PHY 2311

PHYSICS 231Lecture 18: equilibrium & revision

Remco ZegersWalk-in hour: Thursday 11:30-13:30 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

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:

PHY 2314

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 2315

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 2316

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 2317

Tips for study

• Look through the lecture sheets and pick out the summaries to get a good overview• Make an overview for yourself (about 1 Letter

size paper)• Read the chapters in the book to make sure that your overview contains all the main issues.• Study the examples given in the lectures• Study the problems in LON-CAPA• Study the worked-out examples in the book• Practice the previous midterm exams. • Practice problems from the book

PHY 2318

Revision: chapter 5 Work: W=Fcos()x Energy transfer Power: P=W/t Rate of energy transfer Potential energy (PE) Energy associated with

position. Gravitational PE: mgh Energy associated with

position in grav. field. PE stored in a spring: 1/2kx2 x is the compression of the spring

k is the spring constant Kinetic energy KE: 1/2mv2 Energy associated with

motion Conservative force: Work done does not depend on path Non-conservative force: Work done does depend on path Mechanical energy ME: ME=KE+PE

Conserved if only conservative forces are present KEi+PEi=KEf+PEf

Not conserved in the presence of non-conservative forces (KEi+PEi)-(KEf+PEf)=Wnc

PHY 2319

example

k=100 N/m

m=1 kg

1 cm

A pendulum is pushed with initialvelocity 0.1 m/s from a height of1 cm. How far does it compress thespring? (assume m does not risesignificantly after hitting the spring)

Conservation of ME:(mgh+1/2mv2+1/2kx2)initial= (mgh+1/2mv2+1/2kx2)final

1*9.8*0.01+0.5*1*0.12+0.=0.+0.+0.5*100*x2 so x=0.045 m

PHY 23110

Saving electricity

A ‘smart’ student decides to save energy by connecting his exercise treadmill to his laptop battery. If it takes 70 J to move the belt on the treadmill by 1 meter and 50% of the generated energy is stored in the battery, how‘far’ must the student run to use his 100 W laptop for freefor 2 hours?

Work done by student: W=70*d JEnergy given to the battery 0.5W=35*d J100 W laptop for 2 hours: 100 J/s*3600*2 s=7.2E+5 J

720 KJ

7.2E+5=35*d so d=(7.2E+5)/35=2.1E+4 m =21km!!!

PHY 23111

Non-conservative vs conservative case

450

A block of 1 kg is pushed up a 45o

slope with an initial velocity of 10 m/s.How high does the block go if:a) there is no frictionb) if the coefficient of kinetic friction is 0.5.

A) Conservation of ME: (mgh+1/2mv2)initial= (mgh+1/2mv2)final

0.+0.5*1*102=1*9.8*h+0. So h=5.1 m

B) Energy is lost to friction: (mgh+1/2mv2)initial= (mgh+1/2mv2)final+Wfriction

[W=Fx=nx=mgcos(45o)h/sin(45o)=0.5*1*9.8*h=4.9h] 0.+0.5*1*102=1*9.8*h+0.+4.9h so h=3.4 m

h

PHY 23112

Chapter 6Momentum p=mvF=p/tImpulse (the change in momentum) p= Ft

Inelastic collisions Elastic collisions

•Momentum is conserved•Some energy is lost in the collision: KE not conserved•Perfectly inelastic: the objects stick together.

•Momentum is conserved•No energy is lost in the collision: KE conserved

(v1i-v2i)=(v2f-v1f)

Conservation of momentum: m1v1i+m2v2i=(m1+m2)vf

Conservation of momentum: m1v1i+m2v2i=m1v1f+m2v2f

Conservation of KE: ½m1v1i

2+½m2v2i2=½m1v1f

2+½m2v2f

2

PHY 23113

Inelastic collision

5 m/s5 kg

V m/s0.1 kg

An excellent, but somewhat desperate sharp shooter triesto stop a cannon ball aimed directly at him by shooting a bulletfrom his gun against it. With what velocity does he need toshoot the bullet to stop the cannon ball assuming that the bullet gets stuck in the ball? How much energy is released?

Inelastic: only conservation of momentum.m1v1i+m2v2i=(m1+m2)vf so 0.1V-5*5=0v=25/0.1=250 m/sChange in kinetic energy:Before: ½m1v1i

2+½m2v2i2=3125+62.5=3187.5 J After: 0 J

Release: 3187.5 J

PHY 23114

Elastic collisionh

=10

0 m

Two balls (m1=1 kg, m2=2 kg) are released on a slope and collide inthe valley. How far does each goback up?

