYSICS 231Lecture 20: equilibrium and more

rotations

Remco ZegersWalk-in hour: Monday 9:15-10:15 am

Helproom

PHY 2312

gravitationOnly if an object is near the surface of earth one can use:Fgravity=mg with g=9.81 m/s2

In all other cases:

Fgravity= GMobjectMplanet/r2 with G=6.67E-11 Nm2/kg2

This will lead to F=mg but g not equal to 9.8 m/s2 (see Previous lecture!)

If an object is orbiting the planet:

Fgravity=mac=mv2/r=m2r with v: linear velocity =angular vel.

So: GMobjectMplanet/r2 = mv2/r=m2r

Kepler’s 3rd law: T2=Ksr3 Ks=2.97E-19 s2/m3 r: radius of planetT: period(time to make one rotation) of planet

Our solar system!

PHY 2313

A child of 40 kg is sitting in a Ferris wheel, rotating with an angular velocity of 0.4 rad/s. The radius of the wheelis 9 m. What is the force exerted by the seat on the childat the top and at the bottom.

top: Fcenter=0 -mg+n+mac=0 n=mg-m2r=392-58=334bottom: Fcenter=0 -mg+n-mac=0 n=mg+m2r=392+58=450

The seat is fixed to the wheel. Its centripetal accelerationis directed towards the center.Top: the seat is accelerated away from the childBottom: the seat is accelerated towards the child

PHY 2314

Previously

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

PHY 2315

examples: A lot more in the book!

Where is the center of gravity?

0067.018

12.0

1116

)53cos(1.01)53cos(1.01016 00

ii

iii

CG m

xmx

01116

)53sin(1.01)53sin(1.01016 00

ii

iii

CG m

ymy

PHY 2316

quiz (extra credit)

0-2 1 2 m

20N 40N

40N

A wooden bar is initially balanced. Suddenly, 3 forces areapplied, as shown in the figure. Assuming that the bar canonly rotate, what will happen (what is the sum of torques)?a) the bar will remain in balanceb) the bar will rotate counterclockwisec) the bar will rotate clockwise

Torque: =Fd

PHY 2317

Weight of board: wWhat is the tension in each of thewires (in terms of w)?

w

T1 T2

0

Translational equilibriumF=ma=0T1+T2-w=0 so T1=w-T2

Rotational equilibrium

=0T10-0.5*w+0.75*T2=0T2=0.5/0.75*w=2/3w T1=1/3w

T2=2/3w

PHY 2318

s=0.5 coef of friction between the wall andthe 4.0 meter bar (weight w). What is the minimum x where you can hang a weight w for which the bar does not slide?

w

sn

n

w

T Ty

Tx

(x=0,y=0)

Translational equilibrium (Hor.)Fx=ma=0n-Tx=n-Tcos37o=0 so n=Tcos37o

Translational equilibruim (vert.)Fy=ma=0sn-w-w+Ty=0sn-2w+Tsin37o=0sTcos370-2w+Tsin370=01.00T=2w Rotational equilibrium:

=0xw+2w-4Tsin370=0 so w(x+2-4.8)=0x=2.8 m

PHY 2319

Demo: fighting sticks

PHY 23110

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 23111

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 23112

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 23113

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 23114

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 23115

More general.

=IMoment of inertia I:I=(miri

2)

compare with:F=maThe moment of inertiain rotations is similar tothe mass in Newton’s 2nd law.

PHY 23116

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 23117

The rotation axis matters!

I=(miri2)

=0.2*0.52+0.3*0.52+ 0.2*0.52+0.3*0.52

=0.5 kgm2

I=(miri2)

=0.2*0.+0.3*0.52+ 0.2*0+0.3*0.52

=0.3 kgm2