1 2

Step 1: calculate their velocities just before the collision: ball 1: cons. of ME: mgh=0.5mv2 9.8*100=0.5 v2 v=44. m/s ball 2: cons. of ME: 2*9.8*100=0.5*2*v2 v=-44. m/sStep 2: Collision, use cons. of P and KE (simplified). m1v1i+m2v2i=m1v1f+m2v2f so 44-88=v1f+2v2f

(v1i-v2i)=(v2f-v1f) so 88=v2f-v1f 3v2f=44 v2f=15 v1f=-73Step 3: Back up the ramp: cons. of ME: ball 1: 0.5mv2=mgh 0.5*732=9.8h h=272 m ball 2: 0.5mv2=mgh 0.5*152=9.8h h=11.5 m

PHY 23115

Bouncing ball

A 0.5 kg ball is dropped to the floor from a height of 2 m.If it bounces back to a height of 1.8 m, what is the magnitude of its change in momentum?

Some energy is lost in the bounce. Just before it hits the ground, its velocity is:(use conservation of ME)mgh=1/2mv2 so v=(2gh)=(2*9.8*2)= 6.26 m/s

After the bounce it goes back up 1.8 m.Just after it bounces back it velocity is:(use conservation of ME)mgh=1/2mv2 so v=(2gh)=(2*9.8*1.8)=5.93 m/sMust be negative!! So -5.93 m/s

p=m[6.26-(-5.93)]=0.5*12.2=6.1 kgm/s

PHY 23116

Chapter 7

t

ttt

t

if

if

0lim

Average angularvelocity (rad/s)

Instantaneous Angular velocity

t

ttt

t

if

if

0lim

Average angularacceleration (rad/s2)

Instantaneous angularacceleration

2 rad 3600

10 2/360 rad1 rad 360/2 deg

Be aware that sometimes rev/s or rev/min is asked

PHY 23117

Angular vs linear/tangential

r

v

r

a

linear

angular

r

See e.g. the bike example (lecture 16)

PHY 23118

Rotational motion

Angular motion(t)= (0)+(0)t+½t2

(t)= (0)+t

ac=v2/r directed to the center of the circular motion

Also v=r, so ac=2r

Centripetal acceleration

Fto center=mac=mv2/r

This acceleration is caused by ‘known’ force (gravitation, friction, tension…)

Make sure you understand how to use Kepler’s 3rd lawand the general definition of gravitational PE.

PHY 23119

Whirling ball

vA ball of mass 2 kg isattached to a string of 1mand whirled around a smooth horizontal table.If the tension in the stringexceeds 200 N, it will break. What will bethe speed at the momentthe string breaks?

Fto center=mac=mv2/rT=2*v2/1=200 Nv=10 m/s

PHY 23120

Consider...…a child playing on a swing. As she reaches the lowestpoint in her swing, which of the following is true?

A) The tension in the robe is equal to her weightB) The tension in the robe is equal to her mass timesher accelerationC) Her acceleration is downward and equal to g (9.8 m/s2)D) Her acceleration is zeroE) Her acceleration is equal to her velocity squared divided by the length of the swing.

Fto center=T-mg=mac=mv2/LLowest point, so no linear acceleration!!!

PHY 23121

Two objects...

Are rotating. One starts with an initial linear velocityof 1 m/s and rotates with a radius of 2 m. The other startsfrom rest and undergoes a constant angular acceleration overa circle with radius 3 m.What should its angular acceleration be, so that it overtakes(for the first time) the first object after 10 revolutions?

1=v1/r1=1/2=0.5 rad/s1(t)= 1(0)+1(0)t+½1t2=0.5t1(t)=10*2=0.5t so t=40

2(t)= 2(0)+2(0)t+½2t2=½2t2

20=½2(40)2 so 2=8E-03 rad/s

PHY 23122

Conical motion

mg

T

If the mass of the swinging object is 1 kg,and =30o what should the velocity of theobject be so that does not sink or rise?The length of the robe is 2 m.

Tcos

Tsin

Vertical direction:F=maTcos-mg=0So T=mg/cos=1*9.8/0.866=11.3 N

Horizontal direction:F=mac

Tsin=mac

11.3*0.5=mac=1v2/rv2=11.3*0.5*2 so v=3.4 m/s

PHY 23123

Chapter 8

Summary: see beginning of this lecture!

PHY 23124

Opening a hatch door.?N

1 m

A person is trying to open a1 meter, 20 kg, trap door by pullinga rope attached to its non rotatingend at an angle of 30o.With what force should he at least pull?30o

Center of gravity of the door: halfway the door’s length: 0.5m

=-mdoorgdCG+Fpull,perpendicularl 0=-20*9.8*0.5+Fpullsin(30o)*1 Fpull=196 N